Sketch Solutions for Exercises

in the Main Text of A First Look at Vectors

T. W. Körner

Introduction

Here are what I believe to be sketch solutions to the bulk of exercises in the main text the book (i.e. those not in the “Further Exercises”). I have written in haste in the hope that others will help me correct at leisure. I am sure that they are stuffed with errors ranging from the TEXtual through to the arithmetical and not excluding serious mathematical mistakes. I would appreciate the opportunity to correct at least some of these problems. Please tell me of any errors, unbridgeable gaps, misnumberings etc. I welcome suggestions for additions.

ALL COMMENTS GRATEFULLY RECEIVED. If you can, please use L^{A}TEX 2ε or its relatives for mathematics. If not,

please use plain text. My e-mail is twk@dpmms.cam.ac.uk. You may safely assume that I am both lazy and stupid so that a message saying ‘Presumably you have already realised the mistake in Exercise Z’ is less useful than one which says ‘I think you have made a mistake in Exercise Z because you have have assumed that the sum is necessarily larger than the integral. One way round this problem is to assume that f is decreasing.’

When I was young, I used to be surprised when the answer in the back of the book was wrong. I could not believe that the wise and gifted people who wrote textbooks could possibly make mistakes. I am no longer surprised.

It may be easiest to navigate this document by using the table of contents which follow on the next few pages. To avoid disappointment, observe that those exercises marked ⋆ have no solution given.

Contents

Introduction 2 Exercise 1.1.2 13 Exercise 1.2.1 14 Exercise 1.2.2 15 Exercise 1.2.3 16 Exercise 1.2.5 17 Exercise 1.2.6 18 Exercise 1.2.7 19 Exercise 1.2.8 20 Exercise 1.2.9⋆ 20 Exercise 1.2.10 21 Exercise 1.3.5 22 Exercise 1.3.9 23 Exercise 1.3.10 25 Exercise 1.4.3 26 Exercise 2.1.3 27 Exercise 2.1.4 28 Exercise 2.1.6⋆ 28 Exercise 2.1.7⋆ 28 Exercise 2.1.9 29 Exercise 2.2.4⋆ 29 Exercise 2.2.5⋆ 29 Exercise 2.2.8 30 Exercise 2.2.10 31 Exercise 2.3.3 32 Exercise 2.3.4⋆ 32 Exercise 2.3.11 33 Exercise 2.3.12 34 Exercise 2.3.13 35 Exercise 2.3.14 36 Exercise 2.3.16 37 Exercise 2.3.17 38 Exercise 2.3.18 39 Exercise 2.4.1 40 Exercise 2.4.2 41 Exercise 2.4.4⋆ 41 Exercise 2.4.6 42 Exercise 2.4.9 43 Exercise 2.4.10 44 Exercise 2.4.11 45 Exercise 2.4.12 46 Exercise 3.2.1 47 Exercise 3.3.4 48 Exercise 3.3.6 49 Exercise 3.3.9 50 Exercise 3.3.10 52 Exercise 3.3.12 53 Exercise 3.3.13 54 Exercise 3.3.14 55 Exercise 3.4.5 56 Exercise 3.4.7 57 Exercise 3.5.1 58 Exercise 3.5.2 59 Exercise 4.1.1 60 Exercise 4.1.2 61 Exercise 4.1.3 62 Exercise 4.2.1 63 Exercise 4.2.2 64 Exercise 4.2.3⋆ 64 Exercise 4.3.2 65 Exercise 4.3.3 66 Exercise 4.3.10 67 Exercise 4.3.13 68 Exercise 4.3.14 69 Exercise 4.3.15 70 Exercise 4.3.16⋆ 70 Exercise 4.4.1 71 Exercise 4.4.4 72 Exercise 4.4.6 73 Exercise 4.4.7 74 Exercise 4.4.8 75 Exercise 4.4.9 76 Exercise 4.5.1 78 Exercise 4.5.3 79 Exercise 4.5.3 80 Exercise 4.5.5 81 Exercise 4.5.6 82 Exercise 4.5.7 83 Exercise 4.5.8 84 Exercise 4.5.10 85 Exercise 4.5.13 86 Exercise 4.5.14 87 Exercise 4.5.15 88 Exercise 4.5.16 89 Exercise 4.5.17 90 Exercise 5.1.1 91 Exercise 5.1.2⋆ 91 Exercise 5.2.7 92 Exercise 5.2.8 93 Exercise 5.2.11 94 Exercise 5.3.2 95 Exercise 5.3.3 96 Exercise 5.3.10 97 Exercise 5.3.11 98 Exercise 5.3.14 99 Exercise 5.3.16 100 Exercise 5.4.3 101 Exercise 5.4.11 102 Exercise 5.4.12 104 Exercise 5.4.13 105 Exercise 5.4.14 106 Exercise 5.5.3 107 Exercise 5.5.5 108 Exercise 5.5.6 109 Exercise 5.5.12⋆ 109 Exercise 5.5.13 110 Exercise 5.5.14 111 Exercise 5.5.16 113 Exercise 5.5.17 114 Exercise 5.5.18 115 Exercise 5.5.19 116 Exercise 5.6.2 118 Exercise 5.6.3 119 Exercise 5.6.4 120 Exercise 5.6.5 121 Exercise 5.6.6 122 Exercise 6.1.2 123 Exercise 6.1.3 124 Exercise 6.1.5 125 Exercise 6.1.7 126 Exercise 6.1.8 127 Exercise 6.1.9 128 Exercise 6.1.12 129 Exercise 6.2.3 130 Exercise 6.2.4 131 Exercise 6.2.9 132 Exercise 6.2.11 133 Exercise 6.2.14 134 Exercise 6.2.15 135 Exercise 6.2.16 136 Exercise 6.3.2 137 Exercise 6.4.2 138 Exercise 6.4.5 139 Exercise 6.4.7 140 Exercise 6.4.8 141 Exercise 6.4.9 142 Exercise 6.5.5 143 Exercise 6.5.2 144 Exercise 6.6.2 145 Exercise 6.6.7 146 Exercise 6.6.8 147 Exercise 6.6.9 148 Exercise 6.6.10 149 Exercise 6.6.11 150 Exercise 6.6.12 151 Exercise 6.6.13 153 Exercise 6.7.1 154 Exercise 6.7.2 155 Exercise 6.7.3 156 Exercise 6.7.6 157 Exercise 6.7.7 158 Exercise 6.7.9 159 Exercise 6.7.10 160 Exercise 6.7.11 161 Exercise 6.7.12 162 Exercise 7.1.4 163 Exercise 7.1.6 164 Exercise 7.1.8 165 Exercise 7.1.9 166 Exercise 7.1.10 167 Exercise 7.2.2 168 Exercise 7.2.6 169 Exercise 7.2.10 170 Exercise 7.2.13 171 Exercise 7.3.2 172 Exercise 7.3.4 173 Exercise 7.3.6 174 Exercise 7.3.7 175 Exercise 7.4.3 176 Exercise 7.4.6 177 Exercise 7.4.9 178 Exercise 7.5.1 179 Exercise 7.5.2 180 Exercise 7.5.3 181 Exercise 7.5.4 182 Exercise 7.5.5 183 Exercise 7.5.6 185 Exercise 7.5.7 186 Exercise 8.1.7 188 Exercise 8.2.2 189 Exercise 8.2.6 190 Exercise 8.2.7 191 Exercise 8.2.9 192 Exercise 8.2.10 193 Exercise 8.3.2 194 Exercise 8.3.3⋆ 194 Exercise 8.3.4 195 Exercise 8.3.5 196 Exercise 8.4.2 197 Exercise 8.4.3 198 Exercise 8.4.4 199 Exercise 8.4.6 200 Exercise 8.4.7 201 Exercise 8.4.8 202 Exercise 8.4.9 203 Exercise 8.4.10 204 Exercise 8.4.11 206 Exercise 8.4.12 207 Exercise 8.4.13 208 Exercise 8.4.14 209 Exercise 8.4.15 210 Exercise 8.4.16 211 Exercise 8.4.17 212 Exercise 8.4.18 213 Exercise 8.4.19 214 Exercise 8.4.20 215 Exercise 9.1.2 216 Exercise 9.1.4 217 Exercise 9.2.1 218 Exercise 9.3.1 219 Exercise 9.3.5 220 Exercise 9.3.6 221 Exercise 9.3.8 222 Exercise 9.4.3 223 Exercise 9.4.5 225 Exercise 9.4.6 226 Exercise 9.4.7 227 Exercise 9.4.9 228 Exercise 9.4.10 229 Exercise 9.4.11 230 Exercise 9.4.12 231 Exercise 9.4.13 232 Exercise 9.4.14 233 Exercise 9.4.15 234 Exercise 9.4.16 235 Exercise 9.4.17 236 Exercise 9.4.19 237 Exercise 9.4.21 238 Exercise 9.4.22 239 Exercise 10.1.4 240 Exercise 10.1.2 241 Exercise 10.1.3 242 Exercise 10.1.6 243 Exercise 10.1.8 245 Exercise 10.2.1 246 Exercise 10.2.2 247 Exercise 10.2.3 248 Exercise 10.2.4 249 Exercise 10.3.2 250 Exercise 10.4.1 251 Exercise 10.4.3 252 Exercise 10.4.6 253 Exercise 11.1.2 254 Exercise 11.1.3 255 Exercise 11.1.4 256 Exercise 11.1.5 257 Exercise 11.1.9 258 Exercise 11.1.15 260 Exercise 11.1.16 261 Exercise 11.2.1 262 Exercise 11.2.3 266 Exercise 11.2.4 267 Exercise 11.2.5 268 Exercise 11.3.1 269 Exercise 11.3.6 270 Exercise 11.3.7 271 Exercise 11.3.8 272 Exercise 11.3.9 273 Exercise 11.3.10 274 Exercise 11.4.2 275 Exercise 11.4.4 276 Exercise 11.4.5 277 Exercise 11.4.6 278 Exercise 11.4.9 279 Exercise 11.4.10 280 Exercise 11.4.12 281 Exercise 11.4.15 282 Exercise 11.4.19 283 Exercise 11.4.20 284 Exercise 12.1.2 285 Exercise 12.1.3 286 Exercise 12.1.4 287 Exercise 12.1.6 289 Exercise 12.1.7 290 Exercise 12.1.11 291 Exercise 12.1.13 292 Exercise 12.1.14 293 Exercise 12.1.15 294 Exercise 12.1.16 295 Exercise 12.1.17 296 Exercise 12.1.18 297 Exercise 12.1.19 298 Exercise 12.2.1 299 Exercise 12.2.2 300 Exercise 12.2.3 301 Exercise 12.2.6 302 Exercise 12.2.7 303 Exercise 12.2.8 304 Exercise 12.2.9 305 Exercise 12.2.11 306 Exercise 12.2.12 307 Exercise 12.3.1 308 Exercise 12.3.3 309 Exercise 12.3.6 310 Exercise 12.3.7 311 Exercise 12.3.9 312 Exercise 12.3.11 313 Exercise 12.3.13 314 Exercise 12.3.14 316 Exercise 12.3.15 317 Exercise 12.4.3 318 Exercise 12.4.4 319 Exercise 12.4.7 320 Exercise 12.4.11 321 Exercise 12.4.12 322 Exercise 12.4.13 323 Exercise 12.4.14 324 Exercise 12.5.1 326 Exercise 12.5.2 327 Exercise 12.5.3 328 Exercise 12.5.4 330 Exercise 12.5.5 331 Exercise 12.5.6 332 Exercise 12.5.7 333 Exercise 12.5.8 334 Exercise 12.5.9 336 Exercise 12.5.10 337 Exercise 12.5.11 338 Exercise 13.2.2 340 Exercise 13.2.3 341 Exercise 13.2.4 342 Exercise 13.2.6 343 Exercise 13.2.7 344 Exercise 13.2.8 346 Exercise 13.2.9 347 Exercise 13.3.1 348 Exercise 13.3.2 349 Exercise 13.3.3 350 Exercise 13.3.4 351 Exercise 13.3.5 352 Exercise 13.3.6 353 Exercise 13.3.7 354 Exercise 13.3.9 355 Exercise 13.3.10 356 Exercise 13.3.12 357 Exercise 13.3.13 358 Exercise 13.3.14 359 Exercise 13.3.15 360 Exercise 13.3.16 361 Exercise 13.3.17 362 Exercise 13.3.18 363 Exercise 14.1.3 364 Exercise 14.1.4 365 Exercise 14.1.5 367 Exercise 14.1.11 369 Exercise 14.1.12 370 Exercise 14.1.14 372 Exercise 14.1.16 373 Exercise 14.2.1 374 Exercise 14.2.3 375 Exercise 14.2.5⋆ 375 Exercise 14.2.6 376 Exercise 14.2.7 377 Exercise 14.2.10 378 Exercise 14.2.12 379 Exercise 14.3.2 380 Exercise 14.3.3 381 Exercise 14.3.4 382 Exercise 14.3.5 383 Exercise 14.3.7 384 Exercise 14.3.8 385 Exercise 14.3.9 386 Exercise 14.3.10⋆ 386 Exercise 14.3.11 387 Exercise 14.3.12 388 Exercise 14.3.13 389 Exercise 14.3.14 390 Exercise 14.3.15 391 Exercise 15.1.1 392 Exercise 15.1.3 393 Exercise 15.1.7 394 Exercise 15.1.9 395 Exercise 15.1.10 396 Exercise 15.1.11 397 Exercise 15.1.12 398 Exercise 15.1.13 399 Exercise 15.2.2 400 Exercise 15.2.4 401 Exercise 15.3.2 402 Exercise 15.3.3 403 Exercise 15.3.7 404 Exercise 15.3.8 405 Exercise 15.3.11 406 Exercise 15.4.3 408 Exercise 15.4.6 409 Exercise 15.4.7 411 Exercise 15.4.9 412 Exercise 16.1.5 413 Exercise 16.1.9 414 Exercise 16.1.14 415 Exercise 16.1.16 416 Exercise 16.1.17 417 Exercise 16.1.23 419 Exercise 16.1.25 420 Exercise 16.1.26 421 Exercise 16.1.27 423 Exercise 16.2.4 424 Exercise 16.2.8 425 Exercise 16.2.9 426 Exercise 16.2.10 427 Exercise 16.2.12 429 Exercise 16.2.13 430 Exercise 16.3.2 431 Exercise 16.3.3 432 Exercise 16.3.4 433 Exercise 16.3.5 434 Exercise 16.3.6 435 Exercise 16.3.7 437 Exercise 16.3.11 438 Exercise 16.3.12 439 Exercise 16.3.14 440 Exercise 16.3.15 441 Exercise 16.3.16 442 Exercise 16.3.17 444 Exercise 16.3.19 445 Exercise 16.4.2 447 Exercise 16.4.3 448 Exercise 16.4.6 449 Exercise 16.4.7 450 Exercise 16.4.8 451 Exercise 16.4.9 452

Exercise 1.1.2

Starting with

x + y + z = 1
x + 2y + 3z = 2
x + 4y + 9z = 6.

We subtract the first equation from the second and from the third to get

x + y + z = 1
y + 2z = 1
3y + 8z = 5.

We now solve the equations

y + 2z = 1
3y + 8z = 5,

for z by subtracting 3 times the first equation from the second to obtain

y + 2z = 2
2z = 2.

Thus z = 1 and, working backwards, y = 1 − 2z = −1, whence x = 1 − y − z = 1.

Exercise 1.2.1

STEP 1 If a_{ij} = 0 for all i and j, then our equations have the form

0 = y_{i} [1 ≤ i ≤ m].

Our equations are inconsistent unless y_{1} = y_{2} = … = y_{m} = 0. If y_{1} = y_{2} = … = y_{m} = 0 the equations impose no constraints on x_{1}, x_{2}, …, x_{n} which can take any value we want.

STEP 2 If the condition of STEP 1 does not hold, we can arrange,by reordering the equations and the unknowns, if necessary, that a_{11} ≠ 0. We now subtract a_{i1}/a_{11} times the first equation from the ith equation [2 ≤ i ≤ m] to obtain

⋆⋆  ∑_{j=2}^{n} b_{ij}x_{j} = z_{i} [2 ≤ i ≤ m].

where

b_{ij} = \frac{a_{11}a_{ij} − a_{i1}a_{1j}}{a_{11}} and z_{i} = \frac{a_{11}y_{i} − a_{i1}y_{1}}{a_{11}} .

STEP 3 If the new set of equations ⋆⋆ has no solution, then our old set ⋆ has no solution. If our new set of equations ⋆⋆ has a solution x_{i} = x′_{i} for 2 ≤ i ≤ m, then our old set ⋆ has the solution

x_{1} = \frac{1}{a_{11}} (y_{1} − ∑_{j=2}^{n} a_{1j}x′_{j})
x_{i} = x′_{i} [2 ≤ i ≤ n].

(Continued in Exercise 1.2.2.)

Exercise 1.2.2

(Continued from Exercise 1.2.1.)

This means that, if ⋆⋆ has exactly one solution, then ⋆ has exactly one solution and, if ⋆⋆ has infinitely many solutions, then ⋆ has infinitely many solutions. We have already remarked that, if ⋆⋆ has no solutions, then ⋆ has no solutions.

Exercise 1.2.3

(i)(a) If a_{i} = 0, but y_{i} ≠ 0 for some i, then there can be no solution, since the equation 0 = y_{i} cannot be satisfied.

(b) If y_{r} = 0 whenever a_{r} = 0 and there exists an i such that a_{i} = 0, then there are an infinity of solutions with x_{s} = a_{s}^{−1}y_{s} when a_{s} ≠ 0 and x_{s} chosen freely otherwise.

(c) If a_{i} ≠ 0 for all i, there is a unique solution x_{i} = a_{i}^{−1}y_{i} for all i.

(ii) (a) If a_{j} = 0 for all j then either b = 0 and every choice of x_{j} gives a solution (so there is an infinity of solutions or b ≠ 0 and there is no solution.

(b) If a_{k} ≠ 0 for some k, then, choosing x_{i} freely for i ≠ k and setting

x_{k} = a_{k}^{−1} (b − ∑_{j≠k} a_{j}x_{j})

gives an infinity of solutions.

Exercise 1.2.5

(i) If a = 1, b = 2, c = d = 4 the first and second equations are incompatible, so there is no solution.

(ii) If a = 1, b = 2, c = d = 4, then the third equation gives the same information as the first equation, so the system reduces to

x + y = 2
x + 2y = 4

Subtracting the first equation from the second we see that y = 2. Thus x = 0. By inspection this is a solution.

(iii) If a = b = 2, c = d = 4, then the second and third equation give the same information as the first equation so the system reduces to

x + y = 1

with the infinite set of solutions given by choosing x arbitrarily and setting y = 1 − x.

Exercise 1.2.6

The two equations are incompatible if a = 1. There is then no solution.

If a ≠ 1, then subtracting the first equation from the second we get

x + y + z = 2
(a − 1)z = 2,

so z = 2/(a − 1). Knowing the value of z the system reduces to

x + y = 2 − \frac{2}{a − 1} = 2\frac{a − 2}{a − 1}

so x may be chosen freely and then

y = 2\frac{a − 2}{a − 1} − x.

There are an infinity of solutions.

Exercise 1.2.7

(i) Observe that, if n ≥ 4

n^{3} ≥ ∑_{r=1}^{n} n^{2} ≥ ∑_{r=1}^{n} r^{2} ≥ ∑_{n≥r≥n/2}^{n}(n/2)^{2} ≥ (n/4) × (n/2)^{2} = n^{3}/16.

(ii) Let f(x) = x^{2}. Then f is increasing so

f(r) ≤ f(x) ≤ f(r + 1)

for r ≤ x ≤ r + 1.

Integrating we get f(r) = ∫_{r}^{r+1} f(r) dx ≤ ∫_{r}^{r+1} f(x) dx ≤ ∫_{r}^{r+1} f(r + 1) dx = f(r + 1) Thus, summing,

∑_{r=1}^{n−1} f(r) ≤ ∫_{1}^{n} f(x) dx ≤ ∑_{r=2}^{n} f(r).

In other words

∑_{r=1}^{n−1} r^{2} ≤ (n^{3} − 1)/3 ≤ ∑_{r=2}^{n} r^{2}

so

(∑_{r=1}^{n} r^{2}) − n^{2} ≤ (n^{3} − 1)/3 ≤ (∑_{r=1}^{n} r^{2}) − 1

Thus

(n^{3} − 1)/3 ≤ (∑_{r=1}^{n} r^{2}) − 1 ≤ (n^{3} − 1)/3 + n^{2}.

Dividing by n^{3} and allowing n → ∞, we obtain

n^{−3} ∑_{r=1}^{n} r^{2} → 1/3.

Exercise 1.2.8

If we have the triangular system of size n, one operation is needed to get x_{n} from the nth equation. Substitution of the value of x_{n} in the remaining n − 1 equations reduces them to a triangular system of size n − 1 in about 2(n − 1) operations. We can repeat this process to obtain the complete solution in about

∑_{r=1}^{n} 2(r − 1) + 1 ≈ 2 ∑_{r=1}^{n−1}(r − 1) ≈ n^{2}

operations.

Exercise 1.2.9⋆

Rather you than me.

Exercise 1.2.10

The number of operations is about An^{3}. I reckon A ≈ 1, so I need about 1000 operations. Reckoning about 100 operations an hour and a 5 hour day gives 2 days. The reader may disagree about everything, but we should still have about the same order of magnitude for the task.

Exercise 1.3.5

Row operations

Subtract twice first row from second row.

(\begin{matrix}1 & −1 & 3 \\ 2 & 5 & 2 \end{matrix}) → (\begin{matrix}1 & −1 & 3 \\ 0 & 6 & −4 \end{matrix})

Divide second row by 6.

(\begin{matrix}1 & −1 & 3 \\ 0 & 6 & −4 \end{matrix}) → (\begin{matrix}1 & −1 & 3 \\ 0 & 1 & −\frac{4}{3} \end{matrix})

Add second row to first row.

(\begin{matrix}1 & −1 & 3 \\ 0 & 1 & −\frac{4}{3} \end{matrix}) → (\begin{matrix}1 & 0 & \frac{5}{3} \\ 0 & 1 & −\frac{4}{3} \end{matrix})

Column operations.

Add first column to second. Subtract three times first column from third.

(\begin{matrix}1 & −1 & 3 \\ 2 & 5 & 2 \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & 7 & −10 \end{matrix})

Divide second column by 7.

(\begin{matrix}1 & 0 & 0 \\ 0 & 7 & −10 \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & −\frac{10}{7} \end{matrix})

Add 10/7 times second column to third.

(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & −\frac{10}{7} \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix})

Exercise 1.3.9

(i) Subtract 2 times first row from second. Subtract 4 times first row from third.

(\begin{matrix}1 & −1 & 3 \\ 2 & 5 & 2 \\ 4 & 3 & 8 \end{matrix}) → (\begin{matrix}1 & −1 & 3 \\ 0 & 7 & −4 \\ 0 & 7 & −4 \end{matrix})

Subtract second row from third.

(\begin{matrix}1 & −1 & 3 \\ 0 & 7 & −4 \\ 0 & 7 & −4 \end{matrix}) → (\begin{matrix}1 & −1 & 3 \\ 0 & 7 & −4 \\ 0 & 0 & 0 \end{matrix})

Add first column to second. Subtract 3 times first column from third.

(\begin{matrix}1 & −1 & 3 \\ 0 & 7 & −4 \\ 0 & 0 & 0 \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & 7 & −4 \\ 0 & 0 & 0 \end{matrix})

Add 4/7 times second column to third. Divide second column by 7.

(\begin{matrix}1 & 0 & 0 \\ 0 & 7 & −4 \\ 0 & 0 & 0 \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix})

(ii) Subtract first row from second and interchange first and second row.

(\begin{matrix}2 & 4 & 5 \\ 3 & 2 & 1 \\ 4 & 1 & 3 \end{matrix}) → (\begin{matrix}2 & 4 & 5 \\ 1 & −2 & −4 \\ 4 & 1 & 3 \end{matrix}) → (\begin{matrix}1 & −2 & −4 \\ 2 & 4 & 5 \\ 4 & 1 & 3 \end{matrix})

Subtract 2 times first row from second and 4 times first row from third.

(\begin{matrix}1 & −2 & −4 \\ 2 & 4 & 5 \\ 4 & 1 & 3 \end{matrix}) → (\begin{matrix}1 & −2 & −4 \\ 0 & 8 & 13 \\ 0 & 9 & 19 \end{matrix})

Subtract second row from third, interchange second and third.

(\begin{matrix}1 & −2 & −4 \\ 0 & 8 & 13 \\ 0 & 9 & 19 \end{matrix}) → (\begin{matrix}1 & −2 & −4 \\ 0 & 8 & 13 \\ 0 & 1 & 6 \end{matrix}) → (\begin{matrix}1 & −2 & −4 \\ 0 & 1 & 6 \\ 0 & 8 & 13 \end{matrix})

Add twice second row to first and subtract 8 times second row from third.

(\begin{matrix}1 & −2 & −4 \\ 0 & 1 & 6 \\ 0 & 8 & 13 \end{matrix}) → (\begin{matrix}1 & 0 & 8 \\ 0 & 1 & 6 \\ 0 & 0 & −35 \end{matrix})

Divide third row by −35. Subtract 6 times third row from second and 8 times third row from first.

(\begin{matrix}1 & 0 & 8 \\ 0 & 1 & 6 \\ 0 & 0 & −35 \end{matrix}) → (\begin{matrix}1 & 0 & 8 \\ 0 & 1 & 6 \\ 0 & 0 & 1 \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

Subtract first row from second and interchange first and second row.

2x + 4y + 5z = −3 2x + 4y + 5z = −3 x − 2y − 4z = 5
3x + 2y + z = 2  x − 2y − 4z = 5  2x + 4y + 5z = −3
4x + y + 3z = 1  4x + y + 3z = 1  4x + y + 3z = 1

Subtract 2 times first row from second and 4 times first row from third. Subtract second row from third, interchange second and third

x − 2y − 4z = 5  x − 2y − 4z = 5  x − 2y − 4z = 5
8y + 13z = −13  8y + 13z = −13  y + 6z = −6
9y + 19z = −19  y + 6z = −6  8y + 13z = −13

Add twice second row to first and subtract 8 times second row from third. Divide third row by −35. Subtract 6 times third row from second and 8 times third row from first.

x + 8z = −7  x + 8z = −7  x = 1
y + 6z = −6  y + 6z = −6  y = 0
−35z = 35  z = −1  z = −1

Exercise 1.3.10

By subtracting the second row from the first and the third from the second, we see that the matrix

(\begin{matrix}1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{matrix})

has rank 3. By adding the first row to the second and the first row to the third, we obtain a matrix of the same rank 3 with all entries non-zero.

(\begin{matrix}1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 1 & 2 & 2 \end{matrix}) .

The same arguments, starting with

(\begin{matrix}1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}) and (\begin{matrix}1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) ,

yields matrices of rank 2 and 1 with all entries non-zero

(\begin{matrix}1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 1 & 1 & 1 \end{matrix}) and (\begin{matrix}1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{matrix}) .

Not possible. All elementary operations (which change a matrix) involve a non-zero row or column which remains non-zero (though possibly moved) after the operation. Thus a rank 0 matrix must be the zero matrix.

Exercise 1.4.3

(i) Observe that π_{i}((x + y) + x) = (x_{i} + y_{i}) + z_{i} = x_{i} + (y_{i} + z_{i}) = π_{i}(x + (y + x)). (ii) Observe that

π_{i}(x + y) = x_{i} + y_{i} = y_{i} + x_{i} = π_{i}(y + x).

(iii) Observe that

π_{i}(x + 0) = x_{i} + 0 = x_{i} = π_{i}(x).

(v) Observe that π_{i}((λ + μ)x) = (λ + μ)x_{i} = λx_{i} + μx_{i} = π_{i}(λx + μy).

(vi) Observe that

π_{i}((λμ)x) = (λμ)x_{i} = λ(μx_{i}) = π_{i}(λ(μy)).

(vii) Observe that

π_{i}(1x) = 1x_{i} = x_{i} = π_{i}(x)

and

π_{i}(0x) = 0x_{i} = 0 = π_{i}(0).

(viii) Can do as above or x − x = 1x + (−1)x = (1 + (−1))x = 0x = 0.

Exercise 2.1.3

If c ≠ 0, then u = (a/c, 0) and v = (0, b/c) are distinct vectors representing points on the line.

If c = 0, then suppose, without loss of generality that a ≠ 0. If b = 0, then u = (0, 0) and v = (0, 1) are distinct vectors representing points on the line. If b ≠ 0, then u = (0, 0) and v = (1, −a/b) are distinct vectors representing points on the line.

Exercise 2.1.4

(i) Suppose that

{v + λw : λ ∈ ℝ} ∩ {v′ + μw : μ ∈ ℝ} ≠ ∅.

Then there exist λ_{0}, μ_{0} ∈ ℝ such that

v + λ_{0}w = v′ + μ_{0}w

and so

v + λw = v′ + (λ − λ_{0} + μ_{0})w.

Thus

{v + λw : λ ∈ ℝ} ⊆ {v′ + μw : μ ∈ ℝ}

and, similarly

{v′ + μw : μ ∈ ℝ} ⊆ {v + λw : λ ∈ ℝ}

so

{v + λw : λ ∈ ℝ} = {v′ + μw : μ ∈ ℝ}.

(ii) The line joining u to u′ is

{u + λw : λ ∈ ℝ}

with w = u − u′. Since w = (σ/τ)v − v′, the line joining v to v′ is

{v + μw : μ ∈ ℝ}.

Thus part (ii) follows from part (i).

(iii) Observe that

u′′ = −\frac{1}{μ′′} (μu + μ′u′) = \frac{1}{μ + μ′} (μu + μ′u′) = u + \frac{μ′}{μ + μ′} (u′ − u).

Exercise 2.1.6⋆ Exercise 2.1.7⋆

Exercise 2.1.9

\frac{λ}{1 − λ} = −\frac{α}{β} ⇒ αλ = −α + λα ⇒ λ = \frac{α}{α − β} .

Interchanging α and −(1 − α) and β and −(1 − β), yields

λ′ = \frac{1 − α}{β − α}.

Exercise 2.2.4⋆ Exercise 2.2.5⋆

Exercise 2.2.8

Let y be the centroid of x_{1}, x_{2}, …, x_{q}. Observe that

\frac{q − 1}{q}y_{j} + \frac{1}{q}x_{j} = \frac{1}{q} ∑_{i≠j} x_{i} + \frac{1}{q}x_{j} = y,

so y lies on the line joining y_{j} and x_{j}.

Exercise 2.2.10

(i) Let y be centre of mass of all the points x_{1}, x_{2}, …, x_{q} and let M = ∑_{j=1}^{q} m_{j}.

Observe that

\frac{m_{j}}{M} y_{j} + \frac{M − m_{j}}{M} x_{j} = \frac{1}{M} ∑_{i≠j} m_{i}x_{i} + \frac{m_{j}}{M} x_{j} = y

so y lies on the line joining y_{j} and x_{j}.

(ii) We might have m_{1} + m_{2} + … + m_{q} = 0 and we cannot divide by

Exercise 2.3.3

(i) We take the positive square root. (ii) x = 0 ⇒ ∥x∥^{2} = 0 ⇒ ∑_{j=1}^{n} x_{j}^{2} = 0 ⇒ x_{j} = 0 ∀j⇒ x = 0. (iii) ∥λx∥^{2} = ∑_{j=1}^{n}(λx_{j})^{2} = λ^{2} ∑_{j=1}^{n} x_{j}^{2}, so ∥λx∥ = |λ|∥x∥.

Exercise 2.3.11

The result is trivial if a = b, or b = c, so we assume this is not the case.

Setting x = a − b, y = b − c, we see that Theorem 2.3.10 gives

∥a − b∥ + ∥b − c∥ = ∥x∥ + ∥y∥
≥ ∥x + y∥
= ∥a − c∥

with equality if and only if λ(a − b) = μ(b − c) for some λ, μ > 0 i.e. if and only if a, b, c lie in order along a line.

The length of one side of a triangle AB is less than or equal to the sum of the lengths of the other two sides BC and CA with equality if and only if the triangle is degenerate with A, B, C lying in order along a straight line.

Exercise 2.3.12

If x lies on the line through a and b then

x = ta + (1 − t)b

for some t ∈ ℝ. The condition

∥x − a∥ = ∥x − b∥

yields

∥(1 − t)(a − b)∥ = ∥t(a − b)∥

that is to say

|1 − t|∥b − a∥ = |t|∥b − a∥

so |t| = |1−t| so t = 1−t or t = t−1. The second equation is insoluble so t = 1/2 and x = \frac{1}{2}(a + b).

We have proved uniqueness. Direct substitution shows we have a solution.

Exercise 2.3.13

(i) Choose X on BC so that AX is perpendicular to BC. By the theorem of Pythagoras and the geometric definition of the cosine and sine

|AC|^{2} = |AX|^{2} + |CX|^{2} = |AX|^{2} + (|BC| − |BX|)^{2}
= |BA|^{2} sin^{2} θ + (|BC| − |BA| cos θ)^{2}
= |BA|^{2}(cos^{2} θ + sin^{2} θ) + |BC|^{2} − 2|BA||BC| cos θ
= |BC|^{2} + |BA|^{2} − 2|BC| × |BA| cos θ

(If the reader is a very careful mathematician, she will not be entirely happy with this argument. She will ask, for example, how we choose the sign of some of the terms. However in this book the algebra is primary and the geometry is illustrative.)

(ii) We have

∥a − c∥^{2} = (a − c) · (a − c) = ∥a∥^{2} + ∥c∥^{2} − 2a · c.

(iii) The formula in (i) can be rewritten

∥a∥^{2} + ∥c∥^{2} − 2∥a∥ × ∥c∥ cos θ = ∥a − c∥^{2}

giving

a · c = ∥a∥∥c∥ cos θ.

Exercise 2.3.14

Since

\frac{v · u}{∥u∥∥v∥} = \frac{u · v}{∥u∥∥v∥},

the angle between v and u is the same as the the angle between u and v

If θ is the angle between u and v and ϕ is the angle between u and −v, then

cos ϕ = \frac{u · (−v)}{∥u∥∥v∥} = −\frac{u · v}{∥u∥∥v∥} = − cos θ

and 0 ≤ θ, ϕ ≤ π, so ϕ = π − θ

Exercise 2.3.16

Let us put the vertices at

a = (a_{1}, a_{2}, …, a_{n}).

The diagonals join vertices a and −a (call this diagonal d_{a}).

Diagonals d_{a}, d_{b} are perpendicular if and only if

0 = a · b = ∑_{i=1}^{n} a_{i}b_{i}

Since a_{i}b_{i} = ±1, ∑_{i=1}^{n} a_{i}b_{i} is odd if n is odd and even if n is even. (Proof by induction or modular arithmetic.) Thus if n is odd, a · b ≠ 0 and no diagonals are perpendicular.

Now suppose n = 4. In finding the possible angles θ we may suppose without loss of generality that a = (1, 1, 1, 1). The possible angles are given by

cos θ = \frac{a · b}{∥a∥∥b∥} ∈ \{r/4 : −4 ≤ r ≤ 4\}

and all the possibilities occur (note that we may get different angles according to the direct assigned to the diagonals). The possible angles are

0, cos^{−1} \frac{1}{4}, \frac{π}{6}, cos^{−1} \frac{3}{4}, \frac{π}{2}, π − cos^{−1} \frac{3}{4}, \frac{5π}{6}, π − cos^{−1} \frac{1}{4}, π

the angle 0 and π being the angle between the diagonal and itself and the angle between the diagonal and itself reversed.

Exercise 2.3.17

(i) Always true u ⊥ v ⇒ u · v = 0 ⇒ v · u = 0 ⇒ v ⊥ u.

(ii) Sometimes false. Take u = w = (1, 0, 0) and v = (0, 1, 0). (iii) Always true

u ⊥ u ⇒ u · u = 0 ⇒ u = 0.

Exercise 2.3.18

(i) We have

∥u + v∥^{2} = (u + v) · (u + v)

= u · u + u · v + v · u + v · v

= ∥u∥^{2} + 0 + 0 + ∥v∥^{2} = ∥u∥^{2} + ∥v∥^{2}

Consider the right angled triangle OUV with O at 0, U at u and V at v. Then ∥u + v∥ is the length of the hypotenuse.

(ii) We have

∥u + v + w∥^{2} = (u + v + w) · (u + v + w)

= u · u + v · v + w · w + 2u · v + 2v · w + 2w · u = ∥u∥^{2} + ∥v∥^{2} + ∥w∥^{2} + 0 + 0 + 0 = ∥u∥^{2} + ∥v∥^{2} + ∥w∥^{2}

(iii) If u_{j} ∈ ℝ^{4} and u_{j} ⊥ u_{k} for k ≠ j, we have

∥∑_{j=1}^{4} u_{j}∥^{2} = (∑_{j=1}^{4} u_{j}) · (∑_{j=1}^{4} u_{j})

= ∑_{j=1}^{4} ∑_{k=1}^{4} u_{j} · u_{k}

= ∑_{j=1}^{4} u_{j} · u_{j} = ∑_{j=1}^{4} ∥u_{j}∥^{2}.

Exercise 2.4.1

∥a + b∥^{2} + ∥a − b∥^{2} = (a + b) · (a + b) + (a − b) · (a − b)

= (∥a∥^{2} + 2a · b + ∥b∥^{2}) + (∥a∥^{2} − 2a · b + ∥b∥^{2})

= 2(∥a∥^{2} + ∥b∥^{2})

The sum of the squares of the lengths of the two diagonals of a parallelogram equals the sum of the lengths of the four sides.

Exercise 2.4.2

(i) Consider the parallelogram OACB with vertex 0 at 0, vertex A at a, vertex C at a+b and vertex B at b. The midpoint of the diagonal OC is at

x = \frac{1}{2}0 + \frac{1}{2}(a + b) = \frac{1}{2}(a + b).

The midpoint of the diagonal AB is at

y = \frac{1}{2}a + \frac{1}{2}b = \frac{1}{2}(a + b).

Thus x = y and we are done.

The midpoint M of the side CA opposite 0 is at

m = \frac{1}{2}(a + b) + \frac{1}{2}b = \frac{1}{2}a + b.

The point Z given by

z = \frac{1}{3}0 + \frac{2}{3}m = \frac{1}{3}a + \frac{2}{3}b

trisects OM and AB, so we are done.

Exercise 2.4.6

(i) If (n, m) is a unit vector perpendicular to (u, v) we have nu = −mv so n = −vx, m = ux and 1 = (u^{2} + v^{2})x^{2} = x^{2} thus

(n, m) = (−v, u) or (n, m) = (v, −u).

(ii) Let n be a unit vector perpendicular to ∥c∥^{−1}c as found in (i). Let p = n · a.

Then

x = a + tc ⇒ x · n = p

and if x · n = p we have (x − a) · n = 0 so x − a = tc for some t. (iii) Let a = pn and let c = u be a unit vector perpendicular to n.

x = pn + tu ⇒ x · n = p

and, if x · n = p, we have (x − pn) · u = 0 so x − pn = tu for some t. (Or use a geometric argument.)

Exercise 2.4.9

We have

π_{1} = {(0, x_{2}, x_{3}) : x_{2}, x_{3} ∈ ℝ},
π_{2} = {(x_{1}, 0, x_{3}) : x_{2}, x_{3} ∈ ℝ},
π_{3} = {(x_{1}, x_{2}, x_{3}) : x_{1} + x_{2} = 1} x_{j} ∈ ℝ}.

π_{1} and π_{2} meet in the line x_{1} = x_{2} = 0. π_{1} and π_{3} meet in the line x_{1} = 0, x_{2} = 1. π_{2} and π_{3} meet in the line x_{1} = 1, x_{2} = 0. The lines have no point in common.

(i) Now let π_{j} be given by

n_{j} · x = p_{j},

with

n_{1} = n_{2} = n_{3} = (1, 0, 0)

and p_{1} = −1, p_{2} = 0, p_{3} = 1. By inspection no two planes meet.

(ii) Finally let π_{j} be given by

n_{j} · x = p_{j},

with

n_{2} = n_{3} = (1, 0, 0), n_{1} = (0, 1, 0)

and p_{1} = p_{2} = 0, p_{3} = 1. We have that π_{1} and π_{2} meet in the line given by x_{1} = x_{2} = 0 and π_{1} and π_{2} meet in the line given by x_{1} = 1, x_{2} = 0, but π_{2} and π_{3} do not meet.

Exercise 2.4.10

{x ∈ ℝ^{2} : ∥x − a∥ = r}

is the singleton {a} if r = 0 and the empty set ∅ if r < 0.

Exercise 2.4.11

(i) We have

y(y(x)) = \frac{1}{∥y(x)∥^{2}} y(x) = \frac{∥x∥^{4}}{∥x∥^{2}} \frac{1}{∥x∥^{2}} x = x.

(ii) Observe that, if ∥a∥ ≠ r, then, if x ≠ 0, setting C = (∥a∥^{2} − r^{2}),

∥x − a∥^{2} = r^{2} ⇔ ∥x∥^{2} − 2x · a + (∥a∥^{2} − r^{2}) = 0
↔ 1 − 2y · a + (∥a∥^{2} − r^{2})∥y∥^{2}
⇔ ∥y∥^{2} − 2y · (C^{−1}a) + C^{−1} = 0
⇔ ∥y − C^{−1}a∥ = C^{−1}(C^{−2}∥a∥^{2} − 1)

so we transform the circle into a circle centre C^{−1}a radius

((C^{−1}(C^{−2}∥a∥^{2} − 1))^{1/2}.

(Note that we are taking the square root of a positive number.)

(iii) Observe that, if ∥a∥ = r > 0, then, if x ≠ 0,

∥x − a∥^{2} = r^{2} ⇔ ∥x∥^{2} − 2x · a = 0
⇔ 1 − 2y · a = 0

so we transform the circle into a line perpendicular to a closest distance to the origin (2∥a∥)^{−1}.

(iv) Goes through without change, replacing ‘circle’ by ‘sphere’ and ‘line’ by plane’

Exercise 2.4.12

∥\frac{x}{∥x∥^{2}} − \frac{y}{∥y∥^{2}} ∥^{2} = (\frac{x}{∥x∥^{2}} − \frac{y}{∥y∥^{2}} ) · (\frac{x}{∥x∥^{2}} − \frac{y}{∥y∥^{2}} )

= \frac{x · x}{∥x∥^{4}} − 2\frac{x · y}{∥x∥^{2}∥y∥^{2}} + \frac{y · y}{∥y∥^{4}}

= \frac{1}{∥x∥^{2}} − 2\frac{x · y}{∥x∥^{2}∥y∥^{2}} + \frac{1}{∥y∥^{2}}

= \frac{∥x∥^{2} − 2x · y + ∥y∥^{2}}{∥x∥^{2}∥y∥^{2}}

= (\frac{∥x − y∥}{∥x∥∥y∥})^{2} .

Thus

∥\frac{x}{∥x∥^{2}} − \frac{y}{∥y∥^{2}} ∥ = \frac{∥x − y∥}{∥x∥∥y∥}.

Using this result, we have, by the triangle inequality, (if x, y, z ≠ 0)

(∥x∥∥y∥∥z∥)^{−1}(∥y∥∥z − x∥ + ∥x∥∥y − z∥) = \frac{∥z − x∥}{∥z∥∥x∥} + \frac{∥y − z∥}{∥y∥∥z∥}

= ∥\frac{z}{∥z∥^{2}} − \frac{x}{∥x∥^{2}} ∥ + ∥\frac{y}{∥y∥^{2}} − \frac{z}{∥z∥^{2}} ∥

≥ ∥\frac{x}{∥x∥^{2}} − \frac{y}{∥y∥^{2}} ∥

= (∥x∥∥y∥∥z∥)^{−1}∥z∥∥y − x∥

with equality if and only if ∥x∥^{−2}x − ∥y∥^{−2}y and ∥z∥^{−2}z − ∥y∥^{−2}y are scalar multiples of each other (i.e. ∥x∥^{−2}, ∥y∥^{−2}y ∥z∥^{−2}z lie on a straight line). Thus

∥z∥∥x − y∥ ≤ ∥y∥∥z − x∥ + ∥x∥y∥

if x, y, z ≠ 0. If at least one of the vectors is zero the result is trivial.

To obtain the Euclidean result place D at the origin , let A have position vector x, B have position vector y and C position vector z.

We now use the notation of Exercise 2.4.11. The condition ∥x∥^{−2}x, ∥y∥^{−2}y ∥z∥^{−2}z are scalar multiples of each other is the same as saying that f(x), f(y), f(y) lie on a straight line so x = f^{2}(x), y = f^{2}(y) z = f^{2}(z) lie on a circle through the origin i.e. A, B, C, D are concyclic.

Exercise 3.2.1

z_{i} = a_{ik}y_{k} = a_{ik}(b_{kj}x_{j}) = (a_{ik}b_{kj})x_{j} = c_{ij}x_{j}

Exercise 3.3.4

There are many proofs. We can work coordinatewise e.g.

a_{ij} + (b_{ij} + c_{ij}) = (a_{ij} + b_{ij}) + c_{ij}

so A + (B + C) = (A + B) + C

Or direct from definition.

(A + (B + C))x = Ax + (B + C)x = Ax + (Bx + Cx)

= (Ax + Bx) + Cx) = (A + B)x + Cx = ((A + B) + C))x

for all x, so (A + B) + C = A + (B + C).

Exercise 3.3.6

Suppose Cx = x for all x. Taking x to be the column vector with 1 in k th place and zero elsewhere, we get

c_{ik} = δ_{ik}.

Conversely, if y ∈ ℝ^{n}

∑_{j=1}^{n} δ_{ij}y_{i} = y_{j}

so Iy = y.

Exercise 3.3.9

(iii) By definition. Observe that

((B + C)A)x = (B + C)(Ax) = B(Ax) + C(Ax)

= (BA)x + (CA)x = (BA + CA)x

for all x so (B + C)A = BA + CA. (iii) By calculation. We have

∑_{j=1}^{n}(b_{ij} + c_{ij})a_{jk} = ∑_{j=1}^{n} b_{ij}a_{jk} + c_{ij}a_{jk} = ∑_{j=1}^{n} b_{ij}a_{jk} + ∑_{j=1}^{n} c_{ij}a_{jk}

(iii) By summation convention. Observe that (b_{ij} + c_{ij})a_{jk} = b_{ij}a_{jk} + c_{ij}a_{jk}

(iv) By definition. Observe that

((λA)B)x = (λA)(Bx) = λ(A(Bx))

= λ((AB)x) = (λ(AB))x

for all x so (λA)B = λ(AB).

Again

(A(λB))x = A((λB)x) = (A(λ(Bx)))
= λ(A((Bx)) = λ((AB)x) = (λ(BA))x

for all x so A(λB) = λ(AB). (iv) By calculation. We have

∑_{j=1}^{n}(λa_{ij})b_{jk} = λ ∑_{j=1}^{n} a_{ij}b_{jk} = ∑_{j=1}^{n} a_{ij}(λb_{jk})

(iv) By summation convention. Observe that (λa_{ij})b_{jk} = λ(a_{ij}b_{jk}) = a_{ij}(λb_{jk})

(v) By definition. Observe that (IA)x = I(Ax) = Ax

and

(AI)x = A(Ix) = Ax

for all x so IA = AI.

(v) By calculation. We have

∑_{j=1}^{n} δ_{ij}a_{jk} = a_{ik} = ∑_{j=1}^{n} a_{ij}δ_{jk} = ∑_{j=1}^{n} b_{ij}a_{jk} + c_{ij}a_{jk}.

(v) By summation convention. Observe that δ_{ij}a_{jk} = a_{ik} = a_{ij}δ_{jk}.

Exercise 3.3.10

We have BA = A ∀A ⇒ BI = I ⇒ B = I.

Exercise 3.3.12

If A is an m × n matrix and λ ∈ ℝ, then λA = C where C is the m × n matrix such that

λ(Ax) = Cx

for all x ∈ ℝ^{n}.

If A and B are m × n matrices, then A + B = C where C is the m × n matrix such that

Ax + Bx = Cx

If A is an m × n matrices and B is an m × p matrix, then AB = C where C is the n × p matrix such that

A(Bx) = Cx

for all x ∈ ℝ^{n}

Remark The matrices defined above are certainly unique. The definitions do not by themselves show existence, but the existence may be easily checked.

In (i) take c_{ij} = λa_{ij}. In (ii) take c_{ij} = λa_{ij} + b_{ij}. In (iii) take c_{ij} = ∑_{r=1}^{m} a_{ir}b_{rj}.

Exercise 3.3.13

(i) Observe that

((A + B) + C)x = (A + B)x + Cx = (Ax + Bx) + Cx
= Ax + (Bx + Cx) = Ax + (B + C)x
= ((A + (B + C))x

for all x ∈ ℝ^{n} so (A + B) + C = A + (B + C).

(ii) Observe that

(A + B)x = Ax + Bx = Bx + Ax = (B + A)x

for all x ∈ ℝ^{n} so A + B = B + A).

(iii) Observe that

(A + 0)x = Ax + 0x = Ax + 0 = Ax

for all x ∈ ℝ^{n} so A + 0 = A.

(iv) Observe that

(λ(A + B))x = λ((A + B)x) = λ(Ax + Bx)
= λ(Ax) + λ(Bx) = (λA)x + (λB)x
= (λA + λB)x

for all x ∈ ℝ^{n} so λ(A + B) = λA + λB.

(v) Observe that

((λ + μ)A)x = (λ + μ)(Ax) = λ(Ax) + μ(Ax)
= (λA)x + (μA)x = (λA + μA)x

for all x ∈ ℝ^{n} so (λ + μ) = λA + μA.

(vi) Observe that

((λμ)A)x = (λμ)Ax = λ(μAx) = (λ(μA))x

for all x ∈ ℝ^{n} so (λμ)A = λ(μA).

(vii) Observe that

(0A)x = 0(Ax) = 0 = \underline{0}x

for all x ∈ ℝ^{n} so 0A = \underline{0}.

(ix) We have

A − A = (1 − 1)A = 0A = \underline{0}.

Exercise 3.3.14

(i) Observe that ((AC)F )x = (AC)(F x) = A(C(F x)) = A((CF)x)) = ((ACF))x

for all x ∈ ℝ^{n}, so (AC)F = A(CF).

(ii) Observe that

(G(A + B))x = G((A + B)x) = G(Ax + Bx)
= G(Ax) + G(Bx) = (GA)x + (GB)x
= (GA + GB)x

for all x ∈ ℝ^{n}, so G(A + B) = GA + GB.

(iii) Observe that

((A + B)C)x = (A + B)(Cx) = A(Cx) + A(Bx)
= (AC)x + (BC)x
= (AC + BC)x

for all x ∈ ℝ^{n}, so (A + B)C = AC + BC. (iv) Observe that ((λA)C)x = (λA)(Cx) == λ(A(Cx)) = λ((AC)x)) = (λ(AC))x

for all x ∈ ℝ^{n}, so (λA)C = λ(AC).

Observe that

(A(λC))x = A((λC)x) = A(λ(Cx)) = λ(A(Cx))
= λ((AC)x)) = (λ(AC))x

for all x ∈ ℝ^{n} so A(λC) = λ(AC).

Exercise 3.4.5

The conjecture is false. The matrices A and B given by

A = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) and B = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix})

are shear matrices, but

AB = (\begin{matrix}2 & 1 \\ 1 & 1 \end{matrix})

is not.

Exercise 3.4.7

Since r ≠ s, we have δ_{rs} = 0 and

∑_{j=1}^{n}(δ_{ij} + λδ_{ir}δ_{js})(δ_{jk} + μδ_{jr}δ_{ks})

= ∑_{j=1}^{n}(δ_{ij}δ_{jk} + λδ_{ir}δ_{js}δ_{jk} + μδ_{jr}δ_{ks}δ_{ij} + λμδ_{ir}δ_{js}δ_{jr}δ_{ks})

= δ_{ik} + λδ_{ir}δ_{ks} + μδ_{ir}δ_{ks} + λμδ_{sr}δ_{ir}δ_{ks}

= δ_{ik} + (λ + μ)δ_{ir}δ_{ks}

The result is also obvious geometrically.

Exercise 3.5.1

(L_{k}L_{k−1} … L_{1}I)A = L_{k}L_{k−1} … L_{1}A = I,

so A is invertible with

L_{k}L_{k−1} … L_{1}I = A^{−1}.

Thus the same set of elementary operations applied in the same order which reduce A to I will transform I to A^{−1}.

Exercise 3.5.2

If we use row operations and column exchanges we can reduce A to I. If we omit column exchanges then the same row operations in the same order produce a matrix with exactly one 1 in each row and column and the remaining entries 0. By exchanging rows we can ensure that the first row is (1, 0, 0, …, 0), the second row (0, 1, 0, …, 0) and so on, i.e. we can produce I.

Exercise 4.1.1

(i) The vertices ABC, BCA, CAB are described in one sense (say anti clockwise) and the same vertices ACB, BAC, CBA in the opposite sense (say clockwise).

(ii) (Of course a bit hand waving.) OAXB, 0P QB, BXQ and 0AP anticlockwise but AXP Q clockwise. Writing | area Σ| for the absolute value of the area we have

| area OP QB|

= | area OAXB| − | area AP QX| + | area OAP | − | area BXQ|

= | area OAXB| − | area AP QX|.

Exercise 4.1.2

D(a, b) + D(b, a) = D(a + b, b) + D(a + b, a)

(By adding second entry to first.)

D(a + b, b) + D(a + b, a) = D(a + b, a + b)

(Since first entry the same, we can add as shown.)

D(a + b, a + b) = D(a + b − (a + b), a + b)

(By subtracting second entry from first entry.)

D(a + b − (a + b), a + b) = D(0, a + b)

(Just do the calculation.)

D(0, a + b) = 0

(Area degenerate parallelogram.) D(a, b) is the area of the parallelogram with vertices 0, a, a + b, b described in that order. D(b, a) is the area of the same parallelogram but with the vertices laid out in the opposite sense 0, b, a + b, a.

Exercise 4.1.3

D ((\begin{matrix}a_{1} \\ a_{2} \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix})) = D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix})) + D ((\begin{matrix}0 \\ a_{2} \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix}))

(Since second column the same, can add first columns.)

D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix})) + D ((\begin{matrix}0 \\ a_{2} \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix}))

= D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix}) − \frac{b_{1}}{a_{1}} (\begin{matrix}a_{1} \\ 0 \end{matrix})) + D ((\begin{matrix}0 \\ a_{2} \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix}) − \frac{b_{2}}{a_{2}} (\begin{matrix}0 \\ a_{2} \end{matrix}))

(Since we may subtract multiples of the first column from the second.)

D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix}) − \frac{b_{1}}{a_{1}} (\begin{matrix}a_{1} \\ 0 \end{matrix})) + D ((\begin{matrix}0 \\ a_{2} \end{matrix}) , (\begin{matrix}b_{1} \\ b_{2} \end{matrix}) − \frac{b_{2}}{a_{2}} (\begin{matrix}0 \\ a_{2} \end{matrix}))

= D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}0 \\ b_{2} \end{matrix})) + D ((\begin{matrix}0 \\ a_{2} \end{matrix}) , (\begin{matrix}b_{2} \\ 0 \end{matrix}))

(Just doing the calculation.)

D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}0 \\ b_{2} \end{matrix})) + D ((\begin{matrix}0 \\ a_{2} \end{matrix}) , (\begin{matrix}b_{2} \\ 0 \end{matrix}))

= D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}0 \\ b_{2} \end{matrix})) − D ((\begin{matrix}b_{2} \\ 0 \end{matrix}) , (\begin{matrix}0 \\ a_{2} \end{matrix}))

(Interchanging columns.)

D ((\begin{matrix}a_{1} \\ 0 \end{matrix}) , (\begin{matrix}0 \\ b_{2} \end{matrix})) − D ((\begin{matrix}b_{2} \\ 0 \end{matrix}) , (\begin{matrix}0 \\ a_{2} \end{matrix}))

= (a_{1}b_{2} − a_{2}b_{1})D ((\begin{matrix}1 \\ 0 \end{matrix}) , (\begin{matrix}0 \\ 1 \end{matrix})) = a_{1}b_{2} − a_{2}b_{1}.

(Using the rules D(λa, b) = D(a, λb) = λD(a, b) and the fact that that the area of a unit square is 1.)

Exercise 4.2.1

(Not a good idea, but possible as follows.)

D(AB) = D (\begin{matrix}a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{matrix})

= (a_{11}b_{11} + a_{12}b_{21})(a_{21}b_{12} + a_{22}b_{22})

− (a_{11}b_{12} + a_{12}b_{22})(a_{21}b_{11} + a_{22}b_{21}) = a_{11}a_{22}(b_{11}b_{22} − b_{12}b_{21}) − a_{12}a_{21}(b_{11}b_{22} − b_{12}b_{21}) = D(A)D(B).

Exercise 4.2.2

(i) We have

DI = D (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) = 1 × 1 − 0 × 0 = 1.

The area of a unit square is 1.

(i) We have

DE = D (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) = 0 × 0 − 1 × 1 = −1,

and

E (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}0x + 1y \\ 1x + 0y \end{matrix}) = (\begin{matrix}y \\ x \end{matrix}) .

We observe that

E (\begin{matrix}cos t \\ sin t \end{matrix}) = (\begin{matrix}sin t \\ cos t \end{matrix}) = (\begin{matrix}cos(\frac{π}{2} − t) \\ sin(\frac{π}{2} − t) \end{matrix})

runs clockwise from (0, 1) back to (0, 1) as (cos t, sin t)^{T} runs anticlockwise from (1, 0) back to (1, 0).

(iii) We have

D (\begin{matrix}1 & λ \\ 0 & 1 \end{matrix}) = 1 × 1 − λ × 0 = 1

and

D (\begin{matrix}1 & 0 \\ λ & 1 \end{matrix}) = 1 × 1 − 0 × λ = 1.

(iv) We have

D (\begin{matrix}a & 0 \\ 0 & b \end{matrix}) = a × b − 0 × 0 = ab

corresponding to the fact that a rectangle with sides of length a and b has area ab.

Exercise 4.3.2

As we hoped, ϵ_{11}a_{11}a_{12} + ϵ_{12}a_{11}a_{22}ϵ_{21}a_{21}a_{12} + ϵ_{22}a_{21}a_{22} = a_{11}a_{22} − a_{21}a_{12}.

Exercise 4.3.3

The result is automatic if r, s and t are not distinct.

When r, s andt are distinct, we check the case r = 1 when either s = 2, t = 3 or s = 3, t = 2 in Definition 4.3.1. We now do the same for r = 2 and r = 3.

Exercise 4.3.10

Observe that writing C = AB D = C^{T}, A′ = A^{T}, B′ = B^{T}

d_{ij} = c_{ji} = ∑_{r=1}^{m} a_{jr}b_{ri} = ∑_{r=1}^{m} a′_{rj}b′_{ir} = ∑_{r=1}^{m} b′_{ir}a′_{rj}

(for 1 ≤ i ≤ p, 1 ≤ j ≤ n). Thus (AB)^{T} = B^{T} A^{T}.

Exercise 4.3.13

x ↦ E_{r,s,λ}x is a shear which leaves volume unchanged.

Let D have ith diagonal entry d_{i}. x ↦ Dx (where D is a diagonal matrix) stretches by |d_{i}| in the 0x_{i} direction and reflects if d_{ii} < 0. Thus the volume is multiplied by ∏_{i=1}^{n} d_{i} = det D.

x ↦ P (σ)x exchanges the handedness of coordinates (or equivalently is a reflection in a particular plane) which multiplies volume by −1.

If A is any 3 matrix then A = a_{1}A_{2} … A_{k} with the A_{j} elementary matrices. Since

Ax = A_{1}(A_{2}(A_{3} … (A_{k}x) …))

the transformation x ↦ Ax rescales by ∏_{j=1}^{k} det A_{j} = det A.

Exercise 4.3.14

The map x ↦ Mx changes the length scale by λ and the volume scale by λ^{3}. (This is just a special case of the map x ↦ Dx considered in the previous question.)

Exercise 4.3.15

Let

F (r, s, t) = ϵ_{ijk}a_{ir}a_{js}a_{kt}.

We observe that F (r, s, t) is the determinant of the matrix with first row the rth row of A, second row the sth row of A and third row the tth row of A.

Thus interchanging any two of r, s, t multiplies F (r, s, t) by −1 and so

F (r, s, t) = ϵ_{rst}K

for some constant K.

Since F (1, 2, 3) = det A we have

ϵ_{ijk}a_{ir}a_{js}a_{kt} = F (r, s, t) = ϵ_{rst} det A.

(ii) Writing C = AB, we have

ϵ_{rst} det AB = ϵ_{rst} det C = ϵ_{ijk}c_{ir}c_{js}c_{kt}
= ϵ_{ijk}a_{iu}b_{ur}a_{jv}b_{vs}a_{kw}b_{wt} = ϵ_{ijk}a_{iu}a_{jv}a_{kw}b_{ur}b_{vs}b_{wt}
= ϵ_{uvw} det Ab_{ur}b_{vs}b_{wt} = ϵ_{rst} det det B

Taking r = 1, s = 2, t = 3 we get det AB = det A det B.

Exercise 4.4.1

There is no non-trivial χ. Observe that

χ_{ijkrs} = −χ_{kijrs} = χ_{jkirs} = −χ_{ijkrs},

so χ_{ijkrs} = 0.

Exercise 4.4.4

We have

ζ(σ) = \frac{(σ2 − σ1)(σ3 − σ1)(σ4 − σ1)(σ3 − σ2)(σ4 − σ2)(σ4 − σ3)}{(2 − 1)(3 − 1)(4 − 1)(3 − 2)(4 − 2)(4 − 3)},

ζ(τ) = \frac{(3 − 2)(1 − 2)(4 − 2)(1 − 3)(4 − 3)(4 − 1)}{(2 − 1)(3 − 1)(4 − 1)(3 − 2)(4 − 2)(4 − 3)} = 1,

ζ(ρ) = \frac{(3 − 2)(4 − 2)(1 − 2)(4 − 3)(1 − 3)(1 − 4)}{(2 − 1)(3 − 1)(4 − 1)(3 − 2)(4 − 2)(4 − 3)} = −1.

Exercise 4.4.6

(i) If t ∉ {1, 2, i}

αρα1 = αρ2 = α1 = i = τ1
αρα2 = αρ1 = α2 = 2 = τ2
αραi = αρi = αi = 1 = τi
αραt = αρt = αt = t = τt

(ii) If j = 1, then the result follows from part (iv) of the lemma. If not, let τ be as in part (iv) of the lemma and β ∈ S_{n} interchange 1 and j leaving the remaining integers unchanged. Then, by inspection,

βτβ(r) = κ(r)

for all 1 ≤ r ≤ n and so βτβ = κ. It follows that

ζ(κ) = ζ(β)ζ(τ)ζ(β) = ζ(β)^{2}ζ(τ) = −1.

Exercise 4.4.7

If all the suffices i, j, k, l are distinct, then σ(1) = i, σ(2) = j, σ(3) = k, σ(4) = l gives a unique σ ∈ S_{4}, so

ϵ_{ijkl} = ϵ_{σ(1)σ(2)σ(3)σ(4)} = ζ(σ)

does define ϵ_{ijkl}. If τ interchanges two suffices and leaves the rest unchanged

ϵ_{τ(i)τ(j)τ(k)τ(l)} = ϵ_{τσ(1)τσ(2)τσ(3)τσ(4)} = ζ(τσ) = ζ(τ)ζ(σ) = −ζ(σ) = −ϵ_{ijkl}

If i, j, k, l are not all distinct and i′, j′, k′, l′ is some rearrangement

ϵ_{ijkl} = 0 = −ϵ_{i′j′k′l′}.

Finally ϵ_{1234} = ζ(ι) = 1.

Exercise 4.4.8

Observe that 1^{−1} = 1 and (−1)^{−1} = −1. Since ζ(σ)ζ(σ^{−1}) = ζ(σσ^{−1}) = ζ(ι) = 1, it follows that ζ(σ) =

ζ(σ^{−1}).

Thus

det A^{T} = ∑_{σ∈S_{n}} ζ(σ)a_{1σ(1)}a_{2σ(2)} … a_{nσ(n)}

= ∑_{σ∈S_{n}} ζ(σ)a_{σ^{−1}(1)1}a_{σ^{−1}(2)2} … a_{σ^{−1}(n)n}

= ∑_{σ∈S_{n}} ζ(σ^{−1})a_{σ^{−1}(1)1}a_{σ^{−1}(2)2} … a_{σ^{−1}(n)n}

= ∑_{τ∈S_{n}} ζ(τ)a_{τ(1)1}a_{τ(2)2} … a_{τ(n)n}

= det A

Exercise 4.4.9

(i) We have

det (\begin{matrix}1 & 1 \\ x & y \end{matrix}) = y − x.

(ii) Each term ζ(σ)a_{σ(1)1}a_{σ(2)2}a_{σ(3)3} in the standard expansion is a multinomial of degree 3.

F (x, x, z) = det (\begin{matrix}1 & 1 & 1 \\ x & x & z \\ x^{2} & x^{2} & z^{2} \end{matrix}) = 0,

because two columns of the matrix are identical. Thus y − x must be be a factor of F (x, y, z). Similarly z − y and z − x are factors. Since F has degree 3 and (y − x)(z − y)(z − x) has degree 3,

F (x, y, z) = A(y − x)(z − y)(z − x)

for some constant A. By considering the terms in the standard determinant expansion, we know that the coefficient of yz^{2} is 1. Thus A = 1 and

F (x, y, z) = (y − x)(z − y)(z − x).

(iii) Since each term in the standard expansion of det V is a multinomial of degree 0 + 1 + … + (n − 1) = (n − 1)n/2, F is a multinomial of degree (n − 1)n/2.

F (x_{1}, x_{1}, x_{3}, x_{4}, …, x_{n}) = 0

because two columns of the associated matrix are identical. Thus x_{2} − x_{1} must be be a factor of F. Similarly x_{j} −x_{k} is a factor for each j > k. Since F has degree (n − 1)n/2 and ∏_{i>j}(x_{i} − x_{j}) has the same degree,

F (x_{1}, x_{2}, …, x_{n}) = A ∏_{i>j}(x_{i} − x_{j})

for some constant A.

By considering the terms in the standard determinant expansion, we know that the coefficient of ∏_{j=2}^{n} x_{j}^{j−1} is 1. Thus A = 1 and

F (x_{1}, x_{2}, …, x_{n}) = ∏_{i>j}(x_{i} − x_{j}).

(iv) We have

ζ̃_{x}(σ) = \frac{F (x_{σ(1)}, x_{σ(2)}, …, x_{σ(n)})}{F (x_{1}, x_{2}, …, x_{n})} = ∏_{i>j} \frac{x_{σ(i)} − x_{σj}}{x_{i} − x_{j}} .

Now

∏_{i>j}(x_{σ(i)} − x_{σj}) = Bσ ∏_{i>j}(x_{i} − x_{j})

where B depends only on σ (and not on x). Thus

ζ̃_{x}(σ) = ζ̃_{y}(σ)

with y_{j} = j and so

ζ̃_{x}(σ) = ζ(σ).

Thus ζ = ζ̃.

Exercise 4.5.1

Each of the n! terms involves n multiplications (in addition to finding the appropriate sign) so we need n × n! multiplications. We can either use a Stirling approximation or an electronic calculator (which probably uses some version of Stirling’s formula) to obtain an estimate of about 36 000 000 for n = 10 and about 4.9 × 10^{19} for n = 20.

Exercise 4.5.3

If A = (a_{ij}) is an n × n matrix which is both upper and lower triangular then a_{ij} = 0 if i < j or if j < i so A is diagonal.Each of the n! terms involves n multiplications (in addition to finding the appropriate sign) so we need nn! multiplications. We can either use a Stirling approximation or an electronic calculator (which probably uses some version of Stirling’s formula) to obtain an estimate of about 36 000 000 for n = 10 and about 4.9 × 10^{19} for n = 20.

Exercise 4.5.3

If A = (a_{ij}) is an n × n matrix which is both upper and lower triangular then a_{ij} = 0 if i < j or if j < i so A is diagonal.

Exercise 4.5.5

(i) By row and column operations on the first r rows and columns we can reduce

C = (\begin{matrix}A & 0 \\ 0 & B \end{matrix}) to C′ = (\begin{matrix}A′ & 0 \\ 0 & B \end{matrix})

with det C = K det C′, det A′ = K det A and A′ lower triangular.

By row and column operations on the last s rows and columns we can reduce C′ to C′′ with

C′′ = (\begin{matrix}A′ & 0 \\ 0 & B′ \end{matrix})

with det C′′ = K′ det C′, det B′ = K′ det B and B′ lower triangular.

We now have

det C = KK′ det C′′ = KK′ det A′ det B′ = det A det B.

(ii) FALSE. Consider

F = (\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix})

With the suggested notation

det A = det B = det C = det D = 0

but det F = 1 ≠ 0 = det A det D − det B det C.

Exercise 4.5.6

Dividing the first row by 2,

det (\begin{matrix}2 & 4 & 6 \\ 3 & 1 & 2 \\ 5 & 2 & 3 \end{matrix}) = 2 det (\begin{matrix}1 & 2 & 3 \\ 3 & 1 & 2 \\ 5 & 2 & 3 \end{matrix}) .

Subtracting multiples of the first row from the second and third row,

2 det (\begin{matrix}1 & 2 & 3 \\ 3 & 1 & 2 \\ 5 & 2 & 3 \end{matrix}) = 2 det (\begin{matrix}1 & 2 & 3 \\ 0 & −5 & −7 \\ 0 & −8 & −12 \end{matrix}) .

Expanding by first row,

2 det (\begin{matrix}1 & 2 & 3 \\ 0 & −5 & −7 \\ 0 & −8 & −12 \end{matrix}) = 2 det (\begin{matrix}−5 & −7 \\ −8 & −12 \end{matrix}) .

Dividing the first row by −1 and the second row by −4,

(\begin{matrix}−5 & −7 \\ −8 & −12 \end{matrix}) = 8 det (\begin{matrix}5 & 7 \\ 2 & 3 \end{matrix}) .

Subtracting two times the second row from the first (hardly necessary, we could have finished the calculation here),

8 det (\begin{matrix}5 & 2 \\ 2 & 1 \end{matrix}) = 8 det (\begin{matrix}1 & 0 \\ 2 & 1 \end{matrix}) .

From the definition,

8 det (\begin{matrix}1 & 0 \\ 2 & 1 \end{matrix}) = 8.

Exercise 4.5.7

(i) Observe that

det (\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}) = det (\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix})^{T}

= ∑_{σ∈S_{3}} ∏_{i=1}^{3} a_{iσ(i)}

= a_{11}a_{22}a_{33} − a_{11}a_{23}a_{32} + a_{12}a_{23}a_{31} − a_{12}a_{21}a_{33}

+ a_{13}a_{21}a_{32} − a_{13}a_{22}a_{31}

= a_{11} det (\begin{matrix}a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix}) − a_{12} det (\begin{matrix}a_{21} & a_{23} \\ a_{31} & a_{33} \end{matrix}) + a_{31} det (\begin{matrix}a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix}) .

(ii) We have

det (\begin{matrix}2 & 4 & 6 \\ 3 & 1 & 2 \\ 5 & 2 & 3 \end{matrix}) = 2 det (\begin{matrix}1 & 2 \\ 2 & 3 \end{matrix}) − 4 det (\begin{matrix}3 & 2 \\ 5 & 3 \end{matrix}) + 6 det (\begin{matrix}3 & 1 \\ 5 & 2 \end{matrix})
= 2 × (−1) − 4(−1) + 6 × 1 = 8.

Exercise 4.5.8

(i) We look at each of the four terms a_{r1} det B_{r1} say. If r ≠ 1, j then det B_{r1} changes sign to give −a_{r1} det B_{r1}. a_{11} det B_{11} changes to −a_{j1} det B_{j1} and a_{j1} det B_{j1} changes to −a_{11} det B_{11}.

Thus

F(Ã) = −F(A).

(iii) The result is already proved if i or j takes the value 1. If not, interchanging row 1 with row i, then interchanging row 1 of the new matrix with row j of the new matrix and finally interchanging row 1 and i again transforms A to Ã. By part (i)

F(Ã) = −F(A).

If rows i and j are the same this gives F(A) = −F(A) and so F(A) = 0.

(iv) By inspection

F(Ã) = F(A) + F(B)

where B is the matrix A with the first row replaced by the ith row. By part (iii), F(B) = 0 so

F(Ã) = F(A).

By considering the effect of interchanging the first and jth row, we get the more general result

F(Ā) = F(A).

(v) If we now carry out the diagonalisation procedure of Theorem 3.4.8 on A using the rules above and observe that

F(I) = 1 = det I,

we get

F(A) = det A.

Exercise 4.5.10

(i) The argument goes through essentially word for word (replacing 4 by n). The formula

∑_{j=1}^{n} a_{1j}A_{1j} = det A.

is the row expansion formula so obtained.

(ii) If i = 1 there is nothing to prove. If i ≠ 1, we argue as follows.

If B is the matrix obtained from A by interchanging row 1 and row i.

det A = − det B = − ∑_{j=1}^{n} b_{1j}B_{1j} = ∑_{j=1}^{n} a_{ij}A_{ij}.

(iii) If i ≠ k, then

∑_{j=1}^{n} a_{ij}A_{kj} = det C

where C is an n × n matrix with ith and kth rows the same. Thus

∑_{j=1}^{n} a_{ij}A_{kj} = 0.

(iv) Conditions (ii) and (iii) together give

∑_{j=1}^{n} a_{ij}A_{kj} = δ_{kj} det A

Exercise 4.5.13

We have

det A det A^{−1} = det AA^{−1} = det I = 1

so det A^{−1} = (det A)^{−1}.

Exercise 4.5.14

The formula of Exercise 4.5.10 (v) shows that ∑_{k=1}^{n} b_{k}A_{kj} is the determinant of a matrix obtained from A by replacing the jth column of A by b.

Thus

∑_{j=1}^{n} a_{ij} det B_{j} = ∑_{j=1}^{n} a_{ij} ∑_{k=1}^{n} b_{k}A_{kj}

= ∑_{j=1}^{n} ∑_{k=1}^{n} a_{ij}b_{k}A_{kj}

= ∑_{k=1}^{n} ∑_{j=1}^{n} a_{ij}b_{k}A_{kj}

= ∑_{k=1}^{n} b_{k} ∑_{j=1}^{n} a_{ij}A_{kj}

= ∑_{k=1}^{n} b_{k}δ_{ik} det A = b_{i} det A

Thus, if det A ≠ 0,

x_{j} = \frac{det B_{j}}{det A}

gives a solution of Ax = b (and, since there is only one solution, it is the solution).

Exercise 4.5.15

The statement is true but you still need to find the determinant of two n×n matrices. Unless we deal with very small systems of equations (corresponding to a 3×3 matrix, say) the labour involved in computing det A (at least by the methods given in this book or any I am aware of) is, at best, comparable with the effort of finding all the solutions of Ax = b

Exercise 4.5.16

(i) Observe that, since S_{n} is a group

perm A^{T} = ∑_{σ∈S_{n}} ∏_{i=1}^{n} a_{iσ(i)}

= ∑_{σ∈S_{n}} ∏_{i=1}^{n} a_{σ^{−1}(i)i}

= ∑_{σ∈S_{n}} ∏_{i=1}^{n} a_{τ(i)i} = perm A

(ii) Both statements false. If we set

A = (\begin{matrix}1 & 1 \\ 1 & 1 \end{matrix}) , B = (\begin{matrix}1 & 1 \\ 1 & −1 \end{matrix})

then perm A = 4 ≠ 0, but det A = 0, and det B = −2, but perm B = 0.

(iii) Write A_{ji} for the n − 1 × n − 1 matrix obtained by removing the ith row and jth column from A. We have

perm A = ∑_{i=1}^{n} a_{1i} perm A_{i1}

(iv) By (iii),

perm A ≤ nK max_{i} A_{i1}

so, by induction, | perm A| ≤ n!K^{n}.

If we take A(n) to be the n × n matrix with all entries K (where K ≥ 0), then

perm A(n) = nK perm A(n − 1)

so, by induction, perm A(n) = n!K^{n}.

(v) If B is the n × n matrix with b_{ij} = |a_{ij}| | det A| ≤ perm B ≤ n!K^{n}.

Hadamard’s inequality is proved later.

Exercise 4.5.17

(i) True. If A is antisymmetric,

(\begin{matrix}0 & b \\ −b & 0 \end{matrix}) ,

and det A = b^{2} ≠ 0 if A ≠ 0. (ii) False. Consider

(\begin{matrix}0 & 1 & 0 & 0 \\ −1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) .

(iii) True.

det A = det A^{T} = det(−A) = (−1)^{n} det A = − det A.

Exercise 5.1.1

If we work over ℤ we cannot always divide. Thus the equation 2x = 1 has no solution, although the 1 × 1 matrix

(2) has non-zero determinant.

Exercise 5.2.7

I would be inclined to pick (iv) and (vi). (ii) (f + g)(x) = f(x) = g(x) = g(x) + f(x) = (g + f)(x) for all x so

f + g = g + f.

(iii) (f +0)(x) = f(x)+0(x) = f(x)+0 = f(x) for all x so f +0 = f. (iv) (λ(f +g))(x) = λ((f +g)(x)) = λ(f(x)+g(x)) = λf(x)+λg(x) =

(λf)(x) + (λg)(x) = (λf + λg)(x) for all x so λ(f + g) = λf + λg.

(v) ((λ + μ)f)(x) = (λ + μ)(f(x)) = λf(x) + μf(x) = (λf)(x) + (μg)(x) + (λf + μg)(x) for all x so (λ + μ)f = λf + μf.

(vi) ((λμ)f)(x) = (λμ)(f(x)) = (λμ)f(x) = λ(μf(x)) = λ((μf)(x)) = (λ(μf))(x) for all x so (λμ)f = λ(μf).

(vii) (1f)(x) = 1 × f(x) = f(x) and (0f)(x) = 0 × f(x) = 0 = 0(x) for all x and so 1f = f and 0f = 0.

Exercise 5.2.8

The correspondence 𝔽^{X} ↔ 𝔽^{n} given by

f ↔ (f(1), f(2), f(3), …, f(n))

identifies 𝔽^{X} with the known vector space 𝔽^{n}.

Exercise 5.2.11

Observe that all these sets are subsets of the vector space ℝ^{ℝ} so we may use Lemma 5.2.10.

(i) Subspace. If f and g are 3 times differentiable so is λf + μg.

(ii) Not a vector space. Let f(t) = t^{2}. Then f is in the set but (−1)f is not.

(iii) Not a vector space. Let P (t) = t. Then P is in the set but (−1)P is not.

(iv) Subspace. If P, Q in set, then λP + μQ is a polynomial and (λP + μQ)′(1) = λP′(1) + μQ′(1) = 0

(v) Subspace. If P, Q in set, then λP + μQ is a polynomial and

∫_{0}^{1}(λP + μQ)(t) dt = ∫_{0}^{1}(λP (t) + μQ(t) dt
= λ ∫_{0}^{1} P (t) dt + μ ∫_{0}^{1} q(t) dt = 0.

(vi) Not a vector space. Let

h(t) = max{(1 − 10|t|), 0}

and observe that ∫ h(t)^{3} dt = A ≠ 0. If f(t) = h(t) − h(t + 1/5) and g(t) = h(t) − h(t + 2/5), then

∫_{−1}^{1} f(t)^{3} dt = ∫_{−1}^{1} g(t)^{3} dt = A − A = 0

but

∫_{−1}^{1}(f(t) + g(t))^{3} dt = 8A − 2A ≠ 0.

(vii) Not a vector space. If f(t) = −g(t) = t^{3} then f and g have degree exactly 3 but f + g = 0 does not.

(viii) Not a vector space. If P (t) = t^{2} − t, Q(t) = −t^{2}, then P and Q are in the set, but not P + Q.

Exercise 5.3.2

T (0) = T (00) = 0T (0) = 0.

Exercise 5.3.3

Let f, g ∈ D

(i) We have δ(λf + μg) = (λf + μg)(0) = λf(0) + μg(0) = λδf + μδg.

(ii) We have

D(λf + μg) = (λf + μg)′ = λf′ + μg′ = λDf + μDg. (iii) We have

(K(λf + μg))(x) = (x^{2} + 1)(λf + μg)(x) = (x^{2} + 1)(λf(x) + μg(x))
= λ(x^{2} + 1)f(x) + μ(x^{2} + 1)g(x)

= λ(Kf)(x) + μ(Kg)(x) = (λKf + μKg)(x)

for all x and so

K(λf + μg) = λKf + μKg.

(iv) We have

(J(λf + μg))(x) = ∫_{0}^{x}(λf(t) + μg(t)) dt
= λ ∫_{0}^{x} f(t) dt + μ ∫_{0}^{x} g(t) dt = (λJf + μJg)(x)

for all x and so

J(λf + μg) = λJf + μJg.

Exercise 5.3.10

(i) We have

ι(λx + μy) = λx + μy = λιx + μιy

so ι is linear (ii) Just a restatement of definitions. (iii) The fundamental theorem of the calculus states that DJ = ι.

However JD1 = J0 = 0 ≠ 1. To see that J is injective observe that

Jf = Jg ⇒ DJf = DJg ⇒ f = g

However Jf(0) = 0 so 1 ∉ JD so J is not surjective.

To see that D is surjective observe that D(Jf) = f. However D0 = D1 = 0 so D is not injective.

(iv) Observe that

(αβ)(β^{−1}α^{−1}) = α(ββ^{−1})α^{−1} = αια^{−1} = α^{−1}α = ι

and similarly

(β^{−1}α^{−1})(αβ) = ι

Exercise 5.3.11

The only if part is trivial To prove the if part suppose

T u = 0 ⇒ u = 0.

Then

T x = T y ⇒ T x − T y = 0

⇒ T (x − y) = 0 ⇒ x − y = 0 ⇒ x = y.

Exercise 5.3.14

(i) If α, β ∈ GL(U), then α and β are bijective, so αβ is, so αβ ∈ GL(U).

(ii) Observe that if α, β, γ ∈ GL(U), then ((αβ)γ)u = (αβ)(γu) = (α(β(γu))) = α(β(γu)) = α((βγ)u) = (α(βγ))u for all u ∈ U so (αβ)γ = α(βγ).

(iii) ι ∈ GL(U) and αι = ια = α for all α ∈ GL(U). (iv) Use the definition of GL(U). Let α(x, y) = (x + y, x − y), β(x, y) = (y, x). Then αβ(x, y) =

(x + y, y − x), βα(x, y) = (x − y, x + y). (Could use matrices.)

Exercise 5.3.16

(i) A subgroup. If α(a) = βa = a, then ιa = a, α^{−1}(a) = a and

(αβ)a = α(βa) = αa = a.

(ii) Not necessarily a group. Work in ℝ^{2}. Let α(x, y) = (x, 0), a = (1, 0), β(x, y) = (0, y), b = (0, 1).

(iii) Not necessarily a group. Work in ℝ^{3}. Let alpha be a rotation through π about the x axis, β a rotation through π about an axis running through (0, 0, 0) and (1, 1, 1) α^{2} = β^{2} = ι but by considering (αβαβ(1, 1, 1) we see that (αβ)^{2} ≠ ι.

Exercise 5.4.3

Observe that

∑_{j=1}^{n} λ_{j}(e_{j} + y) = 0 ⇒ ∑_{j=1}^{n} λ_{j}e_{j} + ∑_{j=1}^{n} ∑_{k=1}^{n} λ_{j}a_{k}e_{k} = 0

⇒ ∑_{j=1}^{n} λ_{j}e_{j} + ∑_{j=1}^{n} ∑_{k=1}^{n} λ_{k}a_{j}e_{j} = 0

⇒ ∑_{j=1}^{n} (λ_{j} + a_{j} ∑_{r=1}^{n} λ_{r}) e_{j} = 0

⇒ λ_{j} + a_{j} ∑_{r=1}^{n} λ_{r} = 0 ∀j

Thus, if

∑_{j=1}^{n} λ_{j}e_{j} + y = 0

and we write K = − ∑_{r=1}^{n} λ_{r},

λ_{j} = Ka_{j}.

Summing we obtain

−K = ∑_{j=1}^{n} λ_{j} = K ∑_{j=1}^{n} a_{j}

so, a_{1} + a_{2} + … + a_{n} + 1 ≠ 0, K = 0 and λ_{j} = 0 showing that the vectors e_{1} + y are linearly independent.

If a_{1} + a_{2} + … + a_{n} + 1 = 0, then

∑_{j=1}^{n} a_{j}(e_{j} + y) = 0

and (since y ≠ 0) not all the a_{j} are zero, so we do not have linear independence.

Exercise 5.4.11

U and V are the null spaces of linear maps from ℝ^{4} to R^{2}.

The elements of U ∩ V are given by

x + y − 2z + t = 0
−x + y + z − 3t = 0
x − 2y + z + 2t = 0
y + z − 3t = 0.

Subtracting the 4th equation from the 2nd yields x = 0 so the system becomes

y − 2z + t = 0
x = 0
−2y + z + 2t = 0
y + z − 3t = 0.

Adding the 1st and 3rd equation reveals that the 4th equation is superfluous so the system is

y − 2z + t = 0
x = 0
−2y + z + 2t = 0,

or equivalently

x = 0
y − 2z + t = 0
−3z + 6t = 0

so x = 0, z = 2t, y = −2t. Thus U ∩ V has basis e_{1} = (0, −2, 2, 1)^{T}.

The equations for U yield

x + y − 2z + t = 0
2y − z − 2t = 0

so by inspection e_{2} = (1, 1, 2, 0)^{T} is in U. Since e_{1} and e_{2} are linearly independent and dim U = 2 we have basis for U

(To see that dim U = 2 either quote general theorems or observe that x and y determine z and t.)

The equations for v yield

x − 2y + z + 2t = 0
y + z − 2t = 0

so, by inspection, e_{3} = (3, 1, 2, 0)^{T} is in U. Since e_{1} and e_{3} are linearly independent and dim V = 2 (argue as before) we have basis for V.

The proof of Lemma 5.4.10 now shows that e_{1}, e_{2} form a basis for e_{3}

Exercise 5.4.12

(i) Since U ⊇ V + W ⊇ V, W we have

dim U ≥ dim(V + W) ≥ dim V, dim W

By Lemma 5.4.10

dim(V + W) = dim V + dim W − dim(V ∩ W) ≥ dim V + dim W. Putting these two results together, we get min{dim U, dim V + dim W } ≥ dim(V + W) ≥ max{dim V, dim W }.

(ii) Consider a basis e_{1}, e_{2}, …, e_{n} for U.

Let

V = span{e_{1}, e_{2}, …, e_{r}}
W = span{e_{t−s+1}, e_{t−s+2}, …, e_{t}}

Then

V + W = span{e_{1}, e_{2}, …, e_{t}}

and

dim V = r, dim W = s, dim(V + W) = t.

Exercise 5.4.13

We use row vectors. E is the null space of a linear map from ℝ^{3} to ℝ so a subspace. Let e_{1} = (1, −1, 0), e_{2} = (1, 0, −1). We have e_{1}, e_{2} ∈ U.

x_{1}e_{1} + x_{2}e_{2} = 0 ⇒ (x_{1} + x_{2}, −x_{1}, −x_{2}) = 0 ⇒ x_{1} = x_{2} = 0 so e_{1}, e_{2} are independent.

If x ∈ U then

x = (x_{1}, x_{2}, x_{3}) ⇒ x = (−x_{2} − x_{3}, x_{2}, x_{3}) ⇒ x = x_{2}e_{1} + x_{3}e_{2}. Thus e_{1}, e_{2} span U and so form a basis.

I do not think everybody will choose the basis and so there cannot be a ‘standard basis’ in this case.

Exercise 5.4.14

(i) 0 ∈ V for all V ∈ V and so 0 ∈ ⋂_{V ∈V} V. Further, if λ, μ ∈ 𝔽,

u, v ∈ ⋂_{V ∈V} V ⇒ u, v ∈ V ∀V ∈ V

⇒ λu + μv ∈ V ∀V ∈ V

⇒ λu + μv ∈ ⋂_{V ∈V} V.

Thus ⋂_{V ∈V} V is a subspace of U.

(ii) By (i) W is a subspace of U. Since E ⊆ V for all V ∈ V, we have E ⊆ W. If W′ is as stated W′ ∈ V so W′ ⊇ W.

(iii) Let

W′ = {∑_{j=1}^{n} λ_{j}e_{j} : λ_{j} ∈ 𝔽 for 1 ≤ j ≤ n} .

Observe that E ⊆ W′ (take λ_{j} = δ_{ij}). and W is a subspace of U since

λ ∑_{j=1}^{n} λ_{j}e_{j} + μ ∑_{j=1}^{n} μ_{j}e_{j} = ∑_{j=1}^{n}(λλ_{j} + μμ_{j}e_{j}.

Any subspace of U containing E must certainly contain W′. Thus W = W′.

Exercise 5.5.3

Always false. 0 ∉ α^{−1}v.

Exercise 5.5.5

We use the rank-nullity theorem.

α injective ⇔ α^{−1}(0) = 0

⇔ dim α^{−1}(0) = 0 ⇔ n − dim α^{−1}(0) = n ⇔ dim α(U) = n ⇔ α surjective

We now note that bijective means injective and surjective and that a map is bijective if and only if it is invertible.

Exercise 5.5.6

Adding the second and third row to the top row, we get

det (\begin{matrix}a & b & b \\ b & a & b \\ b & b & a \end{matrix}) = det (\begin{matrix}a + 2b & a + 2b & a + 2b \\ b & a & b \\ b & b & a \end{matrix})

= (a + 2b) (\begin{matrix}1 & 1 & 1 \\ b & a & b \\ b & b & a \end{matrix}) = (2a + b) (\begin{matrix}1 & 1 & 1 \\ 0 & a − b & 0 \\ 0 & 0 & a − b \end{matrix}) = (2a + b)(a − b)^{2}.

Thus, if a ≠ b and a ≠ −2b, α has rank 3 α^{−1}(0) = 0, has empty basis and α(U) = U has basis (1, 0, 0)^{T}, (0, 1, 0)^{T}, (0, 0, 1)^{T}.

If a = b, there are two cases. If a = 0, α has rank 3 α(U) = 0 has empty basis and α^{−1}(0) = U has basis (1, 0, 0)^{T}, (0, 1, 0)^{T}, (0, 0, 1)^{T}.

If a = b ≠ 0, then

(x, y, z) ∈ α^{−1}(0) ⇔ x + y + z = 0

so α has rank 3 − 2 = 1 α^{−1}(0) has basis (1, 0, −1)^{T}, (0, 1, −1)^{T}. U has basis (1, 0, −1)^{T}, (0, 1, −1)^{T}, (0, 0, 1)^{T}, so α(U) has basis α(0, 0, 1)^{T} = (a, a, a)^{T} so basis (1, 1, 1)^{T}.

If a = −2b ≠ 0, then

(x, y, z) ∈ α^{−1}(0 ⇔ x − 2y + z = 0, x + y − 2z = 0 ⇔ x = y = z

so α has rank 3 − 1 = 2 α^{−1}(0) has basis (1, 1, 1)^{T}. U has basis (1, 1, 1)^{T}, (0, 1, 0)^{T}, (0, 0, 1)^{T} so α(U) has basis α(0, 1, 0)^{T} = (b, a, b)^{T} α(0, 1, 0)^{T} = (b, b, a)^{T} so basis (1, −2, 1)^{T}, (1, 1, −2).

Exercise 5.5.13

(i) If A is an m × n matrix with rows a_{1}, a_{2}, … a_{m} then the row rank of A is dimension of

span{a_{1}, a_{2}, …, a_{m}}.

(ii) If B is a non-singular m × m matrix then if x is a row vector with m entries

xB = 0 ⇒ xBB^{−1} = 0B^{−1} ⇒ x = 0 ⇒ xB = 0

so

x = 0 ⇔ xB = 0.

It follows that

∑_{j=1}^{m} λ_{j}a_{j}B = 0 ⇔ (∑_{j=1}^{m} λ_{j}a_{j}) B = 0

⇔ ∑_{j=1}^{m} λ_{j}a_{j} = 0,

so

dim span{a_{1}B, a_{2}B, …, a_{m}B} = dim span{a_{1}, a_{2}, …, a_{m}}

and the row rank of A and AB are the same.

Since the elementary column operations correspond to post multiplication by invertible matrices, the row rank is unaltered by elementary column operations. A similar argument with pre-multiplication replacing post-multiplication shows that row rank is unaltered by elementary row operations. Similarly the column rank is unaltered by elementary row and column operations.

(iii) By elementary row and column operations any matrix A can be transformed to a matrix C = (c_{ij}) with c_{ij} = 0 if i ≠ j. By inspection, C has row and column rank equal to the number of non-zero entries c_{ii}. Thus, by (ii), the row rank and column rank of A are the same.

Exercise 5.5.14

If a = b = 0, then r = 0.

If a ≠ 0 and b = 0, the matrix is

(\begin{matrix}a & 0 & 0 & 0 \\ 0 & a & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 & 0 & 0 & a \end{matrix})

The first and third rows are identical, so r ≤ 3. However, the matrix obtained by removing the third row and first column

(\begin{matrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix})

is non-singular, so r = 3.

If a = 0, b ≠ 0, we get

(\begin{matrix}0 & 0 & b & 0 \\ 0 & 0 & 0 & b \\ 0 & 0 & 0 & b \\ 0 & b & 0 & 0 \end{matrix}) .

The second and third rows are identical, so r ≤ 3. However, the matrix obtained by removing the third row and first column

(\begin{matrix}0 & b & 0 \\ 0 & 0 & b \\ b & 0 & 0 \end{matrix})

is non-singular (for example, because the determinant is non-zero), so r = 3.

If a = b ≠ 0, the matrix is

(\begin{matrix}a & 0 & a & 0 \\ 0 & a & 0 & a \\ a & 0 & 0 & a \\ 0 & a & 0 & a \end{matrix}) .

The second and fourth rows are identical so r ≤ 3. However, the matrix obtained by removing the fourth row row and column

(\begin{matrix}a & 0 & a \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix})

is non-singular (for example, because the determinant is non-zero) so r = 3.

If a, b ≠ 0 and a ≠ b, then subtracting the first row from the third row and the a/b times the second row from the fourth

(\begin{matrix}a & 0 & b & 0 \\ 0 & a & 0 & b \\ a & 0 & 0 & b \\ 0 & b & 0 & a \end{matrix}) → (\begin{matrix}a & 0 & b & 0 \\ 0 & a & 0 & b \\ 0 & 0 & −b & b \\ 0 & b & 0 & a \end{matrix}) → (\begin{matrix}a & 0 & b & 0 \\ 0 & a & 0 & b \\ 0 & 0 & −b & b \\ 0 & 0 & 0 & a − b^{2}/a \end{matrix})

yields a non-singular matrix (e.g. by looking at the determinant), so r = 4 and the original matrix was non-singular.

Exercise 5.5.16

Consider the linear maps α and β whose matrices with respect to a given basis are (a_{ij}) with a_{ii} = 1 if 1 ≤ i ≤ r, a_{ij} = 0, otherwise, b_{ii} = 1 if r − t + 1 ≤ i ≤ r + s − t, b_{ij} = 0, otherwise. (Note that we use the facts that r, s ≥ t and r + s − t ≤ n.) Then AB has matrix (c_{ij}) with c_{ii} = 1 if r − t + 1 ≤ i ≤ r, c_{ij} = 0, otherwise. Then α has rank r, β rank s and γ rank t.

Exercise 5.5.17

By noting that (1, −1, 0, 0, …, 0)^{T}, (1, 0, −1, 0, …, 0) …, (1, 0, 0, 0, …, −1)^{T} are n − 1 linearly independent eigenvectors with eigenvalue 0 and (1, 1, …, 1)^{T} is an eigenvector with eigenvalue n for the n × n matrix A with all entries 1, we see that the characteristic polynomial of A is det(tI − A) = t^{n−1}(t − n). Thus A + sI is invertible if and only if s ≠ 0, −n.

Let P be as stated. Then P P^{T} = Q with

q_{ij} = ∑_{r=1}^{n} p_{ir}p_{jr} = \begin{cases}1 & if i ≠ j \\ k & if i = j \end{cases}

Thus P P^{T} = A + (k − 1)I and, since k ≥ 2 we have P P^{T} invertible so P is invertible so of full rank n.

If k = 1, then there can only be one party P = (1 1 1 … 1)^{T} and has rank 1. The proof fails because P P^{T} = A is not of full rank.

Exercise 5.5.18

(i) Observe that (α + β)U ⊆ V, so dim V ≥ dim(α + β)U.

Since (α+β)u = αu+βu ∈ αU +βU, we have (α+β)U ⊆ αU +βU, so, by Lemma 5.4.10,

dim(α + β)U ≤ dim(αU + βU) ≤ dim αU + dim βU.

Thus

min{dim V, rank α + rank β} ≥ rank(α + β).

(ii) By (i),

rank(α+β)+rank β = rank(α+β)+rank(−β) ≥ rank ((α+β)−β) = rank α.

Thus

rank(α + β) ≥ rank α − rank β

and

rank(α + β) = rank(β + α) ≥ rank β − rank α.

Thus

rank(α + β) ≥ | rank α − rank β|.

(iii) Since α + β = β + α, there is no loss of generality in assuming r ≥ s.

Fix bases for U and V. If t ≥ r, consider the linear map α given by the matrix (a_{ij}) with a_{ii} = 1 if 1 ≤ i ≤ r, a_{ij} = 0 otherwise, and the linear map β given by the matrix (b_{ij}) with b_{ii} = 1 if t − s + 1 ≤ i ≤ t, b_{ij} = 0 otherwise. Then α + β has matrix (c_{ij}) with c_{ii} = 1 if 1 ≤ i ≤ t − s c_{ii} = 2 if t − s + 1 ≤ i ≤ r, c_{ii} = 1 if r + 1 ≤ i ≤ t, c_{ij} = 0 otherwise. Thus rank α = r, rank β = s rank(α + β) = t.

If t ≤ r − 1, consider the linear map α given by the matrix (a_{ij}) with a_{ii} = 1 if 1 ≤ i ≤ r, a_{ij} = 0 otherwise, and the linear map β given by the matrix (b_{ij}) with b_{ii} = −1 if 1 ≤ i ≤ r − t, b_{ii} = 1 if r − t + 1 ≤ i ≤ s, b_{ij} = 0 otherwise. Then α + β has matrix (c_{ij}) with c_{ii} = 2 if r − t + 1 ≤ i ≤ s c_{ii} = 1 if s + 1 ≤ i ≤ r, c_{ij} = 0 otherwise. Thus rank α = r, rank β = s rank(α + β) = t.

Exercise 5.5.19

To avoid repetition, we go straight to (iii). We consider n×n matrices and say A ∈ Γ if and only if there is a K such that ∑_{r=1}^{n} a_{rj} = K for all j and ∑_{r=1}^{n} a_{ir} = K for all i.

(i)′ Observe that

A, B ∈ Γ ⇒ ∑_{r=1}^{n} a_{rj} = ∑_{r=1}^{n} a_{ir}, ∑_{i=1}^{n} b_{ij} = ∑_{j=1}^{n} b_{ij} ∀i, j

⇒ ∑_{r=1}^{n} λa_{rj} + μb_{rj} = ∑_{r=1}^{n} λ_{ir} + μb_{ir} ∀i, j

λA + μB ∈ Γ.

Since 0 ∈ Γ, we have γ a vector space.

(ii)′ Let Γ_{0} be the subset of Γ such that

∑_{r=1}^{n} a_{rj} = ∑_{r=1}^{n} a_{ir} = 0.

We observe that Γ_{0} is a subspace of Γ since 0 ∈ Γ_{0} and

A, B ∈ Γ ⇒ ∑_{r=1}^{n} a_{rj} = ∑_{r=1}^{n} a_{ir} = 0, ∑_{i=1}^{n} b_{ij} = ∑_{j=1}^{n} b_{ij} = 0 ∀i, j

⇒ ∑_{r=1}^{n} λa_{rj} + μb_{rj} = ∑_{r=1}^{3} λ_{ir} + μb_{ir} = 0 ∀i, j

λA + μB ∈ Γ_{0}.

We note that if E is the n × n matrix all of whose entries are 1 then any basis of Γ_{0} together with E gives a basis for Γ.

If 1 ≤ i, j ≤ n − 1, let E(i, j) be the n × n matrix with entry 1 in the (i,j)th place and (n, n)th place and entry −1 in the (i, n)th place and (n, j)th place. We observe that E(i, j) ∈ Γ_{0}.

We observe by looking at the (r, s)th entry that

∑_{1≤i,j≤n−1} a_{ij}E(i, j) = 0 ⇒ a_{rs} = 0 ∀1 ≤ r, s ≤ n − 1,

so the E_{rs} are linearly independent.

If A = (a_{ij}) then

A − ∑_{1≤i,j≤n−1} a_{ij}E(i, j) = 0

We check the (n, n)th entry by observing that

∑_{i=1}^{n} ∑_{j=1}^{n} a_{ij} = 0

Thus the E(i, j) form a basis for Γ_{0} and the E_{i,j} [1 ≤ i, j ≤ n − 1] together with E form a basis for Γ. Thus Γ has dimension (n−1)^{2}+1 = n^{2} − 2n + 1

I cannot see any natural basis.

Exercise 5.6.2

We have

P (r) ≡ b_{0} + b_{1}c_{r}
P (r) ≡ 2 + 2c_{r}
P (1) ≡ 2 + 2 × 2 ≡ 6
P (2) ≡ 2 + 2 × 4 ≡ 3
P (3) ≡ 2 + 2 × 5 ≡ 5

(all modulo 7).

(P (1), c(1)) and (P (2), c(2)) yield

6 ≡ b_{0} + 2b_{1}
3 ≡ b_{0} + 4b_{1}

Subtracting the first equation from the second yields 3 ≡ −2b_{1}. Euclid’s algorithm (or inspection) tells us to multiply by 3 to recover b_{1} = 2. Substitution yields b_{0} = 2.

(P (1), c(1)) and (P (3), c(3)) yield

6 ≡ b_{0} + 2b_{1}
5 ≡ b_{0} + 5b_{1}

Subtracting the first equation from the second yields −1 ≡ −3b_{1}. Euclid’s algorithm (or inspection) tells us to multiply by 5 to recover b_{1} = 2. Substitution yields b_{0} = 2.

Exercise 5.6.3

If c_{j} = c_{k} with j ≠ k, then two people have the same secret. More people are needed to find the full secret.

The Vandermonde determinant

det (\begin{matrix}1 & c_{r(1)} & c_{r(1)}^{2} & … & c_{r(1)}^{k−1} \\ 1 & c_{r(2)} & c_{r(2)}^{2} & … & c_{r(2)}^{k−1} \\ 1 & c_{r(3)} & c_{r(3)}^{2} & … & c_{r(3)}^{k−1} \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 1 & c_{r(k)} & c_{r(k)}^{2} & … & c_{r(k)}^{k−1} \end{matrix})

does not vanish (and so the system is soluble) if and only if the c_{r(j)} are distinct.

If c_{j} = 0 then P (j) = b_{0} and j knows the secret. The proof that fewer cannot find the secret depended on

det (\begin{matrix}c_{r(1)} & c_{r(1)}^{2} & … & c_{r(1)}^{k−1} \\ c_{r(2)} & c_{r(2)}^{2} & … & c_{r(2)}^{k−1} \\ c_{r(3)} & c_{r(3)}^{2} & … & c_{r(3)}^{k−1} \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ c_{r(k−1)} & c_{r(k−1)}^{2} & … & c_{r(k−1)}^{k−1} \end{matrix}) ≡ c_{r(1)}c_{r(2)} … c_{r(k−1)} ∏_{1≤j<i≤k−1}(c_{r(i)}−c_{r(j)}) ≢ 0

and this needs the c_{r(j)} distinct and non-zero.

Exercise 5.6.4

P (r) ≡ b_{0} + b_{1}c_{r}
P (r) ≡ 1 + c_{r}
P (1) ≡ 1 + 1 × 1 ≡ 2
P (2) ≡ 1 + 1 × 4 ≡ 5

all modulo 6. The recovery equations are

2 ≡ b_{0} + b_{1}
5 ≡ b_{0} + 4b_{1}

There are two solutions (b_{0}, b_{1})) = (1, 1) and (b_{0}, b_{1})) = (5, 3).

If p is not a prime, mn ≡ k may have more than one solution in n even if k ≢ 0.

Exercise 5.6.5

(i) The secret remains safe. The arguments continue to work if the c_{j} are known.

(ii) If the b_{s} are known for r ≠ 0 the holder of any pair (c_{r}, P (r)) can compute b_{0} directly as

b_{0} ≡ P (r) − ∑_{s=1} b_{s}c_{r}^{s}.

Exercise 5.6.6

With k rings, they can say nothing with high probability since one of the rings may be a fake so they would have only k − 1 truth telling rings.

With k + 1 rings, either every set of k rings gives the same answer in which case all the rings are true and the answer is correct or each set of k rings will give a different answer (one of which corresponding to the true k rings will be correct). All they can say is that one ring is fake but they do not know which one or what the Integer of Power is.

With k + 2 rings, there are (k + 2)(k + 1)/2 different sets of k rings. Either each set tells the same story in which case all the rings are true and the answer is correct or k + 1 sets will tell the same story in which case they are the k + 1 sets containing k true rings and their answer is correct and k(k − 1)/2 sets giving different answers. The heroes can identify the k + 1 correct rings since they belong to the truth telling sets.

One prime to rule them all One prime to bind them In the land of composites Where the factors are

Exercise 6.1.2

Observe that, writing A = (a_{ij}), we have

αe_{j} = ∑_{i=1}^{n} a_{ij}e_{i} = (a_{1j} ; a_{2j} ; ⋮ ; a_{nj}) .

Exercise 6.1.3

Let A = (a_{ij}), B = (b_{ij}) and let αβ have matrix C = (c_{ij}). Then

∑_{i=1}^{n} c_{ij}e_{i} = αβe_{j} = α(βe_{j})

= α (∑_{k=1}^{n} b_{kj}e_{k})

= ∑_{k=1}^{n} b_{kj}αe_{k} = ∑_{k=1}^{n} b_{kj} ∑_{r=1}^{n} a_{rk}e_{r} = ∑_{r=1}^{n} (∑_{k=1}^{n} a_{rk}b_{kj}) e_{r}

so, by the uniqueness of the basis expansion,

c_{ij} = ∑_{k=1}^{n} a_{ik}b_{kj}

i.e. C = AB.

Exercise 6.1.5

b_{ij}f_{i} = α(f_{j}) = α(p_{rj}e_{r})

= p_{rj}αe_{r} = p_{rj}a_{sr}e_{s} = p_{rj}a_{sr}q_{is}f_{i} = q_{is}a_{sr}p_{rj}f_{i}.

Thus b_{ij} = q_{is}a_{sr}p_{rj}.

Exercise 6.1.7

(i) Theorem 6.1.5 shows that, if A and B represent the same map with respect to two bases, then A and B are similar.

If A and B are similar then B = P^{−1}AP. Let α be the linear map with matrix A corresponding to a basis

e_{1}, e_{2}, …, e_{n}.

Let f_{j} = ∑_{i=1}^{n} p_{ij}e_{i}. Since P has rank n, the f_{j} form a basis and Theorem 6.1.5 tells us that α has matrix B with respect to this basis.

(ii) Follows at once from Theorem 6.1.5 and part (i).

(iii) Write A ∼ B if B = P^{−1}AP for some non-singular P.

I^{−1}AI = IAI = A so A ∼ A.

If A ∼ B then B = P^{−1}AP for some non-singular P and writing Q = P^{−1} we have Q non-singular and A = Q^{−1}BQ so B ∼ A.

If A ∼ B, B ∼ C then B = P^{−1}AP, C = Q^{−1}CQ for some nonsingular P, Q. Then P Q is non-singular and C = (P Q)^{−1}A(P Q) so C ∼ A.

(iv) If A represents α with respect to one basis then A represents α with respect to the same basis. Thus A ∼ A.

If A ∼ B then there exist two bases with respect to which A and B represent the same map so B ∼ A.

By (i), if there exist two bases with respect to which A and B represent the same map then given any basis E_{1} there exists a basis E_{2} so that A and B represent the same map with respect to E_{1} and E_{2}. Thus if A ∼ B, B ∼ C we can find bases E_{j} such that A and B represent the same map with respect to E_{1} and E_{2}, and b and C represent the same map with respect to E_{2} and E_{3}, so A and C represent the same map with respect to E_{1} and E_{3} so A ∼ B.

Exercise 6.1.8

We have

f_{j} = ∑_{i=1}^{n} p_{ij}e_{j} = (p_{i1}, p_{i2}, …, p_{in})^{T} .

Exercise 6.1.9

Observe that e_{1} = f_{1}, e_{2} = f_{2} − f_{1}, e_{3} = f_{3} − f_{2},

α(f_{1}) = (\begin{matrix}1 & 1 & 2 \\ −1 & 2 & 1 \\ 0 & 1 & 3 \end{matrix}) (\begin{matrix}1 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix}1 \\ −1 \\ 0 \end{matrix}) = e_{1} − e_{2} = 2f_{1} − f_{2},

α(f_{2}) = (\begin{matrix}1 & 1 & 2 \\ −1 & 2 & 1 \\ 0 & 1 & 3 \end{matrix}) (\begin{matrix}1 \\ 1 \\ 0 \end{matrix}) = (\begin{matrix}2 \\ 1 \\ 1 \end{matrix})

= 2e_{1} + e_{2} + e_{3} = f_{1} + f_{3},

α(f_{3}) = (\begin{matrix}1 & 1 & 2 \\ −1 & 2 & 1 \\ 0 & 1 & 3 \end{matrix}) (\begin{matrix}1 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix}4 \\ 2 \\ 4 \end{matrix})

= 4e_{1} + 2e_{2} + 4e_{3} = 2f_{1} − 2f_{2} + 4f_{3}.

The new matrix is

B = (\begin{matrix}2 & −1 & 0 \\ 1 & 0 & 1 \\ 2 & −2 & 4 \end{matrix}) .

Exercise 6.1.12

Without Theorem 6.1.10, our definition would depend on the choice of basis.

Exercise 6.2.3

(i) Observe that

det (t (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) − (\begin{matrix}a & b \\ c & d \end{matrix})) = det (\begin{matrix}t − a & −b \\ −c & t − d \end{matrix})

= (t − a)(t − d) − (b)(−c) = t^{2} − (a + d) + (ad − bc) = t^{2} − (Tr A)t + det A.

(ii) We have

det(tI − A) = ∑_{σ∈S_{3}} ζ(σ) ∏_{i=1}^{3}(tδ_{iσi} − a_{iσi}) = ut^{3} + vt^{2} + ct + w.

Looking at the coefficient of t^{3}, we have u = 1. Looking at the coefficient of t^{2}, we have v = −a_{11} − a_{22} − a_{33}. Setting t = 0

w = det(−A) = (−1)^{3} det A = − det A.

Exercise 6.2.4

Write ∂P for the degree of a polynomial P. Observe that ∂b_{iσ(i)} = 0 if σ(i) ≠ i and ∂b_{iσ(i)} = 1 if σ(i) = i. Thus if σ ≠ ι we know that σ(i) ≠ i for at least two distinct values

of i so

∂ ∏_{i=1}^{n} b_{iσ(i)} ≤ ∑_{i=1}^{n} ∂b_{iσ(i)} ≤ n − 2.

It follows that

det(tI−A) = ∑_{σ∈S_{n}} ζ(σ)b_{iσ(i)}(t) = ∏_{i=1}^{n} b_{ii}(t) ∑_{σ≠ι} ζ(σ)b_{iσ(i)}(t) = ∏_{i=1}^{n}(t−a_{ii})+Q(t),

where Q is a polynomial of degree at most n − 2.

Thus the coefficient of t^{n−1} in det(tI − A) is the coefficient of t^{n−1} in ∏_{i=1}^{n}(t − a_{ii}), that is to say, − Tr A.

The constant term of a polynomial P is P (0) so the constant term in det(tI − A) is det((−1)A) = (−1)^{n} det A.

Exercise 6.2.9

We show that any linear map α : ℝ^{2n+1} → ℝ^{3} has an eigenvector. It follows that there exists some line l through 0 with α(l) ⊆ l.

Proof. Since det(tι−α) is a real polynomial of odd degree, the equation det(tι − α) = 0 has a real root λ say. We know that λ is an eigenvalue and so has an associated eigenvector u, say. Let

l = {su : s ∈ ℝ}.
□

Exercise 6.2.11

det(tι − ρ_{θ}) = det (\begin{matrix}t − cos θ & sin θ \\ − sin θ & t − cos θ \end{matrix})

= (t − cos θ)^{2} + sin^{2} θ

= t^{2} − 2 cos θ + 1.

The equation t^{2} − 2 cos θ + 1 = 0 has a real root if and only if

4 cos^{2} θ ≥ 4,

so if and only if cos θ = ±1, i.e. if and only if θ ≡ 0 mod π.

If θ ≡ 0 mod 2π, then ρ_{θ} = ι and every non-zero vector is an eigenvector with eigenvalue 1. If θ ≡ π mod 2π, then ρ_{θ} = −ι and every non-zero vector is an eigenvector with eigenvalue −1.

Exercise 6.2.14

(i) P_{A}(t) = det(tI + A) = (−1)^{n} det(−tI − A) = (−1)^{n}P_{A}(−t). (ii) Let δ be as in Lemma 6.2.13. Let t_{n} = δ/2n.

Exercise 6.2.15

χ_{AB}(t) = det(tI − AB) = det(A(tA^{−1} − B))
= det A det(tA^{−1} − B) = det(tA^{−1} − B) det A
= det((tA^{−1} − B)A) = det(tI − BA) = χ_{BA}(t).

In general, we can find s_{n} → 0 such that s_{n}I + A is invertible. Thus if t is fixed

χ_{(s_{n}I+A)B}(t) = χ_{B(s_{n}−A)}(t).

But χ_{(s_{n}I+A)B}(t) → χ_{AB}(t) and χ_{B(s_{n}I+A)}(t) → χ_{BA}(t) so

χ_{AB}(t) = χ_{BA}(t).

Allowing t to vary, we see that χ_{AB} = χ_{BA}.

Observe that − Tr C is the coefficient of t^{n−1} in χ_{C}(t). Thus χ_{A} = χ_{B} ⇒ Tr A = Tr B. By the previous result this gives Tr AB = Tr BA.

(i) (False in general.) We have

A = (\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}) , B = (\begin{matrix}1 & 1 \\ 0 & 0 \end{matrix}) , C = (\begin{matrix}0 & 0 \\ 1 & 1 \end{matrix})

then

ABC = (AB)C = (\begin{matrix}1 & 0 \\ 1 & 0 \end{matrix}) (\begin{matrix}0 & 1 \\ 0 & 1 \end{matrix}) = (\begin{matrix}1 & 1 \\ 0 & 0 \end{matrix})

and

ACB = (AC)B = (\begin{matrix}0 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix}0 & 0 \\ 1 & 1 \end{matrix}) = (\begin{matrix}0 & 0 \\ 1 & 1 \end{matrix}) .

Thus Tr(ABC) = 1 ≠ 0 = Tr(ACB) and χ_{ABC} ≠ χ_{ACB}.

(ii) (True.) χ_{ABC} = χ_{A(BC)} = χ_{(BC)A} = χ_{BCA}.

Exercise 6.2.16

Let e = (1, 1, 1, …, 1)^{T}. Observe that A is magic with constant κ if and only if

Ae = A^{T} e = κe.

In particular e is an eigenvector of A.

If A is magic with constant κ and AB = BA = I, then

e = BAe = κBe,

so κ ≠ 0 and Be = κ^{−1}e. Since B^{T} A^{T} = (AB)^{T} = I^{T} = I, the argument of the previous sentence gives B^{T} e = κ^{−1}e. Thus B is magic with constant κ^{−1}.

Suppose A is magic with constant κ. If A is invertible Adj A = (det A)A^{−1} so, by the previous paragraph with B = A^{−1}, Adj A is magic with constant κ^{−1} det A.

Chose t_{n} → 0 with A + t_{n}I invertible. If we write Adj(A + t_{n}I) = (c_{ij}(n)), Adj A = (c_{ij}) then c_{ij}(n) → c_{ij} as n → 0. Thus

0 = ∑_{r=1}^{n} c_{ir}(n) − ∑_{r=1}^{n} c_{kr}(n) → ∑_{r=1}^{n} c_{ir} − ∑_{r=1}^{n} c_{kr}

and ∑_{r=1}^{n} c_{ir} − ∑_{r=1}^{n} c_{kr} for all i and k. Similarly ∑_{r=1}^{n} c_{ir} = ∑_{r=1}^{n} c_{rk}. Thus Adj A is magic.

Exercise 6.3.2

(i) If

x_{1}e_{1} + x_{2}e_{2} = 0,

then

0 = α(x_{1}e_{1} + x_{2}e_{2}) = x_{1}αe_{1} + x_{2}αe_{2} = λ_{1}x_{1}e_{1} + λ_{2}x_{2}e_{2}

Thus

0 = λ_{1}(x_{1}e_{1} + x_{2}e_{2}) − (λ_{1}x_{1}e_{1} + λ_{2}x_{2}e_{2})
= (λ_{1} − λ_{2})x_{2}e_{2},

so that

(λ_{1} − λ_{2})x_{2} = 0

and x_{2} = 0. Similarly x_{1} = 0, so e_{1} and e_{2} are linearly independent.

(ii) Observe that

0 = (α − λ_{2}ι)(0) = (α − λ_{2}ι)(x_{1}e_{1} + x_{2}e_{2}) = (λ_{1} − λ_{2})x_{2}e_{2} and proceed as before.

(iii) Observe that

0 = (α − λ_{2}ι)(α − λ_{3}ι)0
= (α − λ_{2}ι)(α − λ_{3}ι)(x_{1}e_{1} + x_{2}e_{2} + x_{3}e_{3})
= (α − λ_{2}ι)((λ_{1} − λ_{3})x_{1}e_{1} + (λ_{2} − λ_{3})x_{1}e_{2})
= (λ_{1} − λ_{2})(λ_{1} − λ_{3})x_{1}e_{1}

so that

(λ_{1} − λ_{2})(λ_{1} − λ_{3})x_{1} = 0

and x_{1} = 0. Similarly x_{2} = x_{3} = 0 so e_{1}, e_{2}, e_{3} are linearly independent.

Exercise 6.4.2

(i) Observe that

det(tι − β) = det (\begin{matrix}t & −1 \\ 0 & t \end{matrix}) = t^{2}.

Thus the characteristic equation of β only has 0 as a root and β only has zero as an eigenvalue.

(ii) If x = (x, y)^{T} then

βx = 0x ⇔ (y, 0)^{T} = (0, 0)^{T} ⇔ y = 0

Thus the eigenvectors of β are the non-zero vectors of the form (x, 0)^{T} and these do not span ℝ^{2}.

Exercise 6.4.5

(i) We have

Q(t) = det (\begin{matrix}t − λ & 0 \\ 0 & t − μ \end{matrix}) = (t − λ)(t − μ) = t^{2} − (λ + μ)t + λμ,

so

Q(A) = (\begin{matrix}λ^{2} & 0 \\ 0 & μ^{2} \end{matrix}) − (\begin{matrix}(λ + μ)λ & 0 \\ 0 & (λ + μ)μ \end{matrix}) + (\begin{matrix}λμ & 0 \\ 0 & λμ \end{matrix}) = (\begin{matrix}0 & 0 \\ 0 & 0 \end{matrix}) .

Thus Q(α) = 0.

(ii) We have

Q(t) = det (\begin{matrix}t − λ & −1 \\ 0 & t − λ \end{matrix}) = (t − λ)^{2} = t^{2} − 2λt + λ^{2},

so

Q(A) = (\begin{matrix}λ^{2} & 0 \\ 0 & λ^{2} \end{matrix}) − (\begin{matrix}2λ^{2} & 0 \\ 0 & 2λ^{2} \end{matrix}) + (\begin{matrix}λ^{2} & 0 \\ 0 & λ^{2} \end{matrix}) = (\begin{matrix}0 & 0 \\ 0 & 0 \end{matrix})

Thus q(α) = 0.

(iii) We know that there exists a basis with respect to which α has one of the two forms discussed, so the result of Example 6.4.4 is correct.

Exercise 6.4.7

Let P be as in Theorem 6.4.6. Since P is non-singular, det P ≠ 0, so (working in ℂ) we can find a κ with κ^{2} = det P. Set ν = κ^{−1} and Q = νP.

Then det Q = ν^{2} det P = 1 and

Q^{−1}AQ = ν^{−1}νP^{−1}AP = P^{−1}AP.

Exercise 6.4.8

(i) Since

\frac{d}{dt}(e^{−λt}x(t)) = −λe^{−λt}x(t) + e^{−λt}ẋ(t) = e^{−λt}(−λx(t) + ẋ(t)) = 0,

the mean value theorem tells us that e^{−λt}x(t) = C and so x(t) = Ce^{−λt} for some constant C.

(ii) Since

\frac{d}{dt}(e^{−λt}x(t) − Kt) = e^{−λt}(−λx(t) + ẋ(t) − Ke^{λt}) = 0,

the mean value theorem tells us that e^{−λt}x(t) − Kt = C and so x(t) = (C + Kt)e^{−λt} for some constant C.

Exercise 6.4.9

If x_{1}(t) = x(t) and x_{2}(t) = ẋ(t) then ẋ_{1}(t) = x_{2}(t) automatically and ẋ_{2}(t) = −bx_{1}(t) − ax_{2}(t) ⇔ ẍ = −aẋ − bx ⇔ ẍ + aẋ + bx = 0

det(λI − A) = λ(λ + a) + b = λ^{2} + aλ + b

Exercise 6.5.5

By definition,

Q^{−1}R_{θ}Q = \frac{1}{2} (\begin{matrix}1 & i \\ i & 1 \end{matrix}) (\begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix}) (\begin{matrix}1 & −i \\ −i & 1 \end{matrix})

= \frac{1}{2} (\begin{matrix}cos θ + i sin θ & − sin θ + i cos θ \\ sin θ + i cos θ & cos θ − i sin θ \end{matrix}) (\begin{matrix}1 & −i \\ −i & 1 \end{matrix})

= \frac{1}{2} (\begin{matrix}e^{iθ} & ie^{iθ} \\ ie^{−iθ} & e^{−iθ} \end{matrix}) (\begin{matrix}1 & −i \\ −i & 1 \end{matrix})

= \frac{1}{2} (\begin{matrix}2e^{iθ} & 0 \\ 0 & 2e^{−iθ} \end{matrix}) = (\begin{matrix}e^{iθ} & 0 \\ 0 & e^{−iθ} \end{matrix}) .

Exercise 6.5.2

Since A has distinct eigenvalues, it is diagonalisable and we can find a an invertible matrix B such that

BAB^{−1} = D,

where D is the diagonal matrix with jth diagonal entry λ_{j}. If we set y = Bx (using column vectors) we obtain

ẏ = Bẋ = BAx = BAB^{−1}y = Dy.

Thus ẏ_{j} = λ_{j}y_{j} and y_{j} = c_{j}e^{λ_{j}t} for [1 ≤ j ≤ n] and so, writing F = B^{−1}, F = (f_{ij})

x_{1} = ∑_{j=1}^{n} f_{1j}y_{j} = ∑_{j=1}^{n} f_{1j}c_{j} exp(iλ_{j}t).

If we set μ_{j} = f_{1j}c_{j}, we obtain the required result.

If A = D, then x = y and x_{1} = c_{1} exp(iλ_{1}t) and, with the notation of the question, we have μ_{j} = 0 for 2 ≤ j ≤ n.

Exercise 6.6.2

D^{q} is the diagonal matrix with diagonal entries the qth powers of the diagonal entries of D.

(ii) A and D represent the same linear map α with respect to two bases E_{1} and E_{2} with basis change rule U = P V P^{−1}. A^{q} and D^{q} represent the same linear map α^{q} with respect to the two bases E_{1} and E_{2} so A^{q} = P D^{q}P^{−1}.

Exercise 6.6.7

(i) We have

λ^{−q}α^{q}(x_{1}e_{1} + x_{2}e_{2}) = x_{1}e_{1} + x_{2}(μ/λ)^{n}x_{2}e_{2} → x_{1}e_{1}

coordinate wise.

(i)′ We have

λ^{−q}α^{q}(x_{1}e_{1} + x_{2}e_{2}) = x_{1}e_{1} + x_{2}(μ/λ)^{n}x_{2}e_{2},

but (μ/λ)^{n} does not tend to a limit, so λ^{−q}α^{q}(x_{1}e_{1} + x_{2}e_{2}) converges coordinatewise if and only if x_{2} = 0.

(ii)′ Nothing to prove.

(iii) We have

α((λ^{q}x_{1} + qλ^{q−1}x_{2})e_{1} + λ^{q}x_{2}e_{2}) = λ(λ^{q}x_{1} + qλ^{q−1}x_{2})e_{1} + λ^{q}(x_{2}e_{1} + λe_{2})
= (λ^{q+1}x_{1} + (q + 1)λ^{q}x_{2})e_{1} + λ^{q+1}x_{2}e_{2}

so, by induction

α^{q}(x_{1}e_{1} + x_{2}e_{2}) = (λ^{q}x_{1} + qλ^{q−1}x_{2})e_{1} + λ^{q}x_{2}e_{2}

for all q ≥ 0.

Thus q^{−1}λ^{−q}α^{q}(x_{1}e_{1} + x_{2}e_{2}) = (q^{−1}x_{1} + λ^{−1}x_{2})e_{1} + q^{−1}x_{2}e_{2} → λ^{−1}x_{2})e_{1} coordinatewise.

(iv) Immediate.

Exercise 6.6.8

F_{1} = F_{2} = 1, F_{3} = 2, F_{4} = 3, F_{5} = 5, F_{6} = 8, F_{7} = 13, F_{8} = 21, F_{9} = 34, F_{10} = 55.

F_{0} + 1 = F_{0} + F_{1} = F_{2} = 1,

so F_{0} = 0.

F_{−1} + 0 = F_{−1} + F_{0} = F_{1} = 1,

so F_{−1} = 1.

Write G_{−n} = (−1)^{n+1}F_{n}. Then

G_{−n+1} + G_{−n} = (−1)^{n}(F_{n−1} − F_{n}) = (−1)^{n−1}F_{n−2} = G_{−n+2} and G_{−1} = F_{−1}, G(0) = F (0) so, by induction, F_{−n} = G_{−n} for all n ≥ 0. Thus F_{−n} = (−1)^{n+1}F_{n} for all n.

Exercise 6.6.9

If b = 0, but a ≠ 0, then u_{n} = (−a)^{n}u_{0}. If a = b = 0, then u_{n} = 0 for all n.

Exercise 6.6.10

We look at vectors

u(n) = (\begin{matrix}u_{n} \\ u_{n+1} \end{matrix}) .

We then have

u(n + 1) = Au(n), where A = (\begin{matrix}0 & 1 \\ −a & −b \end{matrix})

and so

(\begin{matrix}u_{n} \\ u_{n+1} \end{matrix}) = A^{n} (\begin{matrix}u_{0} \\ u_{1} \end{matrix}) .

Now A is not a multiple of the identity matrix so (since the roots are equal) there is a basis e_{1} e_{2} with Ae_{1} = λe_{1} = λe_{1} and Ae_{2} = λe_{1}+λe_{2}.

Now

A^{n}(x_{1}e_{1} + x_{2}e_{1}) = (λ^{n}x_{1} + nλ^{n−1}x_{2}) + λ^{n}x_{2}e_{2}

so, since

u(n) = A^{n}u(0),

we have

u_{n} = (c + c′n)λ^{n}

for some constants c and c′ depending on u_{0} and u_{1}.

Exercise 6.6.11

We look at vectors

u(n) = (u_{n}, u_{n+1}, …, u_{n+m−1})^{T} .

We then have

u(n + 1) = Au(n), where A = (\begin{matrix}0 & 1 & 0 … & 0 \\ 0 & 0 & 1 … & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & 1 … & 1 \\ −a_{m−1} & −a_{m−2} & −a_{m−3} & … & −a_{0} \end{matrix})

The characteristic polynomial of A is P (t), so, since all the roots of P are distinct and non-zero, the eigenvectors e_{j} (corresponding to an eigenvalue λ_{j}) of A form a basis.

Setting u_{0} = ∑_{k=1}^{m} b_{k}e_{k} we get

u_{n} = A^{n}u_{0} = ∑_{k=1}^{m} b_{k}λ_{k}e_{k}

and so looking at the first entry in the vectors

u_{n} = ∑_{k=1}^{m} c_{k}λ_{k}^{n}

for some constants c_{j}.

Conversely, if

u_{n} = ∑_{k=1}^{m} c_{k}λ_{k}^{n},

then

u_{r} + ∑_{j=0}^{m−1} a_{j}u_{j−m+r} = ∑_{k=1}^{m} λ_{k}^{n−m}P (λ_{k}) = 0.

Observe that, if |λ_{1}| > |λ_{k}| for 2 ≤ k ≤ m, then

λ_{1}^{−n}u_{n} = ∑_{k=1}^{m} c_{k}(λ_{1}^{−1}λ_{k})^{n} → c_{1}

as n → ∞. Thus, if c_{0} ≠ 0, we have u_{r−1} ≠ 0 for r large and

\frac{u_{r}}{u_{r−1}} → λ_{1}

as r → ∞.

A very similar proof shows that, if |λ_{k}| > |λ_{m}| for 1 ≤ k ≤ m − 1 and c_{m} ≠ 0, then u_{n} ≠ 0 for −n large and

\frac{u_{r−1}}{u_{r}} → λ_{m}.

Exercise 6.6.12

(i) Observe that

\frac{1 + \sqrt{5}}{2} × \frac{−1 + \sqrt{5}}{2} = \frac{(\sqrt{5})^{2} − 1^{2}}{4} = 1.

(ii) We have

det(tI − A) = t(t − 1) − 1 = t^{2} − t − 1 = (t − τ)(t + τ^{−1}), so the eigenvalues are τ and −τ^{−1}.

If (x, y)^{T} is an eigenvector with eigenvalue τ,

y = τx
x + y = τy,

so e_{1} = (1, τ)^{T} is an eigenvector with eigenvalue τ.

A similar argument shows that e_{2} = (−1, τ^{−1})^{T} is an eigenvector with eigenvalue τ^{−1}.

By inspection (or solving two simultaneous linear equations)

(\begin{matrix}0 \\ 1 \end{matrix}) = \frac{1}{\sqrt{5}}(e_{1} + e_{2}).

(iii) It follows that

(\begin{matrix}F_{n} \\ F_{n+1} \end{matrix}) = A^{n} (\begin{matrix}F_{0} \\ F_{1} \end{matrix}) = 5^{−1/2}(τ^{n}e_{1} + (−τ)^{−n}e_{2})

and so

F_{n} = 5^{−1/2}(τ^{n} + (−τ)^{−n})

(vi) Since F_{n} is an integer and

|5^{−1/2}(−τ)^{−n}| ≤ 5^{−1/2} < 1/2

for n ≥ 0 we know that F_{n} is the closest integer to τ^{n}/5^{1/2}. Both methods give F_{20} = 6765.

(v) Observe that A^{n} has first column

A^{n−1}(F_{0}, F_{1})^{T} = (F_{n−1}, F_{n})^{T}

and second column

A^{n−1}(F_{1}, F_{2})^{T} = (F_{n}, F_{n+1})^{T} .

(Or use induction.)

(−1)^{n} = (det A)^{n} = det A^{n} = F_{n−1}F_{n+1} − F_{n}^{2}.

(vi)True for all n because F_{n} = (−1)^{n+1}F_{−n}. (vi) We have A^{2n} = A^{n}A^{n}, so

(\begin{matrix}F_{n−1} & F_{n} \\ F_{n} & F_{n+1} \end{matrix}) (\begin{matrix}F_{n−1} & F_{n} \\ F_{n} & F_{n+1} \end{matrix}) = (\begin{matrix}F_{2n−1} & F_{2n} \\ F_{2n} & F_{2n+1} \end{matrix})

so, evaluating the upper left hand term of the matrix, we get F_{n}^{2} + F_{n−1}^{2} = F_{2n−1}.

Exercise 6.6.13

(i) Let e_{ij}^{(n)} be the number of routes from i to j involving exactly n flights. We claim that e_{ij}^{(n)} = d_{ij}^{(n)}. We use induction on n. The result is clearly true if n = 1. Suppose it is true for n = r.

e_{ij}^{(r+1)} = number of journeys with exactly r + 1 flights from i to j

= ∑_{k=1}^{m} number of journeys with exactly r + 1 flights

from i to j ending with a flight from k to j

= ∑_{d_{kj}≠0} number of journeys with exactly r flights from i to k

= ∑_{k=1}^{m} d_{ik}^{(r)}d_{kj} = d_{ij}^{(r+1)}

The result follows.

In particular, there is a journey from i to j which involve exactly n flights if and only if d_{ij}^{(n)} > 0.

(ii) We can produce an argument like (i) if we make the convention that a flight includes just staying at an airport (so a flight from i to i is allowed).

In particular there is a journey from i to j which involve n flights or less (between different airports) if and only if d̃_{ij}^{(n)} > 0.

(iii) We write down the safe states for example

i = (G|P W C)

goat on one side, peasant, wolf and cabbage on other and write d_{ij} = 1 if the peasant can change the situation from i to j in one crossing. Otherwise d_{ij} = 0. The least number of crossings is the least N for which

d_{(GP W C|∅),(∅|GP W C)}^{N} ≠ 0.

Observe that, since the peasant will not repeat a state, if there are M safe states the problem is soluble, if at all, with at most M − 1 crossings so the method will reveal if the problem is insoluble.

(The cabbage survived but was so distressed by its narrow escape that it turned over a new leaf).

Exercise 6.7.1

(i) Suppose that A = (a_{ij})_{1≤i≤n}^{1≤i≤n} is lower triangular. Then, by row expansion,

det A = a_{11} det B

where B = (a_{i+1,j+1})_{1≤i≤n−1}^{1≤i≤n−1}. Since B is lower triangular with diagonal entries a_{i+1,i+1} [1 ≤ i ≤ n − 1] the required result follows by induction.

(ii) We have

L invertible ⇔ det L ≠ 0 ⇔ ∏_{j=1}^{n} l_{jj} ≠ 0 ⇔ l_{jj} ≠ 0 ∀j

(iii) Since tI − L is lower triangular

det(tI − L) = ∏_{j=1}^{n}(t − l_{jj})

so the roots of the characteristic equation are the diagonal entries l_{jj} (multiple roots occurring with the correct multiplicity).

Observe that (0, 0, 0, …, 0, 1)^{T} is an eigenvector with eigenvalue l_{nn}.

Exercise 6.7.2

Just repeat the earlier discussion mutatis mutandis. The following is merely a variation. If the system is

∑_{j=1}^{n} u_{ij}x_{j} = y_{i}

then setting x′_{j} = x_{n−j}, y′_{i} = y_{n−i} l_{ij} = u_{n−i,n−j} gives the lower triangular system

∑_{j=1}^{n} l_{ij}x′_{j} = y′_{i}.

Exercise 6.7.3

(i) Observe first that the product of two lower triangular matrices is triangular since if a_{ij}, b_{ij} = 0 for i + 1 ≥ j, then

∑_{r=1}^{n} a_{ir}b_{rj} = ∑_{j≤r≤i} a_{ir}b_{rj} = 0

for i + 1 ≥ j.

We use induction to show that the inverse of an invertible n×n lower triangular matrix is triangular.

Since every 1 × 1 matrix is lower triangular, the result is certainly true for n = 1. Suppose it is true for n = m and A is an m + 1 × m + 1 invertible lower triangular matrix. Since A is invertible a_{11} ≠ 0. If B = (b_{ij}) is the m + 1 × m + 1 matrix with b_{11} = a_{11}^{−1}, b_{i1} = −a_{i1}a_{11}^{−1} b_{ii} = 1 for 2 ≤ i ≤ m + 1 and b_{ij} = 0, otherwise, then

BA = (\begin{matrix}1 & 0^{T} \\ 0 & L \end{matrix} )

where L is an m × m invertible lower triangular matrix and 0 ∈ ℝ^{m} is the zero column vector.

If we now set

C = (\begin{matrix}1 & 0^{T} \\ 0 & L^{−1} \end{matrix})

then since L^{−1} is lower triangular (by the inductive hypothesis), we know that C^{−1} is lower triangular so C^{−1}B is lower triangular. But

(C^{−1}B)A = C^{−1}(BA) = C^{−1}C = I

so A has lower triangular inverse and the induction is complete.

(ii) By the first paragraph of (i),

A, B ∈ L ⇒ AB ∈ L.

By inspection I ∈ L and, by part (i),

A ∈ L ⇒ A^{−1} ∈ L.

Exercise 6.7.6

Observe that

LU = (\begin{matrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ −1 & −3 & 1 \end{matrix}) (\begin{matrix}2 & 1 & 1 \\ 0 & −1 & −2 \\ 0 & 0 & −4 \end{matrix})

= (\begin{matrix}2 & 1 & 1 \\ 4 & 2 − 1 & 2 − 2 \\ −2 & −1 + 3 & −1 + 6 − 4 \end{matrix})

= (\begin{matrix}2 & 1 & 1 \\ 4 & 1 & 0 \\ −2 & 2 & 1 \end{matrix}) .

Exercise 6.7.7

Consider an n × n matrix A. The first step will be impossible only if the entire first row consists of zeros in which case it is clear that the n × n matrix A is not invertible. If the first step can be done, then we repeat the process with a n − 1 × n − 1 square matrix B and the same remarks apply to B.

Thus either the process will stop because the matrix with which we deal at the given stage has top row all zeros or the process will not stop in which case we obtain an upper triangular matrix U and a lower triangular matrix L both with all diagonal entries non-zero so invertible. Since the product of invertible matrices is invertible A = LU is invertible.

Thus the process works if and only if A is invertible.

Exercise 6.7.9

Suppose that

(\begin{matrix}a & 0 \\ c & d \end{matrix}) (\begin{matrix}u & v \\ w & 0 \end{matrix}) = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) .

Then

(\begin{matrix}au & av \\ cu & cv + dw \end{matrix}) = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) .

It follows that au = 0 and av = 1, so u = 0, which is incompatible with cu = 1. Our initial formula cannot hold.

Exercise 6.7.10

We use induction on n. The result is trivial for n = 1 since

(1)(u_{1}) = (1)(u_{2}) ⇒ (u_{1}) = (u_{2}) ⇒ u_{1} = u_{2}.

Now suppose the result is true for n = m and L_{1}, L_{2} are lower triangular (with diagonal entries 1) and U_{1}, U_{2} upper triangular m + 1 × m + 1 matrices with L_{1}U_{1} = L_{2}U_{2}. Observe that we can write

L_{j} = (\begin{matrix}1 & 0^{T} \\ l_{j} & L̃_{j} \end{matrix}) and L_{j} = (\begin{matrix}u(j) & u_{j}^{T} \\ 0 & Ũ_{j} \end{matrix} )

where the L̃_{j}, L_{2} lower triangular (with diagonal entries 1) and the Ũ_{j} upper triangular m × m matrices, u(j) ∈ 𝔽 and l_{j}, u_{j} ∈ 𝔽^{m} are column vectors. Equating first rows in the equation L_{1}U_{1} = L_{2}U_{2} we get

(u(1), u_{1}^{T}) = (u(2), u_{2}^{T})

and equating first columns we get

(1, l_{1}^{T})^{T} = (1, l_{2}^{T})^{T}

so u(1) = u(2), u_{1} = u_{2}, l_{1} = l_{2}.

The condition L_{1}U_{1} = L_{2}U_{2} now gives

L̃_{1}Ũ_{1} = L̃_{2}Ũ_{2}

so by the inductive hypothesis

L̃_{1} = L̃_{2}, Ũ_{1} = Ũ_{2}.

We have shown that

L_{1} = L_{2}, U_{1} = U_{2}

and this completes the induction

Exercise 6.7.11

Since det LU = det L det U and det L, det U are the product of the diagonal entries, det LU is cheap to calculate.

By solving the n sets of simultaneous linear equations

∑_{r=1}^{n} l_{ir}x_{rj} = δ_{ij}

each of which requires of the order of n^{2} operations we can compute X = L^{−1} in the order of n^{3} operations and we can compute U^{−1} similarly. We now compute (LU)^{−1} = U^{−1}L^{−1} again in the order of n^{3} operations.

Exercise 6.7.12

(i) It is probably most instructive to do this ab initio, but quicker to use known results.

We know that (after reordering columns) A = LŨ with L lower triangular and all diagonal entries 1 and Ũ upper triangular. Since A is non-singular Ũ is, so all its diagonal entries are non-zero. Let D be the diagonal matrix with diagonal entries the diagonal entries of Ũ. Set U = D^{−1}Ũ. Then U is upper triangular with all diagonal entries 1, D is non-singular and

A = LDU.

Suppose that L_{j} is a lower triangular n × n matrix with all diagonal entries 1, U_{j} is an upper triangular n × n matrix with all diagonal entries 1 and D_{j} is a non-singular diagonal n × n matrix [j = 1, 2]. We claim that, if

L_{1}D_{1}U_{1} = L_{2}D_{2}U_{2},

then L_{1} = L_{2}, U_{1} = U_{2} and D_{1} = D_{2}.

To see this observe that D_{j}U_{j} is non singular upper triangular so, by Exercise 6.7.10, L_{1} = L_{2} and

D_{1}U_{1} = D_{2}U_{2}.

By looking at the diagonal entries of both sides of the equation we get D_{1} = D_{2} so

U_{1} = D_{1}^{−1}(D_{1}U_{1}) = D_{2}^{−1}(D_{2}U_{2}) = U_{2}.

(ii) Yes. Either repeat the argument mutatis mutandis or argue as follows.

If G = (g_{ij}) is an n × n matrix, write G̃ = (g_{n+1−i,n+1−j}). If G̃ = LU with L lower triangular, U upper triangular, then G = L̃Ũ and L̃ is upper triangular whilst Ũ is lower triangular.

(iii) No. Observe that

(\begin{matrix}a & b \\ c & 0 \end{matrix}) (\begin{matrix}x & 0 \\ y & z \end{matrix}) = (\begin{matrix}ax + by & bz \\ cx & 0 \end{matrix}) .

Exercise 7.1.4

(i) Any set of orthonormal vectors is linearly independent. Any set of k linearly independent vectors forms a basis for U.

(ii) By (i), we can find λ_{j} such that

x = ∑_{j=1}^{k} λ_{j}e_{j}.

Now observe that

⟨x, e_{i}⟩ = ∑_{j=1}^{k} λ_{j}⟨e_{j}, e_{i}⟩ = λ_{i}.

Exercise 7.1.6

If we take e_{1} = 3^{−1/2}(1, 1, 1), e_{1} = 2^{−1/2}(1, −1, 0), then, by inspection, they form an orthonormal system.

We now use Gramm-Schmidt for x = (1, 0, 0).

x − ⟨x, e_{1}⟩e_{1} − ⟨x, e_{2}⟩e_{2} = (1, 0, 0) − 3^{−1}(1, 1, 1) − 2^{−1}(1, −1, 0)
= (1/6, 1/6, −1/3) = a,

say, Setting

e_{3} = ∥a∥^{−1}a = 6^{−1/2}(1, 1, −2)

we have e_{1}, e_{2}, e_{3} orthonormal.

It follows at one that e_{1} = 3^{−1/2}(1, 1, 1), 2^{−1/2}(0, −1, 1), 6^{−1/2}(−2, 1, 1) is another orthonormal system.

Exercise 7.1.8

(Dimension 2) If a point B does not lie in a plane π, then there exists a unique line l′ perpendicular to π passing through B. The point of intersection of l with π is the closest point in π to B. More briefly, the foot of the perpendicular dropped from B to π is the closest point in π to B.

(Dimension 1) If a point B does not lie on a line l, then there exists a unique line l′ perpendicular to l passing through B. The point of intersection of l with l′ is the closest point in l to B. More briefly, the foot of the perpendicular dropped from B to l is the closest point in l to B.

Exercise 7.1.9

(i) We have

∥x − ∑_{j=1}^{k} λ_{j}e_{j}∥^{2} ⟨x − ∑_{j=1}^{k} λ_{j}e_{j}, x − ∑_{j=1}^{k} λ_{j}e_{j}⟩

= ∥x∥^{2} − 2 ∑_{j=1}^{k} λ_{j}⟨x, e_{j}⟩ + ∑_{j=1}^{k} λ_{j}^{2}

= ∥x∥^{2} − ∑_{j=1}^{k}⟨x, e_{j}⟩^{2} + ∑_{j=1}^{k}(λ_{j} − ⟨x, e_{j}⟩)^{2}

≥ ∥x∥^{2} − ∑_{j=1}^{k}⟨x, e_{j}⟩^{2}

with equality if and only if λ_{j} = ⟨x, e_{j}⟩.

(ii) It follows from (i) that

∥x∥^{2} ≥ ∑_{j=1}^{k}⟨x, e_{j}⟩^{2}

with equality if and only if λ_{j} = ⟨x, e_{j}⟩ and

∥x − ∑_{j=1}^{k} λ_{j}e_{j}∥ = 0,

ie

x = ∑_{j=1}^{k} λ_{j}e_{j}

and this occurs if and only if x ∈ span{e_{1}, e_{2}, …, e_{k}}.

Exercise 7.1.10

Observe first that

⟨0, u⟩ = 0

for all u so 0 ∈ U^{⊥}.

Next observe that if λ_{1}, λ_{2} ∈ ℝ

v_{1}, v_{2} ∈ U^{⊥} ⇒ ⟨λ_{1}v_{1} + λ_{2}v_{2}u⟩ = λ_{1}⟨v_{1}, u⟩ + λ_{2}⟨v_{2}, u⟩ = 0 + 0 = 0 for all u ∈ U

⇒ λ_{1}v_{1} + λ_{2}v_{2} ∈ U^{⊥}.

If we take a − b = v in Theorem 7.1.7, we see that a ∈ R^{n} can be written in one and only one way as x = u + v with u ∈ U, v ∈ U^{⊥}.

If e_{1}, …, e_{k} is a basis for U and e_{k+1}, …, e_{l} is a basis for U^{⊥}, the previous paragraph tells us that e_{1}, …, e_{k}, e_{k+1}, …, e_{l} is a basis for ℝ^{n} so l = n and

dim U + dim U^{⊥} = n.

Exercise 7.2.2

⟨αx, y⟩ = ∑_{i=1}^{n} (∑_{j=1}^{n} a_{ij}x_{j}y_{i}) = ∑_{j=1}^{n} (∑_{i=1}^{n} a_{ij}x_{j}y_{i})

= ∑_{j=1}^{n} (∑_{i=1}^{n} c_{ji}x_{j}y_{i}) = ⟨x, α^{∗}y⟩.

Exercise 7.2.6

Suppose that A, B represent α, β with respect to some basis. Lemma 7.2.4 enables us to translate Lemma 7.2.5 into matrix form

(i) Since (αβ)^{∗} = β^{∗}α^{∗}, we have (AB)^{T} = B^{T} A^{T}. (ii) Since α^{∗∗} = (α^{∗})^{∗} = α, we have A^{T T} = (A^{T})^{T} + A. (iii) Since (λα+μβ)^{∗} = λα^{∗}+μβ^{∗}, we have (λA+μB)^{T} = λA^{T} +μB^{T}. (vi) Since ι^{∗} = ι, we have I^{T} = I. In coordinates

(i) If AB = C, then

c_{ij} = ∑_{r=1}^{n} a_{ir}b_{rj}

so

c_{ji} = ∑_{r=1}^{n} b_{rj}a_{ir}

and (AB)^{T} = C^{T} = B^{T} A^{T}.

(ii) A^{T T} = (a_{ji})^{T} = (a_{ij}) = A. (iii) (λA + μB)^{T} = (λa_{ji} + μb_{ji}) = λ(a_{ji} + μb_{ji}) = λA^{T} + μB^{T}. (iv) δ_{ij} = δ_{ji} so I = I^{T}.

Exercise 7.2.10

Let the jth row of A be a_{j}.

A ∈ O(ℝ^{n}) ⇔ AA^{T} = I ⇔ a_{i}a_{j}^{T} = δ_{ij} ⇔ a_{i} · a_{j} = δ_{ij} ⇔ the a_{j} are o.n.

Thus (i)⇔(ii). Since

A ∈ O(ℝ^{n}) ⇒ A^{T T} A^{T} = AA^{T} = I ⇒ A^{T} ∈ O(ℝ^{n})

transposition gives (i)⇔(iii).

We turn to the statements.

(i) FALSE. Consider

(\begin{matrix}1 & −1 \\ 2 & 2 \end{matrix}) .

(ii) FALSE. Consider

(\begin{matrix}1 & 2 \\ 1 & 2 \end{matrix}) .

(iii) FALSE If a_{1}, a_{2}, …, a_{n−1}, a_{n} are the rows of an orthogonal matrix, so are a_{1}, a_{2}, …, a_{n−1}, −a_{n} (and a_{n} ≠ 0).

(iv) TRUE If a_{1}, a_{2}, …, a_{n−1} are orthonormal in ℝ^{n} then there are exactly two choices ±a_{n} giving an orthonormal set and of these exactly one will give det A = 1.

Exercise 7.2.13

Take

A = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) , B = (\begin{matrix}1 & 1 \\ 0 & −1 \end{matrix}) .

Note that the columns in each matrix are not orthogonal.

Exercise 7.3.2

Any pair of orthonormal vectors form a basis for ℝ^{2}. We observe that

⟨e_{1}, e_{1}⟩ = 1, ⟨e_{1}, −e_{2}⟩ = −⟨e_{1}, e_{2}⟩ = 0, ⟨−e_{2}, −e_{2}⟩ = ⟨e_{2}, e_{2}⟩ = 1. Observe that

αe_{1} = cos θe_{1} + sin θe_{2} = cos θe_{1} − sin θ(−e_{2})

= α(−e_{2}) = −αe_{2} = sin θe_{1} − cos θe_{2} = sin θe_{1} + cos θ(−e_{2}) so, with respect to the new basis, α has matrix

(\begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix}) = (\begin{matrix}cos(−θ) & − sin(−θ) \\ sin(−θ) & cos(−θ) \end{matrix})

Exercise 7.3.4

(i) We have

(\begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix}) (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}x cos θ − y sin θ \\ x sin θ + y cos θ \end{matrix})

which I was told in school was a rotation through θ

We have

(\begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}−x \\ y \end{matrix})

which certainly looks like a reflection in the y axis to me.

Since

⟨e_{1}, u_{1}⟩^{2} + ⟨e_{1}, u_{2}⟩^{2} = 1,

we may take ⟨e_{1}, u_{1}⟩ = cos ϕ, ⟨e_{1}, u_{2}⟩ = sin ϕ. We have

e_{1} = cos ϕu_{1} + sin ϕu_{2}

so

e_{1} = −αe_{1}
= − cos ϕαu_{1} − sin ϕαu_{2}
= − cos ϕ(cos θu_{1} + sin θu_{2}) − sin ϕ(sin θu_{1} + cos θu_{2})

Taking the inner product with u_{1} we get

cos ϕ = − cos ϕ cos θ − sin ϕ sin θ = − cos(θ − ϕ)

so

ϕ = π + θ − ϕ

and θ = 2ϕ − π.

Exercise 7.3.6

Since a reflection in a plane has perpendicular eigenvectors with eigenvalues −1, 1, 1, α is a reflection only if det α = −1 so, by our discussion, α has matrix

A = (\begin{matrix}−1 & 0 & 0 \\ 0 & cos θ & − sin θ \\ 0 & sin θ & cos θ \end{matrix}) ,

so

det(tι−α) = det (\begin{matrix}t + 1 & 0 & 0 \\ 0 & t − cos θ & sin θ \\ 0 & − sin θ & t − cos θ \end{matrix}) = (t−1)(t^{2}−2t cos θ+1)

and we can only have the correct eigenvalues if cos θ = 1. In this case we clearly have a reflection in the plane spanned by e_{2} and e_{3}.

By definition α is a reflection in the origin if and only if α = −ι so if and only if cos θ = −1.

Exercise 7.3.7

det(tι − α) = det (\begin{matrix}t − cos θ & sin θ \\ − sin θ & t − cos θ \end{matrix}) det (\begin{matrix}t − cos ϕ & sin ϕ \\ − sin ϕ & t − cos ϕ \end{matrix})

= (t^{2} − 2t cos θ + 1)(t^{2} − 2t cos ϕ + 1)

with no real roots unless cos θ = cos ϕ = 1 i.e. θ ≡ ϕ ≡ 0 (mod π).

Exercise 7.4.3

Let u = ⟨x, n⟩n and v = x − u. Then ⟨v, n⟩ = ⟨x − u, n⟩ = ⟨x − ⟨x, n⟩n, n⟩ = ⟨x, n⟩ − ⟨x, n⟩⟨n, n⟩ = 0

as required. We observe that ρ(x) = u + v − 2(⟨u, n⟩ + ⟨v, n)⟩n = u + v − 2⟨x, n⟩n = v − u.

Exercise 7.4.6

Let A be the point given by the vector a and B be the point given by the vector b. Let OM be an angle bisector of the line OA, OB of length 1. and let OM be given by the vector m. Let n be a unit vector perpendicular to m.

Exercise 7.4.9

The product of two reflections in lines OA, OB through the origin making an angle ∠AOB = θ is a rotation through 2θ.

To prove this, observe that, if ρ_{1} and ρ_{2} are reflections, then ρ_{2}ρ_{1} preserves lengths and has determinant 1, so is a rotation. Now let e_{1} and e_{2} be orthonormal and let n = −e_{1}, m = − cos θe_{1} − sin θe_{2}. If

ρ_{1}(x) = x − 2⟨x, n⟩n
ρ_{2}(x) = x − 2⟨x, m⟩m,

then

ρ_{2}ρ_{1}e_{1} = ρ_{2}(e_{1}) = e_{1} − 2⟨e_{1}, m⟩m
= e_{1} − 2 cos θ(cos θ_{1} + sin θ_{2})
= cos 2θ_{1} + sin 2θ_{2}.

Thus the rotations through angle θ are obtained from two reflections in lines at angle θ/2 (in the correct sense).

Exercise 7.5.1

(i) Observe that

∥Ax − b∥^{2} = ∑_{j=1}^{n}(x − b_{j})^{2}

= nx^{2} − 2x ∑_{j=1}^{n} b_{j} + ∑_{j=1}^{n} b_{j}^{2}

= n (x − n^{−1} ∑_{j=1}^{n} b_{j})^{2} + ∑_{j=1}^{n} b_{j}^{2} − n (n^{−1} ∑_{j=1}^{n} b_{j})^{2}

is minimised by taking x = n^{−1} ∑_{j=1}^{n} b_{j}.

In the case of the mountain this amounts to the reasonable choice of the average of the measured heights.

(ii) Observe that

∥Ax − b∥^{2} = f(x_{1}, x_{2})

with

f(x_{1}, x_{2}) = ∑_{i=1}^{n}(x_{1} − v_{i}x_{2} − b_{i})^{2}

We observe that f(x_{1}, x_{2}) → ∞ as |x_{1}|, |x_{2}| → ∞.

Since

\frac{∂f}{∂x_{1}} (x_{1}, x_{2}) = ∑_{i=1}^{n} 2(x_{1} − v_{i}x_{2} − b_{i}) = 2 (nx_{1} − ∑_{i=1}^{n} b_{i})

and

\frac{∂f}{∂x_{2}} (x_{1}, x_{2}) = − ∑_{i=1}^{n} 2v_{i}(x_{1} − v_{i}x_{2} − b_{i}) = −2 (x_{2} ∑_{i=1}^{n} v_{i}^{2} − ∑_{i=1}^{n} b_{i}v_{i}) ,

we see that f and therefore ∥Ax − b∥ has a unique minimum when (x_{1}, x_{2}) = (μ, κ).

We have sought to find a line u+x_{1}v = x_{2} ‘close to’ the points (u_{i}, v_{i}) by minimising

∑_{i=1}^{n}(u_{i} − x_{1} − x_{2}v_{i})^{2}.

Exercise 7.5.2

Observe that

f(x) ∑_{i=1}^{n} (∑_{j=1}^{m} a_{ij}x_{j} − b_{i})^{2}

is a smooth function of x with f(x) → ∞ as ∥x∥ → ∞. Differentiating and solving the resulting n linear equations in n unknowns will, in general (to see possible problems, take a_{ij} = 0 for all i and j) give a unique stationary point which will be a minimum.

The other two penalty functions cannot (in general) be minimised by calculus techniques, but could be attacked as linear programming problems.

Exercise 7.5.3

(i) If we write the columns of A as column vectors a_{1}, a_{2}, … a_{m}, then (possibly after reordering the columns of A), the Gram–Schmidt method gives us orthonormal row vectors e_{1}, e_{2}, …, e_{k} such that

a_{1} = r_{11}e_{1}
a_{2} = r_{12}e_{1} + r_{22}e_{2}
a_{3} = r_{13}e_{1} + r_{23}e_{2} + r_{33}e_{3}
⋮
a_{k} = r_{1k}e_{1} + r_{2k}e_{2} + r_{3k}e_{3} + … + r_{kk}e_{kk}

for some r_{ij} [1 ≤ j ≤ i ≤ k] with r_{ii} ≠ 0 for 1 ≤ i ≤ k. and

a_{p} = r_{1p}e_{1} + r_{2p}e_{2} + r_{3p}e_{3} + … + r_{pk}e_{k}

for r_{ij} [1 ≤ i ≤ k, k + 1 ≤ p ≤ m]. We set r_{ij} = 0 in all the cases with 1 ≤ i ≤ n and 1 ≤ j ≤ m where this has not previously been defined.

Using the Gram–Schmidt method again, we can now find e_{k+1}, e_{k+2}, …, e_{n} so that the vectors e_{j} with 1 ≤ j ≤ n form an orthonormal basis for the space ℝ^{n} of column vectors. If we take Q to be the n × n matrix with jth column e_{j}, then Q is orthogonal and condition ⋆ gives

A = QR.

(ii) We have

r_{ii} ≠ 0 ∀ 1 ≤ i ≤ m ⇔ rank R = m
⇔ rank Q^{T} A = m
⇔ rank A = m

since the linear map corresponding to Q^{T} is an isomorphism.

Exercise 7.5.4

(i) We know that

π = {At : t ∈ ℝ^{n}}

is a subspace of ℝ^{m}. Thus (e.g. by Theorem 7.1.7) it contains a closest point y to b. Choose x_{0} with Ax_{0} = y. to obtain a minimiser for our problem.

Since A has rank strictly less than m we can find a non zero u ∈ ℝ^{m} with Au = 0. Then every choice x = x_{0} + λz minimises ∥Ax − b∥.

(ii) The QL result is just the QR result with a renumbering of the basis.

If A is an m × n matrix, Q an n × n orthogonal matrix and R an n × m upper triangular matrix with A^{T} = QR, then A = R^{T} Q^{T}, Q^{T} is orthogonal and R^{T} a lower triangular matrix. Thus the QR results imply the corresponding LQ results and the QL results imply the corresponding RQ results.

We could have used the QL in the discussion. However, the RQ and LQ results give Q in the wrong place when we derive the opening formulae. In addition (at least if we follow the obvious path) we factorise an n × m matrix with n ≥ m and we are not usually interested in cases where we have more unknowns than equations.

Exercise 7.5.5

Since ρ is norm preserving,

ρ(a) = b ⇒ ∥a∥ = ∥ρ(a)∥ = ∥b∥.

Now take ∥a∥ = ∥b∥ with b ≠ ±a and set c = (a − b)/2. Suppose that

ρx = λx + μ⟨c, x⟩c

describes a reflection with ρa = b and ρx = x whenever ⟨x, a⟩ = ⟨x, b⟩. If x ⊥ c then

x = ρ(x) = λx

so λ = 1.

Since

⟨c, a⟩ = \frac{1}{2}(∥a∥^{2} + ⟨a, b⟩) = \frac{1}{2}(∥b∥^{2} + ⟨b, a⟩),

we have ρ(c) = c, so

c = c + μ⟨c, c⟩c

and μ = −2∥c∥^{2}. Thus

ρx = x − 2\frac{⟨c, x⟩}{∥c∥^{2}}c

Conversely, if

ρx = x − 2\frac{⟨c, x⟩}{∥c∥^{2}}c,

then ρc = −c and

d ⊥ c ⇒ ρd = d,

so ρ is a reflection. Further

⟨x, a⟩ = −⟨x, b⟩ ⇔ ⟨x, c⟩ = 0 ⇔ ρx = −x

and

ρa = a − 2\frac{⟨c, a⟩}{∥c∥^{2}}c

= a − \frac{⟨c, a⟩ + ⟨c, b⟩}{∥c∥^{2}}c

= a − 2\frac{⟨c, c⟩}{∥c∥^{2}}c

= a − 2c = b

and, since ρ is a reflection ρb = ρa.

We have

T_{ij}x_{j} = x_{i} − 2c_{j}x_{j}∥c∥^{−2}c_{i} = (δ_{ij} − 2(c_{k}c_{k})^{−2}c_{i}c_{j})x_{j},

so

T_{ij} = δ_{ij} − 2(c_{k}c_{k})^{−2}c_{i}c_{j}.

If a = (±∥a∥, 0, 0, …, 0), take T_{1} = I. Otherwise, take T_{1} = T with T as above.

If C is an n×m matrix with c_{ij} = 0 for all j > i when i ≤ r, consider the (n − r) × (m − r) matrix A formed by the c_{ij} with n ≥ i ≥ n − r + 1 m ≥ j ≥ m − r + 1. We can find a (n − r) × (n − r) reflection matrix (or the identity) T with T A having first column zero except possibly the first entry. If S is the n × n matrix given by

S = (\begin{matrix}I & 0 \\ 0 & A \end{matrix}) ,

then H = SC is an n × m matrix with h_{ij} = 0 for all j > i when i ≤ r + 1.

Thus we can find reflection (or identity) matrices T_{j} such that

T_{n−1}T_{n−2} … T_{1}A = R

is upper triangular. The matrix Q = T_{1}T_{2} … T_{n−1} is the product of reflection (or identity) so of orthonormal matrices so (since O(ℝ^{n}) is a group under multiplication) an orthonormal matrix.

We have

A = (T_{n−1}T_{n−2} … T_{1})^{−1}R = T_{1}^{−1}T_{2}^{−1} … T_{n}^{−1}R = T_{1}T_{2} … T_{n}R = QR.

Exercise 7.5.6

We are interested in the vector a = (1, 2, 2)^{T} of length 3 which we wish to reflect to the vector b = (3, 0, 0)^{T}. We thus want a reflection in the plane perpendicular to

c = (a − b)/2 = (−1, 1, 1)^{T} .

This reflection is given by

ρ(x) = x − 2∥c∥^{−2}(c · x)c

with matrix

T = I − 2∥c∥^{2}c^{T} c

= (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) − \frac{2}{3} (\begin{matrix}1 & −1 & −1 \\ −1 & 1 & 1 \\ −1 & 1 & 1 \end{matrix})

= (\begin{matrix}\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & −\frac{2}{3} \\ \frac{2}{3} & −\frac{2}{3} & \frac{1}{3} \end{matrix} )

Now

T A = (\begin{matrix}3 & \frac{11}{3} & \frac{4}{3} \\ 0 & 0 & 1 \\ 0 & 1 & −\frac{1}{3} \end{matrix} )

so (interchanging the second and third row) T A is upper triangular.

Exercise 7.5.7

We are interested in the vector a = (2, 2, −1)^{T} of length 3 which we wish to reflect to the vector b = (3, 0, 0)^{T}. We thus want a reflection in the plane perpendicular to

c = (a − b)/2 = (−\frac{1}{2}, 1, −\frac{1}{2})^{T} .

Setting d = 2c = (−1, 2, −1)^{T}, reflection is given by

ρ(y) = y − 2∥d∥^{2}d · yd

with matrix

T = I − 2∥c∥^{2}d^{T} d

= (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) − \frac{1}{3} (\begin{matrix}1 & −2 & 1 \\ −2 & 4 & −2 \\ 1 & −2 & 1 \end{matrix})

= (\begin{matrix}\frac{2}{3} & \frac{2}{3} & −\frac{1}{3} \\ \frac{2}{3} & −\frac{1}{3} & \frac{2}{3} \\ −\frac{1}{3} & \frac{2}{3} & −\frac{1}{3} \end{matrix} )

Setting

Q = (\begin{matrix}1 & 0 & 0 & 0 \\ 0 & \frac{2}{3} & \frac{2}{3} & −\frac{1}{3} \\ 0 & \frac{2}{3} & −\frac{1}{3} & \frac{2}{3} \\ 0 & −\frac{1}{3} & \frac{2}{3} & −\frac{1}{3} \end{matrix} ) .

We see that (since Q is orthogonal) we seek the least squares fit for

Q (\begin{matrix}1 & 3 \\ 0 & 2 \\ 0 & 2 \\ 0 & −1 \end{matrix}) x = Q (\begin{matrix}4 \\ 1 \\ 4 \\ 1 \end{matrix})

that is to say

(\begin{matrix}1 & 3 \\ 0 & 3 \\ 0 & 0 \\ 0 & 0 \end{matrix}) x = (\begin{matrix}4 \\ 3 \\ 0 \\ 3 \end{matrix})

so we require x_{2} = 1, x_{1} = 1.

Let f be the square of the norm of

(\begin{matrix}1 & 3 \\ 0 & 2 \\ 0 & 2 \\ 0 & −1 \end{matrix}) x − (\begin{matrix}4 \\ 1 \\ 4 \\ 1 \end{matrix})

Then f(x_{1}, x_{2}) = (x_{1} +3x_{2} −4)^{2} +(2x_{2} −1)^{2} +(2x_{2} −4)^{2} +(x_{2} +1)^{2} and the unique stationary value of f (which must be minimum) is given by

0 = \frac{∂f}{∂x_{1}} = 2(x_{1} + 3x_{2} − 4)
0 = \frac{∂f}{∂x_{2}} = 2(3(x_{1} + 3x_{2} − 4) + 2(2x_{2} − 1) + 2(2x_{2} − 4) + (x_{2} + 1))

that is to say, by

4 = x_{1} + 3x_{2}
21 = 3x_{1} + 18x_{2},

that is to say, by

4 = x_{1} + 3x_{2}
7 = x_{1} + 6x_{2},

so we require x_{2} = 1, x_{1} = 1, as stated earlier.

Exercise 8.1.7

(i) We have

A^{T} = A ⇒ (P^{T} AP)^{T} = P^{T} A^{T} P^{T T} = P^{T} AP

and

(P^{T} AP)^{T} = P^{T} AP ⇒ P^{T} A^{T} P = P^{T} AP
⇒ P (P^{T} A^{T} P)P^{T} = P ((P^{T} AP)P^{T}
⇒ A^{T} = A.

(ii) If P ∈ SO(ℝ^{n}), then set Q = P.

If P ∉ SO(ℝ^{n}), then set Q = P D. We have

det Q = det P det D = (−1)^{2} = 1

and Q^{T} Q = D^{T} P^{T} P D = DID = I, so Q ∈ SO(ℝ^{n}). Further, since P^{T} AP is diagonal,

Q^{T} AQ = D^{T} P^{T} AP D = D(P^{T} AP)D = P^{T} AP

and Q^{T} AQ is diagonal.

Exercise 8.2.2

If f a_{kj} = a_{jk} for all j and k, ∑_{j=1}^{n} a_{kj}u_{j} = λu_{k} and ∑_{j=1}^{n} a_{kj}v_{j} = μv_{k} for all k, but λ ≠ μ, then

λ ∑_{k=1}^{n} u_{k}v_{k} = ∑_{k=1}^{n} λu_{k}v_{k}

= ∑_{k=1}^{n} ∑_{j=1}^{n} a_{kj}u_{j}v_{k} = ∑_{k=1}^{n} ∑_{j=1}^{n} a_{jk}u_{j}v_{k}

= ∑_{j=1}^{n} ∑_{k=1}^{n} u_{j}a_{jk}v_{k} = ∑_{j=1}^{n} μu_{j}v_{j}

= μ ∑_{j=1}^{n} u_{j}v_{j} = μ ∑_{k=1}^{n} u_{k}v_{k},

so ∑_{k=1}^{n} u_{k}v_{k} = 0.

Exercise 8.2.6

We have (by Cramer’s rule or direct calculation),

P^{−1} = (\begin{matrix}1 & −1 \\ 0 & 1 \end{matrix})

so

P AP^{−1} = P (AP^{−1}) = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix}2 & −2 \\ 0 & 1 \end{matrix}) = (\begin{matrix}2 & −2 \\ 2 & −1 \end{matrix}) .

Observe that P is not orthogonal.

Exercise 8.2.7

Let

A = (\begin{matrix}1 & i \\ i & −1 \end{matrix})

so A^{T} = A.

det(tI − A) = (t − 1)(t + 1) + 1 = t^{2}

so all the eigenvalues of A are 0 and, if A were diagonalisable, we would have A = 0 which is not the case.

Exercise 8.2.9

Interchanging x_{2} and x_{3} coordinates is the same as reflecting in a plane passing through the x_{1} axis and a line bisecting the angle between the x_{2} and x_{3} axes (so perpendicular to (0, 1, −1)^{T}).

Exercise 8.2.10

If there are n distinct eigenvalues λ_{j} with associated eigenvectors e_{j} then

P^{T} AP = D

with P orthogonal and D diagonal if and only if the columns of P consist of the eigenvectors ±e_{j} (with choice of signs) in some order so there are exactly 2^{n}n! such matrices.

If A has less than n distinct eigenvalues then ℝ^{n} has a basis of eigenvectors e_{j} with e_{n−1} and e_{n} having the same eigenvalues, then, if P has rth column e_{r} for 1 ≤ r ≤ n−2 n−1th column cos θe_{n−1}−sin θe_{n} and nth column sin θe_{n−1} + cos θe_{n}, we have P orthogonal and

P^{T} AP = D

so case (ii) occurs.

Exercise 8.3.2

Observe that there exists a diagonal matrix D and an orthogonal matrix P with

D = (\begin{matrix}λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}) and A = P^{T} AP

where λ_{1}, λ_{2} are the eigenvalues of A. Parts (i), (ii) and (iii) can be read off from the observation that det A = det D = λ_{1}λ_{2} and Tr A = Tr D = λ_{1} + λ_{2}.

(iv) det A = uw − v^{2} ≤ uw, so det A > 0 ⇒ u ≠ 0.

If u > 0 and det A > 0, then w > 0 so Tr A > 0, so, by (iii), the eigenvalues of A are strictly positive.

If u < 0 and det A > 0, then w < 0, so Tr A < 0, so, by applying (iii) to A, the eigenvalues of A are strictly positive.

Exercise 8.3.4

By an orthogonal transformation we may reduce the equation ax^{2} + 2bxy + cy^{2} = d to Ax^{2} + By^{2} = K. By multiplying by −1 if necessary we may assume K ≥ 0 and by interchanging x and y if necessary we may assume that A = 0 ⇒ B = 0 and if A and B are non-zero with opposite signs then A > 0 > B

The cases may now be enumerated as follows.

(1) K > 0, A > 0, B > 0 ellipse.

(2) K > 0, A < 0, B < 0 empty set.

(3) K > 0, A > 0, B < 0 hyperbola.

(4) K = 0, A > 0, B > 0 single point {0}.

(5) K = 0, A < 0, B < 0 single point {0}.

(6) K = 0, A > 0, B < 0 pair of lines meeting at 0.

(7) K > 0, A > 0 B = 0 pair of parallel lines.

(8) K > 0, A < 0, B = 0 empty set.

(7) K = 0, A > 0 B = 0 single line through 0.

(8) K = 0, A < 0, B = 0 single line through 0.

(9) K > 0, A = 0 B = 0 empty set.

(10) K = 0 A = 0, B = 0 whole plane.

Exercise 8.3.5

Let

(\begin{matrix}8 & 6^{1/2} \\ 6^{1/2} & 7 \end{matrix})

Then

det(t − A) = (t − 8)(t − 7) − 6 = t^{2} − 15t + 50 = (t − 10)(t − 5) so A has eigenvalues 5 and 10.

A (\begin{matrix}x \\ y \end{matrix}) = 5 (\begin{matrix}x \\ y \end{matrix}) ⇒ \begin{cases}3x & = 6^{1/2}y \\ 2y & = 6^{1/2}x \end{cases}

so the eigenvector (1, (3/2)^{1/2})^{T} gives one axis of symmetry.

A (\begin{matrix}x \\ y \end{matrix}) = 10 (\begin{matrix}x \\ y \end{matrix}) ⇒ \begin{cases}−2x & = 6^{1/2}y \\ −3y & = 6^{1/2}x \end{cases}

so the eigenvector (1, −(2/3)^{1/2})^{T} gives the other axis of symmetry.

Exercise 8.4.2

(i) ⟨z, z⟩ = ∑_{j=1}^{n} |z_{j}|^{2} is always real and positive. (ii) We have

⟨z, z⟩ = 0 ⇔ ∑_{j=1}^{n} |z_{j}|^{2} = 0
⇔ |z_{j}|^{2} = 0 for 1 ≤ j ≤ n
⇔ |z_{j}| = 0 for 1 ≤ j ≤ n
⇔ z = 0.

(iii) We have

⟨λz, w⟩ = ∑_{j=1}^{n}(λz_{j})w_{j}^{∗}

= λ ∑_{j=1}^{n} z_{j}w_{j}^{∗} = λ⟨z, w⟩.

(iv) We have

⟨z + u, w⟩ = ∑_{j=1}^{n}(z_{j} + u_{j})w_{j}^{∗}

= ∑_{j=1}^{n} z_{j}w_{j}^{∗} + u_{j}w_{j}^{∗} = ⟨z, w⟩ + ⟨u, w⟩.

(v) ⟨z, w⟩^{∗} = (∑_{j=1}^{n} z_{j}w_{j}^{∗})^{∗} = ∑_{j=1}^{n} z_{j}^{∗}w_{j}^{∗∗} = ∑_{j=1}^{n} z_{j}^{∗}w_{j} = ⟨w, z⟩.

Exercise 8.4.3

The result is trivial if z = 0 or w = 0, so we need only consider the case when z, w ≠ 0.

Suppose first that ⟨z, w⟩ is real and positive. Then, if λ is real,

⟨(λz + w), (λz + w)⟩ = λ^{2}⟨z, z⟩ + 2λ⟨z, w⟩ + ⟨w, w⟩

= (λ⟨z, z⟩^{1/2} + \frac{⟨z, w⟩}{⟨z, z⟩^{1/2}} ) + ⟨w, w⟩ − \frac{⟨z, w⟩^{2}}{⟨z, z⟩}

Thus, taking λ = λ_{0}, with λ_{0} = (⟨z, w⟩)^{1/2}(⟨z, z⟩)^{−1/2} we see that

⟨w, w⟩ − \frac{⟨z, w⟩^{2}}{⟨z, z⟩} ≥ 0

with equality if and only if

⟨(λ_{0}z + w), (λ_{0}z + w)⟩ = 0

and so if and only if

λ_{0}z + w = 0

In general, we may choose θ so that

e^{iθ}⟨z, w⟩

is real and positive. Then the result above, applied to e^{iθ}z, yields

⟨w, w⟩ − e^{2iθ} \frac{⟨z, w⟩^{2}}{⟨z, z⟩} ≥ 0

so

|⟨z, w⟩| ≤ ∥z∥∥w∥.

with equality only possible if

λ_{1}z + w = 0

for some λ_{1} ∈ ℂ. By inspection, if this last condition holds, we do have equality.

Exercise 8.4.4

(i) ∥z∥ ≥ 0 since we take the positive square root. (ii) ∥z∥ = 0 ⇔ ⟨z, z⟩ = 0 ⇔ z = 0. (iii) ∥λz∥^{2} = ⟨λz, λz⟩ = λλ^{∗}⟨z, z⟩ = (|λ|∥z∥)^{2}, so ∥λz∥ = |λ|∥z∥. (iv) We have

∥z + w∥^{2} = ⟨z + w, z + w⟩
= ∥z∥^{2} + ⟨z, w⟩ + ⟨w, z⟩ + ∥w∥^{2}
= ∥z∥^{2} + 2ℜ⟨z, w⟩ + ∥w∥^{2}
≤ ∥z∥^{2} + 2∥z∥∥w∥ + ∥w∥^{2}
= (∥z∥ + ∥w∥)^{2}

The result follows on taking square roots.

Exercise 8.4.6

(i) Observe that

∑_{j=1}^{n} λ_{j}e_{j} = 0 ⇒ ⟨∑_{j=1}^{n} λ_{j}e_{j}, e_{k}⟩ = 0 ∀k

⇒ ∑_{j=1}^{n} λ_{j}δ_{jk} = 0 ∀k ⇒ λ_{k} = 0 ∀k,

so we have a set of n linearly independent vectors in a space of dimension n which is thus a basis.

(ii) By (i),

z = ∑_{j=1}^{n} λ_{j}e_{j}

for some λ_{j}. We observe that

λ_{j} = ∑_{j=1}^{n} λ_{j}δ_{jk} = ⟨∑_{j=1}^{n} λ_{j}e_{j}, e_{k}⟩ = ⟨z, e_{k}⟩

(iii) False. If n > m, m vectors cannot span ℂ^{m}.

Exercise 8.4.7

If k = q, we are done, since a linearly independent set forms a basis if and only if the number of elements in the basis equals the dimension of the space.

If not, then k < q and there exists a

u ∈ U \ span{e_{1}, e_{2}, …, e_{k}}

so

v = u − ∑_{j=1}^{k}⟨u, e_{j}⟩e_{j}

is a non-zero element of U. Setting e_{k+1} = ∥v∥^{−1}v we have e_{1}, e_{2}, …, e_{k+1} orthonormal in U.

After repeating the process at most q times we obtain a basis for U.

Exercise 8.4.8

(i) Observe that

∥z − ∑_{j=1}^{k} λ_{j}e_{j}, ∥^{2} = ∥z∥^{2} − ∑_{j=1}^{k} λ_{j}⟨e_{j}, z⟩^{∗} − ∑_{j=1}^{k} λ_{j}^{∗}⟨e_{j}, z⟩ + ∑_{j=1}^{k} |λ_{j}|^{2}

= ∑_{j=1}^{k}(λ_{j} − ⟨e_{j}, z⟩)(λ_{j} − ⟨e_{j}, z⟩)^{∗} + ∥z∥^{2} − ∑_{j=1}^{k} |⟨e_{j}, z⟩|^{2}

= ∑_{j=1}^{k} |λ_{j} − ⟨e_{j}, z⟩|^{2} + ∥z∥^{2} − ∑_{j=1}^{k} |⟨e_{j}, z⟩|^{2}

≥ ∥z∥^{2} − ∑_{j=1}^{k} |⟨e_{j}, z⟩|^{2}

with equality if and only if λ_{j} = ⟨z, e_{j}⟩ for all j.

(ii) Since

z = ∑_{j=1}^{k}⟨z, e_{j}⟩e_{j}

if and only if z ∈ span{e_{1}, e_{2}, …, e_{k}}, it follows from (i) that

∥z∥^{2} ≥ ∑_{j=1}^{k}⟨z, e_{k}⟩^{2},

with equality if and only if z ∈ span{e_{1}, e_{2}, …, e_{k}}.

Exercise 8.4.9

We have

∥z + w∥^{2} − ∥z − w∥^{2} = ⟨z + w, z + w⟩ − ⟨z − w, z − w⟩

= ∥z∥^{2} + ⟨z, w⟩ − ⟨z, w⟩^{∗} + ∥w∥^{2} − ∥z∥^{2} + ⟨z, w⟩ + ⟨z, w⟩^{∗} − ∥w∥^{2} = 4ℜ⟨z, w⟩.

Thus

∥z + iw∥^{2} − ∥z − iw∥^{2} = 4ℜ⟨z, iw⟩ = ℑ⟨z, w⟩

and ∥z + w∥^{2} − ∥z − w∥^{2} + i∥z + iw∥^{2} − i∥z − iw∥^{2} = 4⟨z, w⟩.

Exercise 8.4.10

Existence Choose an orthonormal basis e_{j}. If α has matrix A = (a_{ij}) with respect to this basis, let α^{∗} be the linear map with matrix A^{∗} = (b_{ij}) where b_{ij} = a_{ji}^{∗}. Then

⟨α ∑_{j=1}^{n} z_{j}e_{j}, ∑_{r=1}^{n} w_{j}e_{r}⟩ = ⟨∑_{j=1}^{n} z_{j} ∑_{i=1}^{n} a_{ij}e_{i}, ∑_{r=1}^{n} w_{r}e_{r}⟩

= ∑_{j=1}^{n} ∑_{i=1}^{n} ∑_{r=1}^{n} z_{j}w_{r}a_{ij}δ_{ir}

= ∑_{j=1}^{n} ∑_{i=1}^{n} z_{j}w_{i}a_{ij}

= ∑_{i=1}^{n} ∑_{j=1}^{n} z_{j}w_{i}b_{ji}^{∗}

= ⟨∑_{j=1}^{n} z_{j}e_{j}, α^{∗} ∑_{r=1}^{n} w_{j}e_{r}⟩ ,

so

⟨αz, w⟩ = ⟨z, α^{∗}w⟩

for all z, w ∈ ℂ^{n}.

Uniqueness Observe that, if β and γ are linear maps with

⟨αz, w⟩ = ⟨z, βw⟩ = ⟨z, γw⟩

for all z, w ∈ ℂ^{n}, then

(β − γ)w = 0

for all w ∈ ℂ^{n} (set z = (β − γ)w), so β = γ.

The required formula A^{∗} = (b_{ij}) with b_{ij} = a_{ji}^{∗} follows from the results already established in the question.

Finally

det α^{∗} = det A^{∗} = ∑_{σ∈S_{n}} ζ(σ) ∏_{i=1}^{n} a_{σ(i)i}^{∗}

= ∑_{σ∈S_{n}} ζ(σ) ∏_{j=1}^{n} a_{jσ^{−1}(j)}^{∗}

= ∑_{τ∈S_{n}} ζ(σ) ∏_{j=1}^{n} a_{jτ(j)}^{∗}

= ( ∑_{τ∈S_{n}} ζ(σ) ∏_{j=1}^{n} a_{jτ(j)})^{∗}

= (det A)^{∗} = (det α)^{∗}

Exercise 8.4.11

(i)⇒(ii) If (i) holds, then using the polarisation identity

4⟨αz, αw⟩ = ∥αz + αw∥^{2} − ∥αz − αw∥^{2} + i∥αz + iαw∥^{2} − i∥αz − iαw∥^{2}

= ∥α(z + w)∥^{2} − ∥α(z − w)∥^{2} + i∥α(z + iw)∥^{2} − i∥α(z − iw)∥^{2} = ∥z + w∥^{2} − ∥z − w∥^{2} + i∥z + iw∥^{2} − i∥z − iw∥^{2}

= 4⟨z, w⟩,

so (ii) holds.

(ii)⇒(iii) if (ii) holds

⟨α^{∗}αz, w⟩ = ⟨w, α^{∗}αz⟩^{∗} = ⟨αw, αz⟩^{∗} = ⟨αz, αw⟩ = ⟨z, w⟩ for all w so α^{∗}αz = z for all z so α^{∗}α = ι.

(iii)⇒(iv) Immediate. (iv)⇔(v)⇔(vi) Immediate. (iv)⇒(i) If (iv) holds,

∥αz∥^{2} = ⟨αz, αz⟩ = ⟨z, α^{∗}αz⟩ = ⟨z, z⟩ = ∥z∥^{2}.

Exercise 8.4.12

(i) U(ℂ^{n}) is a subset of the group GL(ℂ^{n}). We have

α, β ∈ U(ℂ^{n}) ⇒ (αβ)^{∗}(αβ) = (β^{∗}α^{∗})(αβ)
= β^{∗}(α^{∗}α)β = β^{∗}ιβ = β^{∗}β = ι
⇒ αβ ∈ U(ℂ^{n})

and

α ∈ U(ℂ^{n}) ⇒ (α^{−1})^{∗} = α^{∗∗} = α = (α^{−1})^{−1} ⇒ α^{−1} ∈ U(ℂ^{n}) whilst

ι^{∗}ι = ιι = ι

so ι ∈ U(ℂ^{n}), Thus U(ℂ^{n}) is a subgroup of GL(ℂ^{n}). (ii) 1 = det ι = det αα^{∗} = det α det α^{∗} = det α(det α)^{∗} = | det α|^{2}. The converse is false. If

A = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) ,

then det A = 1, but

A^{−1} = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) ≠ A^{∗}.

Exercise 8.4.13

If D is a real diagonal matrix with diagonal entries d_{j},

D ∈ O(ℝ^{n}) ⇔ DD^{T} = I ⇔ d_{j}^{2} = 1 ∀j ⇔ d_{j} = ±1 ∀j.

If D is a complex diagonal matrix with diagonal entries d_{j},

D ∈ U(ℂ^{n}) ⇔ DD^{∗} = I ⇔ d_{j}d_{j}^{∗} = 1 ∀j ⇔ |d_{j}| = ±1 ∀j. The diagonal matrix with first diagonal entry e^{iθ} and all other diag-

onal entries 1 is unitary with determinant e^{iθ}.

Exercise 8.4.14

SU(ℂ^{n}) is a subset of the group SU(ℂ^{n}) containing ι.

α, β ∈ SU(ℂ^{n}) ⇒ det αβ = det α det β = 1×1 = 1 ⇒ αβ =∈ SU(ℂ^{n}).

α ∈ SU(ℂ^{n}) ⇒ det α^{−1} = (det α)^{−1} = 1 ⇒ α^{−1} =∈ SU(ℂ^{n}). Thus SU(ℂ^{n}) is a subgroup of U(ℂ^{n}).

Exercise 8.4.15

If e_{1}, e_{2}, …, e_{n} is an orthonormal basis for ℂ^{n} and α has matrix A with respect to this basis.

A = A^{∗} ⇔ α = α^{∗} ⇔ ⟨z, αw⟩ = ⟨z, αw⟩ ∀w, z ∈ ℂ^{n}

⇔ ⟨z, α^{∗}w⟩ = ⟨z, αw⟩ ∀w, z ∈ ℂ^{n}

Exercise 8.4.16

If A is Hermitian,

det A = det A^{∗} = (det A)^{∗}

If

A = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) ,

then det A = 1 is real, but A is not Hermitian.

Exercise 8.4.17

(i) If u is an eigenvector with eigenvalue λ then

λ∥u∥^{2} = ⟨λu, u⟩ = ⟨αu, u⟩ = ⟨u, αu⟩ = ⟨u, λu⟩ = λ^{∗}∥u∥^{2} so λ = λ^{∗}.

(ii) If u, v are eigenvectors with distinct eigenvalues λ and μ then λ⟨u, v⟩ = ⟨λu, v⟩ = ⟨αu, v⟩ = ⟨u, αv⟩ = ⟨u, μv⟩ = μ^{∗}⟨u, v⟩ = μ⟨u, v⟩ so ⟨u, v⟩ = 0.

Exercise 8.4.18

Part (ii) is equivalent to part (i) by the basis change rule.

We prove part (i) by induction on n.

If n = 1, then, since every 1 × 1 matrix is diagonal, the result is trivial.

Suppose now that the result is true for n = m and that α : ℂ^{m+1} → ℂ^{m+1} is a symmetric linear map. We know that the characteristic polynomial must have a root and that all its roots are real. Thus we can can find an eigenvalue λ_{1} ∈ ℝ and a corresponding eigenvector e_{1} of norm 1. Consider the subspace

e_{1}^{⊥} = {u : ⟨e_{1}, u⟩ = 0}.

We observe (and this is the key to the proof) that

u ∈ e_{1}^{⊥} ⇒ ⟨e_{1}, αu⟩ = ⟨αe_{1}, u⟩ = λ_{1}⟨e_{1}, u⟩ = 0 ⇒ αu ∈ e_{1}^{⊥}.

Thus we can define α|_{e_{1}^{⊥}} : e_{1}^{⊥} → e_{1}^{⊥} to be the restriction of α to e_{1}^{⊥}. We observe that α|_{e_{1}^{⊥}} is symmetric and e_{1}^{⊥} has dimension m so, by the inductive hypothesis, we can find m orthonormal eigenvectors of α|_{e_{1}^{⊥}} in e_{1}^{⊥}. Let us call them e_{2}, e_{3}, …,e_{m+1}. We observe that e_{1}, e_{2}, …, e_{m+1} are orthonormal eigenvectors of α and so α is diagonalisable. The induction is complete.

Exercise 8.4.19

We note that

det(tI − A) = (t − 5)(t − 2) − 4 = t^{2} − 7t + 6,

so the eigenvalues of A are

\frac{7 ± 5i}{2}.

Eigenvectors (z, w)^{T} corresponding to the eigenvalue \frac{7+5i}{2} are given by

10z + 4iw = (7 + 5i)z
−4iz + 4w = (7 + 5i)w,

that is to say

(3 − 5i)z + 4iw = 0
−4iz − (3 + 5i)w = 0,

so an eigenvector of norm 1 is given by 2^{−1/2}5^{−1}(4i, 5i − 3)^{T}.

Eigenvectors (z, w)^{T} corresponding to the eigenvalue \frac{7−5i}{2} are given by

10z + 4iw = (7 − 5i)z
−4iz + 4w = (7 − 5i)w,

that is to say,

(3 + 5i)z + 4iw = 0
−4iz − (3 − 5i)w = 0,

so an eigenvector of norm 1 is given by 2^{−1/2}5^{−1}(5i − 3, −4i)^{T}.

Thus

U = 2^{−1/2}5^{−1} (\begin{matrix}4i & 5i − 3 \\ 5i − 3 & −4i \end{matrix})

is a unitary matrix with

U^{∗}AU = (\begin{matrix}(7 + 5i)/2 & 0 \\ 0 & (7 − 5i)/2 \end{matrix})

Exercise 8.4.20

We can do this using matrices, but it is nicer to use maps.

If γ is unitary, set α = (γ+γ^{∗})/2, β = (γ−γ^{∗})/(2i). Then γ = α+iβ and

α^{∗} = (γ^{∗} + γ^{∗∗})/2 = (γ^{∗} + γ)/2 = α
β^{∗} = −(γ^{∗} + γ^{∗∗})/(2i) = −(γ^{∗} − γ)/(2i) = β

so α and β are Hermitian.

To prove uniqueness, suppose that γ is unitary, α and β Hermitian and γ = α + iβ.

Then

α − iβ = α^{∗} − iβ^{∗} = (α + iβ)^{∗} = γ^{∗} = γ

so

2α = (α − iβ) + (α + iβ) = 2γ

and α = (γ + γ^{∗})/2, iβ = γ − α.

We observe that

αβ = (γ + γ^{∗})(γ − γ^{∗})/(4i) = (γ^{2} − 2ι + (γ^{∗})^{2})/(4i)
(γ − γ^{∗})(γ + γ^{∗})/(4i) = βα

and α^{2} + β^{2} = 4^{−1}((γ + γ^{∗})^{2} − (γ − γ^{∗})^{2}) = 4^{−1}(2γγ^{∗} + 2γ^{∗}γ) = ι. If we look at 1×1 matrices we get G = (e^{iθ}), A = (cos θ), B = (i sin θ)

and recover Euler’s formula.

Exercise 9.1.2

We have

LL^{T} = I and det L = det (\begin{matrix}2^{−1/2} & −2^{−1/2} \\ 2^{−1/2} & 2^{−1/2} \end{matrix}) = 1

so L ∈ SO(ℝ^{3}).

However, if we set x = (0, 1, 0)^{T}, we have x′ = (0, 2^{−1/2}, 2^{−1/2}), so

x_{2}^{′4} = 4^{−1} ≠ 2^{−1/2} = l_{2}x_{2}^{4} ∑_{j=1}^{3} l_{2j}x_{j}^{4}

Exercise 9.1.4

(i) We have

u̇′_{i} = \frac{d}{dt}u′_{i} = \frac{d}{dt}l_{ij}u_{j} = l_{ij}u̇_{j}

so u̇ is a Cartesian tensor of order 1. (ii) x is a Cartesian tensor of order 1, so ẋ is, so ẍ is, so F = mẍ is.

Exercise 9.2.1

(i) Observe that a_{ij} = u′_{i}v′_{j} = l_{ir}u_{r}l_{js}v_{s} = l_{ir}l_{js}a_{rs}.

(ii) Observe that

a′_{ij} = \frac{∂u′_{j}}{∂x′_{i}} = \frac{∂u′_{j}}{∂x_{r}} \frac{∂x_{r}}{∂x′_{i}} = l_{ir} \frac{∂l_{jk}u_{k}}{∂x_{r}} = l_{ir}l_{jk} \frac{∂u_{k}}{∂x_{r}} = l_{ir}l_{jk}a_{rk}.

(iii) Observe that

a′_{ij} = \frac{∂^{2}ϕ}{∂x′_{i}∂x′_{j}} = \frac{∂^{2}ϕ}{∂x_{r}∂x_{s}} \frac{∂x_{r}}{∂x′_{i}} \frac{∂x_{s}}{∂x′_{j}} = l_{ir}l_{js}a_{rs}.

Exercise 9.3.1

(i) If u_{i}, v_{j} w_{k} are Cartesian tensors of order 1 then

(u_{i}v_{j}w_{k})′ = u′_{i}v′_{j}w′_{k} = l_{ir}l_{js}l_{kp}u_{r}v_{s}w_{p}

so u_{i}v_{j}w_{k} is a Cartesian tensor of order 3.

(ii) If u_{ij} is a Cartesian tensor of order 2 then

\frac{∂u′_{ij}}{∂x′_{k}} = \frac{∂l_{ir}l_{js}u′_{rs}}{∂x_{p}} \frac{∂x_{p}}{∂x′_{k}} = l_{ir}l_{js}l_{kp} \frac{∂u_{rs}}{∂x_{p}}

so

\frac{∂u_{ij}}{∂x_{k}}

is a Cartesian tensor of order 3.

(iii) If ϕ is a smooth tensor of order 0

\frac{∂^{2}ϕ}{∂x′_{i}∂x′_{j}∂x′_{k}} = \frac{∂^{3}ϕ}{∂x_{r}∂x_{s}∂x_{p}} \frac{∂x_{r}}{∂x′_{i}} \frac{∂x_{s}}{∂x′_{j}} \frac{∂x_{p}}{∂x′_{k}} = l_{ir}l_{js}l_{kp} \frac{∂^{3}ϕ}{∂x_{r}∂x_{s}∂x_{p}}

so

\frac{∂^{2}ϕ}{∂x_{i}∂x_{j}∂x_{k}}

is a Cartesian tensor of order 3.

Exercise 9.3.5

(i) Observe that, in our standard notation

l_{ik}a′_{i}b_{k} = a′_{i}b′_{i} = (a_{i}b_{i})′ = a_{i}b_{i} = a_{k}b_{k}.

and so

(l_{ik}a′_{i} − a_{k})b_{k} = 0.

Since we can assign b_{k} any values we please in a particular coordinate system, we must have

l_{ik}a′_{i} − a_{k} = 0,

so

δ_{ir}a′_{i} − l_{rk}a_{k} = l_{rk}(l_{ik}a′_{i} − a_{k}) = 0

and we have shown that a′_{r} = l_{rk}a_{k} and a_{i} is a Cartesian tensor of order 1.

(ii) By (i), a_{ij}b_{i} is a Cartesian tensor of rank 1 whenever b_{i} is a Cartesian tensor of order 1 and so, by Theorem 9.3.4, a_{ij} is a Cartesian tensor of order 2.

(iii) Apply (ii) with u_{ij} = b_{i}c_{j}.

(iv) Observe that, in our standard notation

l_{kr}l_{ms}a′_{ijkm}b_{rs} = a′_{ijkm}b′_{km} = (a_{ijkm}b_{km})′ = l_{ip}l_{jq}(a_{pqkm}b_{km}) = l_{ip}l_{jq}(a_{pqrs}b_{rs})

and so

(l_{kr}l_{ms}a′_{ijkm} − l_{ip}l_{jq}a_{pqrs})b_{rs} = 0.

Since we can assign b_{rs} any values we please in a particular coordinate system, we must have

l_{kr}l_{ms}a′_{ijkm} − l_{ip}l_{jq}a_{pqrs} = 0.

so

l_{er}l_{fs}(l_{kr}l_{ms}a′_{ijkm} − l_{ip}ljq)a_{pqrs} = 0

whence a′_{ijef} = l_{er}l_{fs}l_{ip}ljqa_{pqrs} and we have shown that a_{ijkl} is a Cartesian tensor.

Exercise 9.3.6

(i) We have

c_{ijkm}e_{km} = \frac{1}{2}(c̃_{ijkm} + c̃_{ijmk})e_{km} = \frac{1}{2}(c̃_{ijkm}e_{km} + c̃_{ijmk}e_{km})
= \frac{1}{2}(c̃_{ijkm}e_{km} + c̃_{ijmk}e_{mk}) = p_{ij}

and c_{ijmk} = \frac{1}{2}(c̃_{ijmk} + c̃_{ijkm}) = \frac{1}{2}(c̃_{ijkm} + c̃_{ijmk}) = c_{ijkm}.

(ii) Set e_{mk} = \frac{1}{2}(b_{km} + b_{mk}) and f_{mk} = \frac{1}{2}(b_{km} − b_{mk}).

(iii) We have

c_{ijmk}f_{mk} = c_{ijkm}f_{km} = −c_{ijmk}f_{mk}

so c_{ijmk}f_{mk} = 0 and

c_{ijkm}e_{km} = c_{ijmk}b_{mk}.

Thus c_{ijmk}b_{mk} is a Cartesian tensor of order 2 whenever b_{mk} is, so, by the quotient rule given in Exercise 9.3.5 (iv), c_{ijmk} is a Cartesian tensor of order 4.

Exercise 9.3.8

If we work with a fixed set of coordinates, Cartesian tensors of rank n may be identified with the space of functions

f : ℝ^{3n} → ℝ

with pointwise multiplication and multiplication by scalars. They therefore form a vector space of dimension 3^{n}.

Exercise 9.4.3

Using the Levi-Civita identity, we have ϵ_{ijk}ϵ_{klm}ϵ_{mni} = ϵ_{kij}ϵ_{klm}ϵ_{mni} = (δ_{il}δ_{jm}−δ_{im}δ_{jl})ϵ_{mni} = ϵ_{jnl}−ϵ_{mnm} = ϵ_{jnl} = ϵ_{ljn}.

Exercise 9.4.4 Observe that

ϵ_{ijk...p}ϵ_{ijk...p} = ∑_{σ∈S_{n}} ζ(σ)^{2} = ∑_{σ∈S_{n}} 1 = card S_{n} = n!

(See the definition of the determinant on page 87 if necessary.)

Exercise 9.4.5

(i) We have

λ(ϵ_{ijk}a_{j}b_{k}) = ϵ_{ijk}(λa_{j})b_{k} = ϵ_{ijk}a_{j}(λb_{k}),

so

λ(a × b) = (λa) × b = a × (λb).

(ii) We have

ϵ_{ijk}a_{j}(b_{k} + c_{k}) = ϵ_{ijk}a_{j}b_{k} + ϵ_{ijk}a_{j}c_{k},

so

a × (b + c) = a × b + a × c.

(iii) We have

ϵ_{ijk}a_{j}b_{k} = −ϵ_{ikj}a_{j}b_{k} = −ϵ_{ikj}b_{k}a_{j},

so

a × b = −b × a.

(iv) Putting b = a in (iii), we get

a × a = −a × a

so a × a = 0.

Exercise 9.4.6

There is no identity. For suppose e was an identity. Then

e = e × e = 0,

but

0 × a = 0

for all a so (since ℝ^{3} has more than one element) 0 is not an identity.

Exercise 9.4.7

(i) We have

λ(a_{j}b_{j}) = (λa_{j})b_{j} = a_{j}(λb_{j}),

so

λ(a · b) = (λa) · b = a · (λb).

(ii) We have

a_{k}(b_{k} + c_{k}) = a_{k}b_{k} + a_{k}c_{k},

so

a · (b + c) = a · b + a · c.

(iii) We have

a_{j}b_{j} = b_{j}a_{j},

so

a · b = b · a.

Exercise 9.4.9

(i) We observe that a × (b × c) = (a_{1}, a_{2}, a_{3}) × (b_{2}c_{3} − b_{3}c_{2}, b_{3}c_{1} − b_{1}c_{3}, b_{1}c_{2} − b_{2}c_{1}) = (a_{2}(b_{1}c_{2} − b_{2}c_{1}) − a_{3}(b_{3}c_{1} − b_{1}c_{3}), a_{3}(b_{2}c_{3} − b_{3}c_{2}) − a_{1}(b_{1}c_{2} − b_{2}c_{1}),

a_{1}(b_{3}c_{1} − b_{1}c_{3}) − a_{2}(b_{2}c_{3} − b_{3}c_{2})) = ((a_{2}c_{2} + a_{3}c_{3})b_{1}, (a_{3}c_{3} + a_{1}c_{1})b_{2}, (a_{1}c_{1} + a_{2}c_{2})b_{3})

− ((a_{2}b_{2} + a_{3}b_{3})c_{1}, (a_{3}b_{3} + a_{1}b_{1})c_{2}, (a_{1}b_{1} + a_{2}b_{2})c_{3}) = ((a_{2}c_{2} + a_{3}c_{3} + a_{1}c_{1})b_{1}, (a_{3}c_{3} + a_{1}c_{1} + a_{2}c_{2})b_{2}, (a_{1}c_{1} + a_{2}c_{2} + a_{3}c_{3})b_{3}) − ((a_{2}b_{2} + a_{3}b_{3} + a_{1}b_{1})c_{1}, (a_{3}b_{3} + a_{1}b_{1} + a_{2}b_{2})c_{2}, (a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3})c_{3}) = (a · c)b − (a · b)c.

(ii) ‘Search me, guv’ner!’ There is a geometric proof in the American Mathematical Monthly (Vol 72, Feb 1965) by K. Bishop, but I doubt if the reader will call it simple.

Exercise 9.4.10

We have (a × b) × c = −c × (a × b) = −((c · b)a − (c · a)b) = (a · c)b − (b · c)a. as required.

If x = (1, 0, 0), y = z = (0, 1, 0) then

(x × y) × z = (0, 0, 1) × (0, 1, 0) = (−1, 0, 0) ≠ (0, 0, 0)

= (0, 0, 1) × (0, 0, 0) = x × (y × z).

Exercise 9.4.11

We have

a × (b × c) + b × (c × a) + c × (a × b)

= (a · c)b − (a · b)c + (b · a)c − (b · c)a + (c · b)a − (c · a)b = 0.

Exercise 9.4.12

We have

(a × b) · (a × b) + (a · b)^{2} = ϵ_{ijk}a_{j}b_{k}ϵ_{irs}a_{r}b_{s} + a_{i}b_{i}a_{j}b_{j}

= (δ_{jr}δ_{ks} − δ_{js}δ_{kr})a_{j}b_{k}a_{r}b_{s} + a_{i}b_{i}a_{j}b_{j} = a_{j}a_{j}b_{k}b_{k} − a_{j}b_{j}a_{k}b_{k} + a_{i}b_{i}a_{j}b_{j}

= a_{j}a_{j}b_{k}b_{k} = ∥a∥^{2}∥b∥^{2}.

If we use the geometric formulae for a × b and a · b, our formula becomes

a^{2}b^{2} cos^{2} θ + a^{2}b^{2} sin^{2} θ = a^{2}b^{2}

where θ is the angle between a and b, a = ∥a∥ and b = ∥b∥. We thus recover the formula

cos^{2} θ + sin^{2} θ = 1.

Exercise 9.4.13

(i) We have

(a × b) · a = ϵ_{ijk}a_{j}b_{k}a_{i} = −ϵ_{jik}a_{j}b_{k}a_{i} = −(a × b) · a so

(a × b) · a = 0

(ii) a × b ⊥ a so (a × b) · a = 0 (iii) The volume of a parallelepiped with one vertex at 0 and adjacent

vertices at a, b and a is 0 so (a × b) · a = [a, b, a] = 0.

Exercise 9.4.14

We have

(a × b) × (c × d) = ((a × b) · d)c − ((a × b) · c)d
= [a, b, d]c − [a, b, c]d.

We also have

(a × b) × (c × d) = −(c × d) × (a × b)
= (c × d) · ab − c × d) · ba
= [c, d, a]b − [c, d, b]a
= −[c, a, d]b − [b, c, d]a.

Thus

a, b, d]c − [a, b, c]d = [c, a, d]b − [b, c, d]a

and the required result follows.

Our formulae also give

(a × b) × (a × c) = ((a × b) · c)a − ((a × b) · a)c
= [a, b, c]a − 0 = [a, b, c]a

since (a × b) ⊥ a

Exercise 9.4.15

Since b × c ⊥ b, c we have

x · (b × c = (λa + μb + νc) · (b × c) = λa · (b × c) so

λ = \frac{[x, b, c]}{[a, b, c]}.

Similarly

μ = \frac{[a, x, b]}{[a, b, c]} and ν = \frac{[a, b, x]}{[a, b, c]}.

Exercise 9.4.16

(i) We work with row vectors

[a, b, c]^{2} = det (\begin{matrix}a \\ b \\ c \end{matrix})^{2}

= det (\begin{matrix}a \\ b \\ c \end{matrix}) det (a^{T} b^{T} c^{T}))

= det (\begin{matrix}aa^{T} & ab^{T} & ac^{T} \\ ba^{T} & bb^{T} & bc^{T} \\ ca^{T} & cb^{T} & cc^{T} \end{matrix} )

= det (\begin{matrix}a · a & a · b & a · c \\ b · a & b · b & b · c \\ c · a & c · b & c · c \end{matrix})

(ii) In the case given we first observe that r^{2} = ∥a + b + c∥^{2} = ∥a∥^{2} + ∥b∥^{2} + ∥c∥^{2} + 2(a · b + b · c + a · b

so

a · b + b · c + a · b = −r^{2}

[a, b, c]^{2} = det (\begin{matrix}a · a & a · b & a · c \\ b · a & b · b & b · c \\ c · a & c · b & c · c \end{matrix})

= det (\begin{matrix}r^{2} & a · b & a · c \\ b · a & r^{2} & b · c \\ c · a & c · b & r^{2} \end{matrix})

= r^{2}(r^{4} − (b · c)^{2}) − a · b(r^{2}a · b − (b · c)(a · c)) + a · c((b · a)(b · c) − r^{2}b · a) = 2r^{6} + 2r^{4}(a · b + b · c + a · b) + r^{2}(a · b + b · c + a · b)^{2} − r^{2}((a · b)^{2} + (b · c)^{2} + (c · a)^{2}) + 2(a · b)(b · c)(c · a) = 2(r^{2} + a · b)(r^{2} + b · c)(r^{2} + c · a).

Exercise 9.4.17

Observe, using the summation convention, that

(a × b) · (c × d) = ϵ_{ijk}a_{j}b_{k}ϵ_{irs}c_{r}d_{s} = (δ_{jr}δ_{ks} − δ_{kr}δ_{js})a_{j}b_{k}c_{r}d_{s}

= a_{j}c_{j}b_{k}d_{k} − a_{j}d_{j}b_{k}c_{k} = (a · c)(b · d) − (a · d)(c · d)

(Or we could use the formula for the triple vector product.)

Thus

(a × b) · (c × d) + (a × c) · (d × b) + (a × d) · (b × c)
= (a · c)(b · d) − (a · d)(c · d)
+ (a · d)(b · c) − (a · b)(c · d)
+ (a · b)(c · d) − (a · c)(b · d)
= 0.

Exercise 9.4.19

(i) Using the summation convention, we have

(\frac{d}{dt}ϕa)_{i} = \frac{d}{dt}(ϕa_{i}) = ϕ̇a_{i} + ϕȧ_{i} = (ϕ̇a + ϕȧ)_{i} .

(ii) Using the summation convention, we have

\frac{d}{dt}a · b = \frac{d}{dt}a_{i}b_{i} = ȧ_{i}b_{i} + a_{i}ḃ_{i} = ȧ · b + a · ḃ.

(iii) By Lemma 9.4.18, \frac{d}{dt}a × ȧ = ȧ × ȧ + a × ä = a × ä.

Exercise 9.4.21

(i) Using the summation convention, we have

(∇ × (ϕu))_{i} = ϵ_{ijk} \frac{d}{dx_{j}} ϕu_{k}
= ϵ_{ijk} \frac{dϕ}{dx_{j}} u_{k} + ϵ_{ijk} \frac{du_{k}}{dx_{j}} ϕ = (∇ϕ) × u + ϕ∇ × u)_{i}.

(ii) Using the summation convention, we have

∇ · (∇ × u) = \frac{d}{dx_{i}} ϵ_{ijk} \frac{du_{k}}{dx_{j}} = ϵ_{ijk} \frac{d^{2}u_{k}}{dx_{i}dx_{j}} = ϵ_{ijk} \frac{d^{2}u_{k}}{dx_{j}dx_{i}} = −∇ · (∇ × u)

so

∇ · (∇ × u) = 0.

(ii) Using the summation convention, we have

(∇ × (u × v))_{i} = ϵ_{ijk} \frac{d}{dx_{j}} ϵ_{krs}u_{r}v_{s}

= ϵ_{kij}ϵ_{krs} \frac{d}{dx_{j}} u_{r}v_{s}

= (δ_{ir}δ_{js} − δ_{is}δ_{jr})\frac{d}{dx_{j}} u_{r}v_{s}

= \frac{d}{dx_{j}} u_{i}v_{j} − \frac{d}{dx_{j}} u_{j}v_{i}

= v_{j} \frac{du_{i}}{dx_{j}} + u_{i} \frac{dv_{j}}{dx_{j}} − v_{i} \frac{du_{j}}{dx_{j}} − u_{j} \frac{dv_{i}}{dx_{j}}

= ((∇ · u)v + u · ∇v − (∇ · v)u − v · ∇u)_{i}.

Exercise 9.4.22

(i) Observe that the cancellation law continues to hold, so it is sufficient to show that the analogue of the scalar vector product (think of the 4 dimensional volume) is a scalar.

Let a_{i} = ϵ_{ijkl}x_{j}y_{k}z_{l}. Then if b_{i} is an vector (ie order 1 Cartesian tensor),

(a_{i}b_{i})′ = ϵ_{ijkl}x′_{j}y′_{k}z′_{l}b′_{i} = ϵ_{ijkl}l_{ju}l_{kv}l_{lv}l_{ip}x_{u}y_{v}z_{w}b_{p}
= (det L)ϵ_{puvw}x_{u}y_{v}z_{w}b_{p}
= ϵ_{puvw}x_{u}y_{v}z_{w}b_{p} = a_{i}b_{i}

since L ∈ SO(ℝ^{4}).

Thus, by the cancellation law, x ∧ y ∧ z is a vector

(ii) We have

((λx + μu) ∧ y ∧ z)_{i} = ϵ_{ijkl}(λx_{j} + μu_{j})y_{k}z_{l}
= λϵ_{ijkl}x_{j}y_{k}z_{l} + μϵ_{ijkl}u_{j}y_{k}z_{l}
= (λx ∧ y ∧ z + μu ∧ y ∧ z)_{i}.

(y ∧ x ∧ z)_{i} = ϵ_{ijkl}y_{j}x_{k}z_{l} = −ϵ_{ikjl}x_{k}y_{j}z_{l} = (x ∧ y ∧ z)_{i} so

x ∧ y ∧ z = −y ∧ x ∧ z.

The proof that

x ∧ y ∧ z = −z ∧ y ∧ x

follows the same lines.

(iii) Using the summation convention with range 1, 2, 3, 4, 5, we define

x ∧ y ∧ z ∧ w = a

with

a_{i} = ϵ_{ijklu}x_{j}y_{k}z_{l}w_{u}.

Exercise 10.1.4

Since the material is isotropic so is a_{ijk} thus a_{ijk} = κϵ_{ijk} and

σ_{ij} = a_{ijk}B_{k} = κϵ_{ijk}B_{k} = −κϵ_{jik}B_{k} = −σ_{ji} = −σ_{ij}.

Thus σ_{ij} = 0.

Exercise 10.1.2

It is sufficient to show that δ_{ij}δ_{kl} is isotropic, since it then follows that δ_{ik}δ_{jl} and δ_{il}δ_{jk} and all linear combinations are.

But (δ_{ij}δ_{kl})′ = δ′_{ij}δ′_{kl} = δ_{ij}δ_{kl} so we are done.

If

αδ_{ij}δ_{kl} + βδ_{ik}δ_{jl} + γδ_{il}δ_{jk} = 0,

then

0 = αδ_{ij}δ_{kk} + βδ_{ik}δ_{jk} + γδ_{ik}δjk = (3α + β + γ)δ_{ij}.

Thus 3α + β + γ = 0 and similarly α + 3β + γ = 0, α + β + 3γ = 0. Thus

2α = 2β = 2γ = −(α + β + γ)

and α = β = γ = 0. (Alternatively, just examine particular entries in the tensor αδ_{ij}δ_{kl} + βδ_{ik}δ_{jl} + γδ_{il}δ_{jk}.)

For the last part, observe that, since ϵ_{ijk} and ϵ_{ist} are isotropic, so is ϵ_{ijk}ϵ_{ist}. Thus

ϵ_{ijk}ϵ_{ist} = αδ_{jk}δ_{st} + βδ_{js}δ_{kt} + γδ_{jt}δks.

We observe that

αδ_{jk}δ_{st}+βδ_{js}δ_{kt}+γδ_{jt}δks = ϵ_{ijk}ϵ_{ist} = −ϵ_{ikj}ϵ_{ist} = −αδ_{jk}δ_{st}+γδ_{js}δ_{kt}+βδ_{jt}δks

so, by the linear independence proved earlier, β = −γ.

Next we observe that

0 = ϵ_{ijj}ϵ_{ist} = (3α + β + γ)δ_{st}

so 3α + β + γ = 0 and α = 0 Finally

6 = ϵ_{ijk}ϵ_{ijk} = 3α + 9β + 3γ

so α = 0, β = 1 and γ = −1. We have shown that

ϵ_{ijk}ϵ_{ist} = δ_{js}δ_{kt} − δ_{ks}δ_{jt}.

Exercise 10.1.3

a′_{ij} = l_{ir}l_{js}a_{rs} = l_{ir}l_{js}a_{sr} = a′_{ji}.

Exercise 10.1.6

(i) We have

a′_{ij} = l_{ir}l_{js}a_{rs} = −l_{ir}l_{js}a_{sr} = −a′_{ji}.

(ii) We have

b_{ij} = \frac{1}{2}(b_{ij} + b_{ji}) + \frac{1}{2}(b_{ij} − b_{ji})

and

\frac{1}{2}(b_{ji} + b_{ij}) = \frac{1}{2}(b_{ij} + b_{ji})
\frac{1}{2}(b_{ji} − b_{ij}) = −\frac{1}{2}(b_{ij} − b_{ji}).

(iii) Observe that the zero order two tensor is both symmetric and antisymmetric. Further

a_{ij} = a_{ji}, b_{ij} = b_{ji} ⇒ λa_{ij} + μb_{ij} = λa_{ji} + μb_{ji}

and

a_{ij} = −a_{ji}, b_{ij} = −b_{ji} ⇒ λa_{ij} + μb_{ij} = −(λa_{ji} + μb_{ji})

Thus the symmetric tensors form a subspace and the antisymmetric tensors form a subspace.

We do not use the summation convention for α and β.

Let e(α, β) be the second order Cartesian tensor with

e_{ij}(α, β) = δ_{αi}δ_{βj} + δ_{αi}δ_{βj}

in a particular coordinate system S [1 ≤ α < β ≤ 3] and

e_{ij}(α, α) = δ_{αi}δ_{αj}

for 1 ≤ α ≤ 3

Let f(α, β) be the second orderCartesian tensor with

f_{ij}(α, β) = δ_{αi}δ_{βj} − δ_{αi}δ_{βj}

in the same coordinate system S [1 ≤ α < β ≤ 3].

Then in that coordinate system e(α, β) is symmetric, f(α, β) is antisymmetric. If a is a second order symmetric tensor then in the coordinate system S,

a = ∑_{1≤α≤β≤3} aαβe(α, β)

and if b is an antisymmetric tensor then in the coordinate system S,

b = ∑_{1≤α<β≤3} b_{αβ}f(α, β).

Since the result is true in one system it is true in all.

Working in S we see that

∑_{1≤α≤β≤3} a_{αβ}e(α, β) = 0 ⇒ a_{αβ} = 0 [1 ≤ α ≤ β ≤ 3]

and

∑_{1≤α<β≤3} b_{αβ}e(α, β) = 0 ⇒ b_{αβ} = 0 [1 ≤ α < β ≤ 3].

Thus the e(α, β) give a basis for the subspace of symmetric tensors which thus has dimension 6 the f(α, β) give a basis for the subspace of antisymmetric tensors which thus has dimension 3.

Exercise 10.1.8

Since

ϵ_{ijk}ω_{j}x_{k} = (ω × x)_{i}

and ω × x ⊥ x, the only possible eigenvalue is 0 and this corresponds to eigenvectors tω with t ≠ 0.

Exercise 10.2.1

Observe first that

∑_{α} m_{α}r_{α} = ∑_{α} m_{α}x_{α} − Mx_{G} = 0.

We have

H = ∑_{α} m_{α}x_{α} × ẋ_{α}

= ∑_{α} m_{α}(r_{α} + x_{G}) × (ṙ_{α} + ẋ_{G})

= ∑_{α} m_{α}r_{α} × ṙ_{α} + x_{G} × ∑_{α} m_{α}ṙ_{α} + ∑_{α} m_{α}r_{α} × ṙ_{α}

+ (∑_{α} m_{α}r_{α}) × x_{G}

= ∑_{α} m_{α}r_{α} × ṙ_{α} + x_{G} × \frac{d}{dt} ∑_{α} m_{α}r_{α} + ∑_{α} m_{α}r_{α} × ṙ_{α} + 0 × x_{G}

= ∑_{α} m_{α}r_{α} × ṙ_{α} + x_{G} × \frac{d}{dt}0 + ∑_{α} m_{α}r_{α}

= Mx_{G} × ẋ_{G} + ∑_{α} m_{α}r_{α} × ṙ_{α}

The total angular momentum about the origin is the angular momentum of the system around the centre of mass plus the angular momentum of a point mass equal to the mass of the system at the centre of gravity around the origin.

Exercise 10.2.2

We have

Ḣ = \frac{d}{dt} ∑_{α} m_{α}x_{α} × ẋ_{α}

= ∑_{α} m_{α}(ẋ_{α} × ẋ_{α} + x_{α} × ẍ_{α})

= ∑_{α} m_{α}x_{α} × ẍ_{α}

= ∑_{α} x_{α} × m_{α}ẍ_{α}

= ∑_{α} x_{α} × F_{α}

= −k ∑_{α} m_{α}x_{α} × ẋ_{α}

= −kH

Thus (you can work with the summation convention, if you prefer)

\frac{d}{dt}e^{kt}H(t) = 0

so H(t)e^{kt} is a constant vector H_{0} and we are done.

Exercise 10.2.3

Since the first lump has centre of mass at 0 we have

∫ x_{i}ρ(x) dV (x) = 0

By a simple change of variable

J_{ij} = ∫ ((x_{k} + a_{k})(x_{k} + a_{k})δ_{ij} − (x_{i} + a_{i})(x_{j} + a_{j}))ρ(x) dV (x)

= ∫ (x_{k}x_{k}δ_{ij} − x_{i}x_{j})ρ(x) dV (x) + 2a_{k}δ_{ij} ∫ x_{k}ρ(x) dV (x) + (a_{k}a_{k}δ_{ij} − a_{i}a_{j}) ∫ ρ(x) dV (x) + a_{i} ∫ x_{j}ρ(x) dV (x)

+ a_{j} ∫ x_{i}ρ(x) dV (x)

= I_{ij} + M(a_{k}a_{k}δ_{ij} − a_{i}a_{j}).

Exercise 10.2.4

In the coordinate system chosen, taking x_{1} = x, x_{2} = y, x_{3} = z,

A = I_{11}

= ∫ (x_{k}x_{k}δ_{11} − x_{1}x_{1})ρ(x) dV (x) = ∫ (x_{2}x_{2} + x_{3}x_{3})ρ(x) dV (x) = ∫∫∫ (y^{2} + z^{2})ρ(x, y, z) dx dy dz ≥ 0

We also have

B = ∫∫∫ (z^{2} + x^{2})ρ(x, y, z) dx dy dz ≥ 0,
C = ∫∫∫ (x^{2} + y^{2})ρ(x, y, z) dx dy dz ≥ 0.

Exercise 10.3.2

(i) We have

l_{ir}l_{js}δ_{rs} = l_{ir}l_{jr} = δ_{ij}

since LL^{T} = I.

(ii) If a_{ijk} is an isotropic extended Cartesian tensor, then a_{ijk} is an isotropic Cartesian tensor so a_{ijk} = Aϵ_{ijk} so a_{ijk} = 0.

(iii) We have (ϵ_{rst}a_{r}b_{s}c_{t})′ = l_{ri}l_{sj}l_{tk}ϵ_{rst}a_{i}b_{j}c_{k} = (det L)ϵ_{ijk}a_{i}b_{j}c_{k} = −ϵ_{ijk}a_{i}b_{j}c_{k}. (iv) We have

(a_{r}b_{r})′ = a′_{r}b′_{r} = l_{ri}l_{rj}a_{i}b_{j} = δ_{ij}a_{i}b_{j}.

(v) If a × b was an extended Cartesian tensor then by (iv) so would be (a × b) · c. Now choose c non-zero and orthogonal to a and b.

Exercise 10.4.1

(i) We have (λu^{i} + μv^{i})¯ = (λū^{i} + μv̄^{i}) = (λl_{j}^{i}u^{j} + μl_{j}^{i}v^{j}) = l_{j}^{i}(λu^{j} + μv^{j}).

(ii) We have

(ū′)^{i} = (l_{j}^{i}u^{i})′ = l_{j}^{i}u̇^{j}.

Exercise 10.4.3

m_{j}^{i} = l_{j}^{i} ⇔ L = (L^{T})^{−1} ⇔ L^{T} = L^{−1} ⇔ L ∈ O(ℝ^{3}).

Exercise 10.4.6

(i) We have

l_{u}^{i}m_{j}^{v}δ_{i}^{j} = l_{u}^{i}m_{i}^{v} = δ_{u}^{v} = δ̄_{u}^{v}.

Choose an L such that L^{2} ≠ I, for example

L = (\begin{matrix}1 & 0 & 0 \\ 0 & \frac{4}{5} & −\frac{3}{5} \\ 0 & \frac{3}{5} & \frac{4}{5} \end{matrix} ) .

Then

l_{i}^{u}l_{j}^{v}δ_{uv} = l_{i}^{u}l_{u}^{v} ≠ δ_{ij} = δ̄_{ij}.

(ii) The rule is

ā_{ij}^{kn} = l_{i}^{u}l_{j}^{v}m_{p}^{k}m_{q}^{n}a_{uv}^{pq}.

(iii) We have l_{i}^{u}l_{j}^{v}m_{p}^{k}m_{q}^{n}a_{uv}^{pq} = l_{i}^{u}l_{j}^{v}m_{p}^{k}m_{q}^{n}δ_{i}^{k}δ_{j}^{n} = l_{i}^{u}l_{j}^{v}m_{p}^{i}m_{q}^{j} = l_{i}^{u}m_{p}^{i}l_{j}^{v}m_{q}^{j} = δ_{p}^{u}δ_{q}^{v} = ā_{pq}^{uv}, so a_{ij}^{kn} is a 4th rank tensor.

Next observe that

a_{ij}^{in} = δ_{i}^{i}δ_{j}^{n} = 3δ_{j}^{n}

a second rank tensor.

However

a_{ii}^{kn} = δ_{i}^{k}δ_{i}^{n} = δ_{kn}

is not a second rank tensor . The same argument shows that a_{ij}^{nn} is not.

(iv) The rule is

ā_{ijk}^{n} = l_{i}^{u}l_{j}^{v}l_{k}^{p}m_{q}^{n}a_{uvp}^{q}.

We observe that

ā_{ijn}^{n} = l_{i}^{u}l_{j}^{v}l_{n}^{p}m_{q}^{n}a_{uvp}^{q} = l_{i}^{u}l_{j}^{v}δ_{q}^{p}a_{uvp}^{q} = l_{i}^{u}l_{j}^{v}a_{uvp}^{p}

so a_{ijn}^{n} is a second rank tensor.

Exercise 11.1.2

Observe that

(α + β)(e_{j}) = α(e_{j}) + β(e_{j}) = ∑_{i=1}^{n} a_{ij}f_{i} + ∑_{i=1}^{n} b_{ij}f_{i} = ∑_{i=1}^{n}(a_{ij} + b_{ij})f_{i}

and

λα(e_{j}) = λ ∑_{i=1}^{n} a_{ij}f_{i} = ∑_{i=1}^{n} λa_{ij}f_{i}.

Exercise 11.1.3

If

α(f_{j}) = ∑_{i=1}^{p} a_{ij}g_{i}
β(e_{r}) = ∑_{s=1}^{n} b_{sr}f_{s},

then

αβ(e_{r}) = ∑_{s=1}^{n} b_{sr}αf_{s} = ∑_{s=1}^{n} b_{sr} ∑_{i=1}^{p} a_{is}g_{i} = ∑_{i=1}^{p} (∑_{s=1}^{n} a_{is}b_{sr}) g_{i},

so αβ has matrix AB with respect to the given bases.

Exercise 11.1.4

(i) Observe that, by definition,

(α + β)γu = α(γu) + β(γu) = (αγ)u + (βγ)u = (αβ + αγ)u for all u ∈ U so, by definition

(α + β)γ = αγ + βγ.

(ii) and (iii) follow at once from Exercises 11.1.2 and 11.1.3. (iv) Let U, V and W be (not necessarily finite dimensional) vector

spaces over 𝔽 and let α, β ∈ L(U, V), γ ∈ L(V, W). Then γ(α + β) = γα + γβ.

If A and B are n × m matrices over 𝔽 and C is an p × n matrix over 𝔽, we have

C(A + B) = CA + CB.

Exercise 11.1.5

Since the f_{i} are a basis for V, we can find p_{ij} ∈ 𝔽 such that

f′_{j} = ∑_{i=1}^{n} p_{ij}f_{i} [1 ≤ j ≤ n]

and, similarly, we can find m_{ij} and q_{rs} such that

f_{j} = ∑_{i=1}^{n} m_{ij}f_{i} [1 ≤ j ≤ n]

and

e′_{s} = ∑_{r=1}^{m} q_{rs}e_{r} [1 ≤ s ≤ m].

We have

f_{j} = ∑_{i=1}^{n} m_{ij}f′_{i} = ∑_{i=1}^{n} m_{ij} ∑_{k=1}^{n} p_{ki}f_{k}

so by uniqueness

δ_{kj} = ∑_{i=1}^{n} p_{ki}m_{ij}

that is to say P M = I so P is invertible with M = P^{−1}. Similarly

αe′_{s} = α ∑_{r=1}^{m} q_{rs}e_{r}

= ∑_{r=1}^{m} q_{rs}αe_{r}

= ∑_{r=1}^{m} q_{rs} ∑_{p=1}^{n} a_{pr}f′_{p}

= ∑_{r=1}^{m} q_{rs} ∑_{p=1}^{n} a_{pr} ∑_{i=1}^{n} m_{ip}f′_{i}

∑_{i=1}^{m} (∑_{p=1}^{n} ∑_{r=1}^{m} m_{ip}a_{pr}q_{rs}) f′_{p},

so

B = MAQ = P^{−1}AQ.

Exercise 11.1.9

(i) Since I ∈ G,

A ∈ X ⇒ A, A ∈ X, A = AI ⇒ A ∼_{1} A.

If A ∼_{1} B then A, B ∈ X and we can find a P ∈ G such that B = P A. But P^{−1} ∈ G and A = P^{−1}B so B ∼_{1} A.

If A ∼_{1} B, B ∼_{1} C, then A, B, C ∈ X and we can find a P, Q ∈ G such that B = P A, C = QB. Since QP ∈ G and C = (QP)A we have A ∼_{1} C. Thus ∼_{1} is an equivalence relation on X.

(ii) Since I ∈ G,

A ∈ X ⇒ A, A ∈ X, A = I^{−1}AI ⇒ A ∼_{2} A.

If A ∼_{2} B, then A, B ∈ X and we can find P, Q ∈ G such that B = P^{−1}AQ. But P^{−1}, Q ∈ G and A = (P^{−1})^{−1}BQ^{−1}, so B ∼_{2} A.

If A ∼_{2} B, B ∼_{2} C, then A, B, C ∈ X and we can find P, Q, R, S ∈ G such that B = P^{−1}AQ, C = R^{−1}BS. Since P R, SQ ∈ G and C = (P R)^{−1}A(SQ), we have A ∼_{2} C. Thus ∼_{2} is an equivalence relation on X.

(iii) If A ∼_{2} B, then A, B ∈ X and we can find P, Q ∈ GL(𝔽^{n}) with B = P^{−1}AQ and so rank B = rank A. If rank B = rank A = k, then we can find P, Q, R, S ∈ GL(𝔽^{n}) such that D = P^{−1}AQ and D = R^{−1}BS where D is the n × n diagonal matrix with first k diagonal entries 1 and remaining entries 0 so A ∼_{2} D, D ∼_{2} B and A ∼_{2} B. Thus there are n + 1 equivalence classes corresponding to the matrices of rank k [0 ≤ k ≤ n].

(iv) Since I ∈ G.

A ∈ X ⇒ A, A ∈ X, A = I^{−1}AI ⇒ A ∼_{3} A.

If A ∼_{3} B then A, B ∈ X and we can find P ∈ G such that B = P^{−1}AP. But P^{−1} ∈ G and A = (P^{−1})^{−1}BP^{−1} so B ∼_{3} A.

If A ∼_{3} B, B ∼_{3} C, then A, B, C ∈ X and we can find P, Q ∈ G such that B = P^{−1}AP, C = Q^{−1}BQ. Since P Q ∈ G and C = (P Q)^{−1}A(P Q) we have A ∼_{3} C. Thus ∼_{3} is an equivalence relation on X.

(v) Since A and P^{−1}AP have the same characteristic polynomial

A ∼_{3} B ⇒ χ_{A} = χ_{B}

Since whenever A is symmetric there exists an orthogonal matrix P such that P^{−1}AP = P^{T} AP = D where D is a diagonal matrix with the same characteristic polynomial

χ_{A} = χ_{B} ⇒ A ∼_{3} B.

Thus

A ∼_{3} B ⇔ χ_{A} = χ_{B}.

If D_{λ} is the diagonal matrix with first diagonal entry λ and all others 0 then

D_{λ} ∼_{3} D_{μ} ⇔ χ_{D_{λ}} = χ_{D_{μ}} ⇔ λ = μ.

Thus there are infinitely many equivalence classes for ∼_{3}.

(vi) Since I ∈ G

A ∈ X ⇒ A, A ∈ X, A = I^{T} AI ⇒ A ∼_{3} A.

If A ∼_{4} B, then A, B ∈ X and we can find P ∈ G such that B = P^{T} AP. But P^{−1} ∈ G and A = (P^{−1})^{T} BP^{−1}, so B ∼_{4} A.

If A ∼_{4} B, B ∼_{4} C, then A, B, C ∈ X and we can find P, Q ∈ G such that B = P^{T} AP, C = Q^{T} BQ. Since P Q ∈ G and C = (P Q)^{T} A(P Q) we have A ∼_{4} C. Thus ∼_{4} is an equivalence relation on X.

(vii) If A is a real symmetric n × n matrix we can find P ∈ O(ℝ^{n}) such that D = P^{T} AP with D = (d_{ij}) a diagonal matrix with d_{ii} > 0 for 1 ≤ i < u, d_{ii} < 0 for u ≤ i < v and d_{ii} = 0 otherwise. Let E = (e_{ij}) a diagonal matrix with e_{ii} = d_{ii}^{−1/2} for 1 ≤ i < u, e_{ii} = (−d_{ii})^{−1/2} for u ≤ i < v and e_{ii} = 1 otherwise. Then P E ∈ GL(ℝ^{n}) and (P E)^{T} A(P E) is a diagonal matrix with entries 1, −1 and 0. Thus there can only be finitely many equivalence classes.

[The reason that this does not complete the discussion is that we have not shown (as we will in Section 16.2) that different (u, v) correspond to different equivalence classes.]

Exercise 11.1.15

Let U and V be finite dimensional vector spaces. and let α : U → V be linear.

By Theorem 11.1.13,

dim im α = dim(U/ ker α)

and, by Lemma 11.1.14,

dim U = dim ker α + dim(U/ ker α),

so dim U = dim ker α + dim(im α).

Exercise 11.1.16

Observe that dim H_{j} = dim B_{j} − dim Z_{j} = rank(α_{j+1}) − (dim C_{j} − rank(α_{j}))

and so

dim H_{j} + dim C_{j} = rank(α_{j+1}) + rank(α_{j})

for 1 ≤ j ≤ n Now rank(α_{n+1}) = rank(α_{0}) = 0 since C_{n+1} = C_{0} = {0} so

∑_{j=1}^{n}(−1)^{j}(dim H_{j} + dim C_{j}) = 0.

Exercise 11.2.1

The sum of two polynomials is polynomial and the product of a polynomial with a real number is a polynomial, so we have a subspace of the vector space of functions f : ℝ → ℝ with pointwise addition and scalar multiplication.

(i) Let λ, μ ∈ ℝ, P, Q ∈ P.

We have

D(λP + μQ) = λDP + μDQ,

by the standard rules for differentiation.

We have

M(λP +μQ)(t) = t(λP (t)+μQ(t)) = λtP (t)+μtQ(t) = (λMP +μMQ)(t),

so

M(λP + μQ) = λMP + μMQ.

We have

E_{h}(λP + μQ)(t) = (λP (t + h) + μQ(t + h)) = λE_{h}P (t) + μE_{h}Q(t)
= (λE_{h}P + μE_{h}Q)(t),

so

M(λP + μQ) = λMP + μMQ.

(ii) We have

((DM − MD))P (t) = \frac{d}{dt}tP (t) − tP′(t) = P (t),

so (DM − MD)P = P for all P ∈ P and DM − MD = ι the identity map.

(iii) Well defined since for each P we only consider a finite sum and if α_{j}P = 0 for all j ≥ N

∑_{j=0}^{N} α_{j}P = ∑_{j=0}^{M} α_{j}P

for all M ≥ N.

If λ, μ ∈ ℝ, P, Q ∈ P. we can find an N such that α_{j}P = α_{j}Q = 0 for all j ≥ N, so

∑_{j=0}^{∞} α_{j}(λP + μQ) = ∑_{j=0}^{N} α_{j}(λP + μQ) = ∑_{j=0}^{N}(λα_{j}P + μα_{j}Q)

= λ ∑_{j=0}^{N} α_{j}P + μ ∑_{j=0}^{N} α_{j}Q = λ ∑_{j=0}^{∞} α_{j}P + μ ∑_{j=0}^{∞} α_{j}Q.

Thus ∑_{j=0}^{∞} α_{j} ∈ L(P, P).

(iv) If P is a polynomial of degree N, then D^{j}P = 0 for j ≥ N + 1.

If e_{n} = t^{n}, then M^{j}e_{0} = e_{j} ≠ 0, so M does not have the required property.

Since E_{h}^{j}e_{0} = e_{j} ≠ 0, E_{h} does not have the required property.

(ι − E_{h})e_{k} is a polynomial of degree k − 1 if k ≥ 1. and is zero if k = 0. Thus, if P is a polynomial of degree n, (ι−E_{h})e_{k} is a polynomial of degree at most n − 1 if n ≥ 1. and is zero if n = 0. Thus if P is a polynomial of degree N (ι−E_{h})^{N+1}P = 0 and we have desired property.

(v) Part (iii) tells us that exp(α) and log(ι − α) are well defined members of L(P, P).

Observe that

exp α − ι = α ∑_{j=1}^{∞} α^{j−1}/(j − 1) = αβ

where β is an endomorphism which commutes with α. Thus

(exp α − ι)^{N} = α^{N}β^{N} = 0

if α^{N} = 0. It follows, by part (iii), that log(exp α) is well defined.

By the standard rules on Taylor expansions, we have

log(exp t) = ∑_{j=0}^{∞} c_{j}t^{j}

for |t| small and (using those standard rules to compute c_{j})

log(exp α) = ∑_{j=0}^{∞} c_{j}α^{j}.

However we know from calculus that

log(exp t) = t

for |t| small and so (by the uniqueness of Taylor expansions) c_{1} = 1 and c_{j} = 0 otherwise. Thus

log(exp α) = α.

(vi) We know that polynomials are their own Taylor series so

((exp hD)P ) = ∑_{j=0}^{∞} \frac{h^{j}}{j!} P^{(j)}(t) = P (t + h) = (E_{h}P)(t)

and thus

exp hD = E_{h}.

We know, from part (iv), that, for each P ∈ P, we can find an N(P) such that △_{h}^{j}(P) = 0 for all j > N(P). Thus (v) tells us that

log E_{h} = hD,

that is to say

hD = ∑_{j=1}^{∞} \frac{(−1)^{j+1}}{j} △_{h}^{j}.

(vii If P has degree N,

(ι−λD) ∑_{j=0}^{∞} λ^{j}D^{r}P = (ι−λD) ∑_{j=N+1}^{∞} λ^{j}D^{r}P = (ι−λ^{N+2}D^{N+2})P = P = ιP

Thus

(ι − λD) ∑_{j=0}^{∞} λ^{j}D^{j} = ι

and similarly

(∑_{j=0}^{∞} λ^{j}D^{j}) (ι − λD) = ι

It follows that

(ι − λD)P = Q ⇒ P = (∑_{j=0}^{∞} λ^{j}D^{j}) (ι − λD)P = (∑_{j=0}^{∞} λ^{j}D^{j}) Q.

We also have

(ι − λD) (∑_{j=0}^{∞} λ^{j}D^{j}) P = P.

Thus the unique polynomial solution of

(ι − λD)P = Q

is

Q = ∑_{j=0}^{∞} λ^{j}D^{j}P = ∑_{j=0}^{∞} λ^{j}P^{(j)}.

In particular, the unique polynomial solution of

f′(x) − f(x) = x^{2}

ie of

(ι − D)f = −e_{2}

is

∑_{j=0}^{∞} \frac{d^{j}}{dx^{j}} (−x^{2}) = −x^{2} − 2x − 2.

There is only one polynomial solution. (The general, not necessarily polynomial, solution is Ae^{x} − x^{2} − 2x − 2 with A chosen freely.)

Exercise 11.2.3

If rank α^{n} ≠ 0, then, since rank α^{m} = 0 for some m,

rank α^{n} − rank α^{n+1} ≥ 1

Thus if n ≥ r ≥ 0

rank α^{r} − rank α^{r+1} ≥ rank α^{n} − rank α^{n+1} ≥ 1

so

rank α^{n} ≤ n − ∑_{r=0}^{n−1} rank α^{r} − rank α^{r+1} ≤ 0

which contradicts our original hypothesis. By reductio ad absurdum, rank α^{n} = 0. Now suppose that α has rank r and α^{m} = 0, We have

n − r = n − rank α ≥ rank α^{j} − rank α^{j+1}

so

m(n − r) ≥ ∑_{j=0}^{m} rank α^{j} − rank α^{j+1} = n − rank α^{m} = n

so mn − n ≥ mr and r ≤ n(1 − m^{−1}).

Exercise 11.2.4

Observe that if A_{k} = (a_{ij}(k)) is the k × k matrix with a_{i,i+1}(k) = 1 for 1 ≤ i ≤ k − 1 a_{ij}(k) = 0 otherwise, then the associated linear map α_{k} : 𝔽^{k} → 𝔽^{k} has rank α^{u}(k) = max(k, u − 1). Consider the n × n matrix A which has (s_{j} − s_{j+1}) − (s_{j+1} − s_{j+2}) copies of A_{j} and one s_{m} ×s_{m} identity matrix along the diagonal and all other entries zero. If α has matrix A with respect to some basis, then α will have the desired property.

Exercise 11.2.5

If we write r_{k} = rank α^{k}, we know from general theorems that

r_{0} − r_{1} ≥ r_{1} − r_{2} ≥ r_{2} − r_{3} ≥ … ≥ r_{n−1} − r_{n} ≥ 0.

We have r_{0} − r_{1} = 2, so 2 ≥ r_{j} − r_{j+1} ≥ 0. Since

2n = r_{0} − r_{n} = ∑_{j=0}^{n−1} r_{j} − r_{j+1},

it follows that 2 = r_{j} − r_{j+1} for 0 ≤ j ≤ n − 1 and rank α^{j} = 2n − 2j for 0 ≤ j ≤ n.

Exercise 11.3.1

(i) We have δ_{a}(λ + μg) = (λf + μg)(a) = λf(a) + μg(a) = λδ_{a}f + μδ_{a}g.

(ii) We have δ′_{a}(λ + μg) = −(λf + μg)′(a) = −λf′(a) − μg′(a) = λδ′_{a}f + μδ′_{a}g.

(iii) We have

J(λf + μg) = ∫_{0}^{1}(λf(x) + μg(x)) dx
= λ ∫_{0}^{1} f(x) dx + μ ∫_{0}^{1} g(x) dx
= λJf + μJg.

Exercise 11.3.6

Φ(α) = 0 ⇒ α′ = 0 by definition. α′ = 0 ⇒ α′(v′) = 0 ∀v′ ∈ V′ by definition. α′(v′) = 0 ∀v′ ∈ V′ ⇒ α′(v′)u = 0 ∀v′ ∈ V′ ∀u ∈ U by definition. α′(v′)u = 0 ∀v′ ∈ V′ ∀u ∈ U ⇒ v′αu = 0 ∀v′ ∈ V′ ∀u ∈ U since,

by definition, α′(v′)u = v′α(u). v′αu = 0 ∀v′ ∈ V′ ∀u ∈ U ⇒ αu = 0 ∀u ∈ U since V′ separates V. αu = 0 ∀u ∈ U ⇒ α = 0 by definition.

Exercise 11.3.7

(i) We have

((βα)′w′)u = w′((βα)u)
= w′(β(αu))
= (β′w′)(αu)
= (α′(β′w′))u
= ((α′β′)w′)u

for all u ∈ U, so

(βα)′w′ = (α′β′)w′

for all w′ ∈ W′, so

(βα)′ = α′β′.

(ii) We have

(ι′_{U}u′)v = u′ι_{u}v = u′v

for all v ∈ U ,so

ι′_{U}u′ = u′

for all u′ ∈ U′, so

ι′_{U} = ι_{U′}.

(iii) If α is invertible, then αα^{−1} = ι so, by the previous parts, (α^{−1})′α′ = ι so α′ is invertible and (α′)^{−1} = (α^{−1})′.

Exercise 11.3.8

We have

α′′(Θu)v′ = Θu(α′v′)

= (α′v′)u = v′αu = (Θ(αu))v′

for all v′ ∈ V. Since V′′ is separated by V′,

α′′(Θu) = Θ(αu)

for all u ∈ U.

Exercise 11.3.9

(i) Certainly the zero sequence 0 ∈ c_{00}.

If a, b ∈ c_{00} we can find n and m such that a_{r} = 0 for r ≥ n and b_{r} = 0 for r ≥ m. Thus if λ, μ ∈ ℝ we have λa_{r} + μb_{r} = 0 for all r ≥ max{n, m} so λa + μb ∈ c_{00}. Thus c_{00} is a subspace of s.

(ii) If x ∈ c_{00}, then we can find an n such that x_{r} = 0 for r ≥ n. We have

x = ∑_{r=1}^{n−1} x_{r}e_{r}

and

T x = ∑_{r=1}^{n−1} x_{r}T e_{r} = ∑_{r=1}^{n−1} a_{r}x_{r} = ∑_{r=1}^{∞} a_{r}x_{r}.

(iii) If x ∈ c_{00}, then we can find an n such that x_{r} = 0 for r ≥ n. Thus

T_{a}x = ∑_{r=1}^{n−1} a_{r}x_{r} ∑_{r=1}^{∞} a_{r}x_{r}

is well defined.

If x, y ∈ c_{00} and λ, μ ∈ ℝ we can find an n such that x_{r} = y_{r} = 0 for r ≥ n. Thus

T a(λx+μ↶) = ∑_{r=1}^{n−1} a_{r}(λx_{r}+μy_{r}) = λ ∑_{r=1}^{n−1} a_{r}x_{r}+μ ∑_{r=1}^{n−1} a_{r}y_{r} = λT_{a}x+μT_{a}y.

We have shown that T_{a} ∈ c_{00}.

If x ∈ c_{00} and x ≠ 0 then T_{x}x ≠ 0. Thus c′_{00} separates c_{00}.

(iv) Let θ : s → c′_{00} be given by θa = T_{a}. By (iii), θ is well defined and, by (ii), θ is surjective.

If x ∈ c_{00}, then we can find an n such that x_{r} = 0 for all r ≥ n.

T_{λa+μb}x = ∑_{r=1}^{n} x_{r}(λa_{r} + μb_{r}) = λ ∑_{r=1}^{n} x_{r}a_{r} + μ ∑_{r=1}^{n} x_{r}b_{r}
= λT_{a}x + μT_{b}x = (λT_{a} + μT_{b})x

Thus T_{λa+μb} = λT_{a} + μT_{b} and θ is linear. Further,

T_{a} = 0 ⇔ T_{a}e_{j} = 0 ∀j ⇔ a_{j} = 0 ∀j ⇔ a = 0,

so θ is injective.

Thus c′_{00} is isomorphic to s.

Exercise 11.3.10

(i) If x ∈ c_{00}, then we can find an n such that x_{r} = 0 for r ≥ n. We have

x = ∑_{r=1}^{n−1} x_{r}e_{r}

(ii) Let If w_{j} ∈ ℝ^{n+1} be the row vector whose kth entry is the m + k th entry of f_{j}. Since n vectors cannot span a space of dimension n + 1 there exists a vector u ∈ ℝ^{n+1} with

u ∉ span{w_{1}, w_{2}, …, w_{n+1}}

Set b_{m+r} = u_{r}.

(iii) Let m(1) = 0 and let m(n) = ∑_{r=1}^{n−1}(r + 1) for n ≥ 2. For each n ≥ 1 choose a_{r} with m(n) + 1 ≤ r ≤ m(n + 1) so that

∑_{j=1}^{n} λ_{j}f_{j} = a

has no solution with λ_{j} ∈ ℝ [1 ≤ j ≤ n].

By construction, a does not lie in the span of the f_{j}. Thus s cannot be spanned by a countable set and is not isomorphic to c_{00} which can. Thus c_{00} which is isomorphic to s is not isomorphic to c′_{00}.

Exercise 11.4.2

We use the fact that e_{j}(x_{k}) = δ_{jk} repeatedly. To see that the e_{j} are linearly independent, observe that

∑_{j=0}^{n} λ_{j}e_{j} = 0 ⇒ ∑_{j=0}^{n} λ_{j}e_{j}(x_{k}) = 0 ∀k ⇒ λ_{k} = 0 ∀k.

To see that the e_{j} span, observe that, if P ∈ P_{n}, then

P − ∑_{j=0}^{n} P (x_{j})e_{j}

is a polynomial of degree at most n which vanishes at the n + 1 points x_{k} and is this identically zero. Thus

P = ∑_{j=0}^{n} P (x_{j})e_{j}.

We have e_{j}(x_{k}) = δ_{jk} = ê_{k}e_{j} so

ê_{k} (∑_{j=0}^{n} λ_{j}e_{j}) = ∑_{j=0}^{n} λ_{j}ê_{k}e_{j} = ∑_{j=0}^{n} λ_{j}e_{j}(x_{k})

and ê_{k}P = P (x_{k}).

Exercise 11.4.4

Observe that

e_{j}ê_{k} = ê_{k}e_{j} = δ_{jk}

for all k so ˆ ê_{j} = e_{j}.

Exercise 11.4.5

We have

f_{1} = ae_{1} + be_{2},
f_{2} = ce_{1} + de_{2}

and, since the ê_{j} form a basis,

f̂_{1} = Aê_{1} + Bê_{2},
f̂_{2} = Bê_{1} + Dê_{2}

Now

1 = f̂_{1}f_{1} = (Aê_{1} + Bê_{2})(ae_{1} + be_{2}) = Aa + Bb

and

0 = f̂_{1}f_{2} = (Aê_{1} + Bê_{2})(ce_{1} + de_{2}) = Ac + Bd.

Similarly Ca + Db = 0, Cc + Dd = 0. In other words

(\begin{matrix}A & B \\ C & D \end{matrix}) (\begin{matrix}a & c \\ b & d \end{matrix}) = I

so

Q = (\begin{matrix}A & B \\ C & D \end{matrix}) = (\begin{matrix}a & c \\ b & d \end{matrix})^{−1} = (ad − bc)^{−1} (\begin{matrix}a & −c \\ −b & d \end{matrix}) .

and

f̂_{1} = (ad − bc)^{−1}aê_{1} − b(ad − bc)^{−1}ê_{2},
f̂_{2} = −c(ad − bc)^{−1}aê_{1} + (ad − bc)^{−1}dê_{2}.

Exercise 11.4.6

Using the summation convention,

δ_{js} = f̂_{s}f_{j} = k_{rs}l_{ij}ê_{r}e_{i} = k_{rs}l_{ij}δ_{ir} = k_{rs}l_{rj}

Thus I = KL^{T} and K = (L^{T})^{−1}.

Exercise 11.4.9

Let A and B be n×n matrices and α and β the corresponding linear maps for some fixed basis.

We know that (α+β)′ = α′ +β′, (λα)′ = λα′, α′′ = α (since we work in a finite dimensional space) ι′ = ι, (αβ)′ = β′α′ so

(A+B)^{T} = A^{T} +B^{T} , (λA)^{T} = λA^{T} , A^{T T} = A, I^{T} = I, (AB)^{T} = B^{T} A^{T} .

If A is invertible, then α is, so α′ is with (α′)^{−1} = (α^{−1})′, so A^{T} is invertible with (A^{T})^{−1} = (A^{−1})^{T}.

Exercise 11.4.10

Let α have matrix A with respect to some basis. Then α′ has matrix A^{T} with respect to the dual basis so (by the matrix result)

det α′ = det A^{T} = det A = det α.

It follows that

det(tι − α) = det(tι − α)′ = det(tι − α)

and α and α′ have the same characteristic polynomials and so the same eigenvalues.

The trace is minus the coefficient of t^{n−1} in the characteristic polynomial (supposed of degree n). Thus Tr α′ = Tr α.

Exercise 11.4.12

We certainly have the zero map 0 ∈ W^{0}. If λ, μ ∈ 𝔽

u′, v′ ∈ W^{0} ⇒ (λu′ + μv′)w = λu′w + μv′w = λ0 + μ0 = 0 ∀w ∈ W
⇒ λu′ + μv′ ∈ W^{0}.

Thus W^{0} is a subspace of U′.

Exercise 11.4.15

∑_{j=1} x_{j}e_{j} ∈ W^{00} ⇔ (∑_{j=1}^{n} x_{j}e_{j}) ∑_{r=k+1}^{n} y_{r}e_{r} = 0 ∀y_{r} ∈ 𝔽

⇔ (∑_{j=1}^{n} x_{j}e_{j}) e_{r} = 0 ∀k + 1 ≤ r ≤ n

⇔ x_{r} = 0 ∀k + 1 ≤ r ≤ n

⇔ ∑_{j=1} x_{j}e_{j} ∈ W

Exercise 11.4.19

dim{u ∈ U : αu = λu} = dim(λι − α)^{−1}0

= dim im(λι − α) = dim im(λι − α)′ = dim im(λι − α′) = dim{u′ ∈ U′ : α′u = λu′.}

The eigenvalues and the dimensions of the spaces spanned by the corresponding eigenvectors are the same for α and α′.

Exercise 11.4.20

v′ ∈ (αU)^{0} ⇒ v′(αu) = 0 ∀u ∈ U

⇒ (α′v′)u = v′(αu) = 0 ∀u ∈ U

⇒ α′v′ ∈ U^{0}

so α′(αU^{0}) is a subspace of U. (Remark that (αU)^{0} is a subspace of V′ so α′(αU)^{0} is a subspace of U′.)

Let V have basis e_{1}, e_{2} and V′ dual basis ê_{1} ê_{2}. Let α be the endomorphism given by α(e_{1}) = e_{1}, α(e_{2}) = 0. We observe that α′(ê_{1}) = ê_{1}, α′ˆ(e_{2}) = 0.

If U = span{e_{1}}, then αU = U, (αU)^{0} = span{ê_{2}} and

α′(αU)^{0} = {0} ≠ U^{0}.

If If W = span{e_{2}}, then αW = {0}, (αW)^{0} = V and

α′(αU)^{0} = span{ê_{1}} = U^{0}

Exercise 12.1.2

If U = U_{1} ⊕ U_{2} ⊕ … ⊕ U_{m} and u ∈ U then, by Definition 12.1.1 (i), we can find u_{j} ∈ U_{j} such that

u = u_{1} + u_{2} + … + u_{m}.

If v_{j} ∈ U_{j} and

u = v_{1} + v_{2} + … + v_{m},

then u_{j} − v_{j} ∈ U_{j} and

0 = u − v = (u_{1} − v_{1}) + (u_{2} − v_{2}) + … + (u_{m} − v_{m})

so, by Definition 12.1.1 (ii), u_{j} − v_{j} = 0 and u_{j} = v_{j} ∀j.

If the equation

u = u_{1} + u_{2} + … + u_{m}

has exactly one solution with u_{j} ∈ U, then the conditions of Definition 12.1.1 can be read off.

Exercise 12.1.3

(i) If u ∈ U then we can find u_{j} ∈ U_{j} such that

u = u_{1} + u_{2} + … + u_{m}.

For each j we can find λ_{jk} ∈ 𝔽 such that

u_{j} = ∑_{k=1}^{n(j)} λ_{jk}e_{jk}

and so

u = ∑_{j=1}^{m} ∑_{k=1}^{n(j)} λ_{jk}e_{jk}.

Thus the e_{jk} span U.

If

0 = ∑_{j=1}^{m} ∑_{k=1}^{n(j)} λ_{jk}e_{jk}

then setting u_{j} = ∑_{k=1}^{n(j)} λ_{jk}e_{jk} we have u_{j} ∈ U_{j} and, applying the definition of direct sum, we obtain

0 = u_{j} = ∑_{k=1}^{n(j)} λ_{jk}e_{jk},

so λ_{jk} = 0 for all k and j. Thus the e_{jk} are linearly independent and so form a basis.

(ii) and (iii) Since every subspace of a finite dimension subspace is finite dimensional

U finite dimensional ⇒ U_{j} finite dimensional

If the U_{j} are finite dimensional the we can choose bases as in (i) and observe that U is finite dimensional with

dim U = ∑_{j=1}^{m} n(j) = ∑_{j=1}^{m} dim U_{j}.

Exercise 12.1.4

Observe that B = U ∩V is a subspace of a finite dimensional space so has a basis e_{1}, e_{2}, …, e_{k}. Since U is a subspace of a finite dimensional space containing B we can extend the basis of B to a basis of U

e_{1}, e_{2}, …, e_{k}, e_{k+1}, e_{k+2}, …, e_{k+l}.

Similarly we can extend the basis of B to a basis of v

e_{1}, e_{2}, …, e_{k}, e_{k+l+1}, e_{k+l+2}, …, e_{k+l+r}.

Let

C = span{e_{k+1}, e_{k+2}, …, e_{k+l}, e_{k+l+1}, e_{k+l+2}, …, e_{k+l+r}}.

We show that the e_{j} are linearly independent as follows. If

∑_{j=1}^{k+l+r} λ_{j}e_{j} = 0

then

∑_{j=1}^{k+l} λ_{j}e_{j} = − ∑_{j=k+l+1}^{k+l+r} λ_{j}e_{j} ∈ U ∩ V = B

Thus

∑_{j=1}^{k+l} λ_{j}e_{j} = ∑_{j=1}^{k} μ_{j}e_{j}

for some μ_{j}. Thus λ_{j} = 0 for 1 ≤ j ≤ k. Thus

∑_{j=1}^{k} λ_{j}e_{j} + ∑_{j=k+l+1}^{k+l+r} λ_{j}e_{j} = 0

so λ_{j} = 0 for k + l + 1 ≤ j ≤ k + l + r and for 1 ≤ j ≤ k.

We can now read off the desired results,

U = A ⊕ B, W = B ⊕ C and U + W = U ⊕ C = W ⊕ A.

Observe that if

U = A′ ⊕ B′, W = B′ ⊕ C′, U + W = U ⊕ C′ = W ⊕ A′

we have B′ ⊆ U, V so

B′ ⊆ U ∩ V.

If v ∈ U ∩ W then we can find b_{1}, b_{2} ∈ B′ and a ∈ A, c ∈ C such that

v = a + b_{1} = c + b_{2}

so

a − c = b_{1} − b_{2} ∈ B′.

Thus

a = c + (b_{1} − b_{2})

with a ∈ A′ ⊆ U and b_{1} − b_{2} ∈ U ∩ W ⊆ U. Since U + W = U ⊕ C′ we have c = 0. so

v = c + b_{2} = b_{2} ∈ B′.

Thus

B′ = U ∩ W

and we have shown B unique.

Not unique. Take V = ℝ^{3} (using row vectors)

U = {(x, y, 0) : x, y ∈ ℝ}, V = {(x, 0, z) : x, z ∈ ℝ}

B = {(x, 0, 0) : x ∈ ℝ},
A = {(0, y, 0) : y ∈ ℝ}, A′ = {(y, y, 0) : y ∈ ℝ}
C = {(0, 0, z) : y ∈ ℝ}, C′ = {(z, 0, z) : z ∈ ℝ}.

Then

U = A ⊕ B, W = B ⊕ C and U + W = U ⊕ C = W ⊕ A and

U = A′ ⊕ B, W = B′ ⊕ C and U + W = U ⊕ C′ = W ⊕ A′.

Exercise 12.1.6

(i) Can occur. Take V = V_{1} = V_{2} = 𝔽^{2}. (ii) Cannot occur. If e_{1}, e_{2}, … e_{k} is a basis for V_{1} and e_{k+1}, e_{k+2},

… e_{k+m} is a basis for V_{2} then e_{1}, e_{2}, … e_{k+m} span V_{1} + V_{2} so

dim(V_{1} + V_{2}) ≤ k + m = dim V_{1} + dim V_{2}

(ii) Can occur. Let e_{1}, e_{2} be a basis for V = 𝔽^{2} and V_{1} = V_{2} = span{e_{1}}.

Exercise 12.1.7

We know from Theorem 11.2.2 that there exists an m such that α^{m+1}U = α^{m}U. We thus have α^{m}U an invariant subspace.

If W is an invariant subspace

W = α^{m}V ⊆ α^{m}U.

Thus α^{m}U is the unique maximal invariant subspace.

(ii) If x ∈ M ∩ N then

α^{m}x = 0

But α|_{M} : M → M is surjective so invertible so α|_{M}^{m} is so x = 0. By the rank-nullity theorem dim M + dim N = dim U so U + M ⊕ V.

(iii) α(M) = M ⊆ M, α^{m}α(N) = αα^{m}N = 0 so αN ⊆ N.

(iv) As we observed in (ii), β : M → M is surjective, so β is invertible, so an isomorphism. We know that γ^{m} = 0.

(v) Observe that M′ is an invariant subspace so M′ ⊆ M. If b ∈ N′ then

α^{r}(b) = (γ′)^{r}b = 0

for r large so b ∈ N. Thus N′ ⊆ N. We have dim M′ ≤ dim M, dim N′ ≤ dim N

dim U = dim M′ + dim N′ ≤ dim M + dim N = dim U

so dim M′ = dim M, dim N′ = dim N and M = M′, N = N′. By considering b = 0, we have β = β′ and similarly, by considering a = 0, we have γ = γ′.

(vi) Let A correspond to a linear map α for some choice of basis for a space U of appropriate dimension. Choose a basis M for M defined as above and a basis N for N. Then E, N is a basis for U with respect to which α has the form

(\begin{matrix}B & 0 \\ 0 & C \end{matrix})

with B an invertible r × r matrix and C a nilpotent n − r × n − r matrix, The result now follows from the change of basis formula.

Exercise 12.1.11

Let dim U = n and dim V = r. If U is neither V nor {0}, then 0 < r < n. Choose a basis

e_{1}, e_{2}, …, e_{r}

and extend it to a basis

e_{1}, e_{2}, …, e_{n}

of V.

If

W = span{e_{r+1}, e_{2}, …, e_{n}}
W′ = span{e_{1} + e_{r+1}, e_{2}, …, e_{n}},

then V = U ⊕ W = U ⊕ W′ but e_{r+1} ∈ W \ W′ so W ≠ W.

Exercise 12.1.13

(i) It will remain true θ has an n^{2} × n^{2} diagonal matrix with dim U diagonal entries taking the value 1 and dim V diagonal entries taking the value −1, since θ|_{U} = ι|_{U} and since θ|_{V} = −ι|_{V}.

(ii) det θ = (−1)^{dim V} = (−1)^{n(n−1)/2}. Now

\frac{(n + 4)((n + 4) − 1)}{2}−\frac{n(n − 1)}{2} = \frac{n^{2} + 8n + 16 − n − 4 − n^{2} + n}{2} = 4n+6

is divisible by 2 so det θ depends only on the value of n modulo 4. By inspection

n ≡ 0 ⇒ det θ = 1,
n ≡ 1 ⇒ det θ = 1,
n ≡ 2 ⇒ det θ = −1,
n ≡ 3 ⇒ det θ = −1.

Exercise 12.1.14

Automatically 0 ∈ V and, if λ, μ ∈ ℝ,

f, g ∈ V ⇒ (λf + μg)(−x) = λf(−x) + μg(−x)
= −λf(x) − μg(x) = −(λf + μg)(x) ∀x
⇒ λf + μg ∈ V,

so V is subspace. A similar argument shows that U is a subspace,

If f ∈ C(ℝ), then u(x) = (f(x)+f(−x))/2, v(x) = (f(x)−f(−x))/2 define u ∈ U, v ∈ V. Since f = u + v we have U + V = C(ℝ).

Now if f ∈ U ∩ V, then f(x) = f(−x) = −f(x) so f(x) = 0 for all x ∈ ℝ and f = 0. Thus U ∩ V = {0} and U and V are complementary subspaces.

Exercise 12.1.15

(i)⇒(ii) Let U = αV and W = (ι − α)V. If u ∈ U, then u = αv for some v ∈ V so

αu = α^{2}v = αv = u

so α|_{U} = ι|_{U}.

If u ∈ W, then u = (ι − α)v for some v ∈ V, so

αu = (α − α^{2})v = αv − αv = 0,

so α|_{W} = 0|_{W} Since v = αv + (ι − α)v for all v ∈ V so V = U + W. If u ∈ U ∩ W, then

u = α|_{U}u = αu = α|_{W} u = 0.

Thus U ∩ W = {0} and V = U ⊕ W.

(ii)⇒(iii) Let e_{1}, e_{2}, …, e_{r} be a basis of U and e_{r+1}, …, e_{n} be a basis of W. Then with respect to the basis e_{1}, e_{2}, …, e_{n} of V, α has a matrix of stated type.

(iii)⇒(i) Since A^{2} = A we have α^{2} = α. We have α_{1}(x, y) = (x, 0) = α_{1}^{2}(x, y) so α_{1} is a projection. We have α_{2}(x, y) = (0, y) ≠ (0, 0) = α_{2}^{2}(x, y) if y = 1, so α_{2} is not a

projection.

We have α_{3}(x, y) = (x, y) ≠ (y, x) = α_{3}^{2}(x, y) if x = 0, y = 1, so α_{3} is not a projection.

We have α_{4}(x, y) = (x + y, 0) = (x + y, 0) = α_{4}^{2}(x, y), so α_{4} is a projection.

We have α_{5}(x, y) = (x + y, x + y) ≠ (2(x + y), 2(x + y)) = α_{5}^{2}(x, y) if x = 0, y = 1, so α_{5} is not a projection.

We have α_{6}(x, y) = (\frac{1}{2}(x + y), \frac{1}{2}(x + y)) = α_{6}^{2}(x, y) so α_{6} is a projection.

Exercise 12.1.16

If α is a projection, then

(ι − α)^{2} = ι − 2α + α^{2} = ι − 2α + α = ι − α

so ι − α is a projection.

If ι − α is a projection, then the first paragraph tells us that α = ι − (ι − α) is a projection.

Exercise 12.1.17

If α and β are projections and αβ = βα, then

(αβ)^{2} = (αβ)(αβ) = α(βα)β = α(αβ)β = α^{2}β^{2} = αβ. We work in ℝ^{2} with row vectors. If

β(x, y) = (x + y, 0), α(x, y) = (0, y)

then α^{2} = α, β^{2} = α

βα(x, y) = (y, 0), αβ(x, y) = (0, y)

so (αβ)^{2} = (αβ) but (βα)^{2} ≠ (βα).

If

β(x, y) = (x + y, 0), α(x, y) = (0, x + y)

then α^{2} = α, β^{2} = α

βα(x, y) = (x + y, 0), αβ(x, y) = (0, x + y)

so (αβ)^{2} = (αβ) and (βα)^{2} = (βα), but (looking e.g. at (x, y) = (1, 1)) αβ ≠ αβ.

Exercise 12.1.18

(i) If αβ = −βα, then

αβ = ααβ = α(−βα) = (−αβ)α = βαα = βα

Thus αβ = 0 and βα = 0.

(ii) If (α + β) is a projection, then

α + αβ + βα + β = (α + β)^{2} = α + β,

so αβ = −βα and αβ = βα = 0.

If αβ = βα = 0, then

(α + β)^{2} = α + αβ + βα + β = α + β,

so α + β is a projection.

(ii) If (α − β) is a projection, then

α − αβ − βα + β = (α − β)^{2} = α − β,

so −αβ − βα = 2β whence (ι − α)β = −β(ι − α) so, since ι − α is a projection, (ι − α)β = β(ι − α) = 0 and αβ = βα = β.

If αβ = βα = β, then

(α − β)^{2} = α − αβ − βα + β = α − β,

so α − β is a projection.

Exercise 12.1.19

If α is diagonalisable with distinct eigenvalues λ_{1}, λ_{2}, …, λ_{m}, then U is the direct sum of the spaces

E_{j} = {e : αe = λ_{j}e}.

If u ∈ U, we can write u uniquely as

u = e_{1} + e_{2} + … + e_{m}

with e_{j} ∈ E_{j} If we set π_{j}(e) = e_{j} for 1 ≤ j ≤ m then direct verification shows that π_{j} is linear and π_{j} is a projection. By inspection π_{k}π_{j} = 0 when k ≠ j and

α = λ_{1}π_{1} + λ_{2}π_{2} + … + λ_{m}π_{m}.

Conversely, if the stated conditions hold and we write E_{j} = π_{j}U, we see that E_{j} is a subspace and π_{i}E_{j} = π_{i}π_{j}E_{j} = 0 for i ≠ j. Since iota = π_{1} + π_{2} + … + π_{m} we have

u = ιu = ∑_{j=1}^{m} π_{j}u = ∑_{j=1}^{m} u_{j}

with u_{j} = π_{j}u ∈ E_{j}. On the other hand, if e_{j} ∈ E_{j} and

∑_{j=1}^{m} e_{j} = 0

then applying π_{i} to both sides we get e_{i} = 0 for all i. Thus

U = E_{1} ⊕ E_{2} ⊕ … ⊕ E_{m}.

Let E_{j} be a basis for E_{j}. The set E = ⋃_{j=1}^{m} E_{j} is a basis for U. The conditions π_{k}π_{j} = 0 when k ≠ j and

α = λ_{1}π_{1} + λ_{2}π_{2} + … + λ_{m}π_{m}

show that E is a basis of eigenvectors so α is diagonalisable.

Exercise 12.2.1

Let E(rs) = (δ_{ir}δ_{js})_{1≤i≤n}^{1≤j≤n} Then, if A = (a_{ij}),

A = ∑_{r=1}^{n} ∑_{s=1}^{n} a_{rs}E(rs)

so the E(r, s) span M_{n}(𝔽).

Also

∑_{r=1}^{n} ∑_{s=1}^{n} b_{rs}E(rs) = 0 ⇒ B = 0 ⇒ b_{rs} = 0 ∀r ∀s

so the E(r, s) form a basis for M_{n}(𝔽). Thus dim M_{n}(𝔽) = n^{2}.

It follows that the n^{2} + 1 elements A^{j} with 0 ≤ j ≤ n^{2} must be linearly dependent, ie we can find a_{j} ∈ ℂ, not all zero, such that ∑_{j=0}^{n^{2}} a_{j}A^{j} = 0.

In other words, there is a non-trivial polynomial P of degree at most n^{2} such that P (A) = 0.

Exercise 12.2.2

(i) Observe that

χ_{D}(t) = ∏_{j=1}^{n}(t − λ_{j}).

Let D_{j} be the diagonal matrix with the same entries as D except that the jth diagonal entry is 0. Then

∑_{k=0}^{n} b_{k}D^{k} = ∏_{j=1}^{n}(D − λ_{j}I) = ∏_{j=1}^{n} D_{j} = 0.

(ii) We have A = P^{−1}DP for some invertible matrix P.

χ_{D} = χ_{A}

and

χ_{A}(A) = χ_{D}(A) = χ_{D}(P^{−1}DP) = P^{−1}χ_{D}(D)P = P^{−1}0P = 0 as stated.

Exercise 12.2.3

(i) Let C = B^{−1}. Using the summation convection, Tr B^{−1}AB = Tr CAB = c_{ij}a_{jk}b_{ki} = b_{ki}c_{ij}a_{jk} = δ_{kj}a_{jk} = a_{kk} = Tr A. (Lots of other proofs available.)

(ii) Q_{A}(tI −A) = tδ_{ii}−a_{ii} = nt−Tr A. Part (i) yields Q_{B^{−1}AB} = Q_{A}. (iii) Q_{A}(A) = nA−(Tr A)I. If a_{11} = −a_{22} = 1 and a_{ij} = 0 otherwise,

the. Q_{A}(A) ≠ 0.

If n = 1, Q_{(a_{11})}(t) = t − a_{11} and Q_{(a_{11})}((a_{11})) = (a_{11}) − a_{11}(1) = (0).

Exercise 12.2.6

The roots of the characteristic polynomial of a triangular matrix are the diagonal entries. If every real n × n was triangularisable by change of basis, then, since the characteristic equation is unaltered by change of basis, every root of the characteristic equation would be real. Taking A = (a_{ij}) with a_{rr} = 1 for r ≠ 1, 2 a_{12} = −a_{21} = 1, a_{ij} = 0, otherwise, we see that this is not the case.

If dim V = 1 every matrix is triangular.

Exercise 12.2.7

(i) Choose a basis such that α has an upper triangular matrix with respect to that basis. The eigenvalues of α are the diagonal entries.

With respect to the chosen basis α^{r} has matrix A^{r} which is upper triangular with diagonal entries the rth powers of the diagonal entries of A. Thus α^{r} has an eigenvalue μ if and only if α has an eigenvalue λ with λ^{r} = μ.

(ii) If α is invertible then α^{r} has an eigenvalue μ if and only if α has an eigenvalue λ with λ^{r} = μ.

To prove this observe that if β is invertible with eigenvalue λ and associated eigenvector e

β^{−1}βe = e

so λ ≠ 0 and β^{−1} has eigenvalue λ^{−1} and associated eigenvector e.

(iii) False. Let

A = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix}) .

Then A has no real eigenvalues but A^{2} = I has 1 as eigenvalue.

(\begin{matrix}0 & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33} \end{matrix}) (\begin{matrix}b_{11} & b_{12} & b_{13} \\ 0 & 0 & b_{23} \\ 0 & 0 & b_{33} \end{matrix}) (\begin{matrix}c_{11} & c_{12} & c_{13} \\ 0 & c_{22} & c_{23} \\ 0 & 0 & 0 \end{matrix}) = (\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) .

Exercise 12.2.8

(i) We have

(\begin{matrix}0 & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33} \end{matrix}) (\begin{matrix}b_{11} & b_{12} & b_{13} \\ 0 & 0 & b_{23} \\ 0 & 0 & b_{33} \end{matrix}) (\begin{matrix}c_{11} & c_{12} & c_{13} \\ 0 & c_{22} & c_{23} \\ 0 & 0 & 0 \end{matrix}) = (\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

= (\begin{matrix}0 & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33} \end{matrix}) (\begin{matrix}b_{11}c_{11} & b_{11}c_{12} + b_{12}c_{22} & b_{11}c_{13} + b_{12}c_{23} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

= (\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

(ii) Let T_{j} be an n×n upper triangular matrix with the jth diagonal entry 0. Then T_{1}T_{2} … T_{n} = 0

We can prove this by induction on n using matrices or as follows. Let τ_{j} be the linear map on an n dimensional vector space U which has matrix T_{j} with respect to some basis e_{1}, e_{2}, …, e_{n}. Then writing

E_{j} = span{e_{1}, …, e_{j}}

for 1 ≤ j ≤ n and E_{0} = {0} we have τ_{j}E_{j} ⊆ E_{j−1}. Thus

τ_{1}τ_{2} … τ_{n}U = τ_{1}τ_{2} … τ_{n}E_{n} = E_{0}

and τ_{1}τ_{2} … τ_{n} = 0 so T_{1}T_{2} … T_{n} = 0

(iii) Not necessarily true.

(\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix}1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix}0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix}0 \\ 0 \\ 1 \end{matrix})

= (\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix}1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix}1 \\ 1 \\ 1 \end{matrix})

= (\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix}3 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix}5 \\ 2 \\ 1 \end{matrix})

so the matrix product is not zero.

Exercise 12.2.9

By induction on r, βα^{r} = α^{r}β, so, by induction on s, β^{s}α^{r} = α^{r}β^{s}. Thus

∑_{j=0}^{n} a_{j}α^{j} ∑_{k=0}^{m} a_{k}β^{k} = ∑_{j=0}^{n} ∑_{k=0}^{m} a_{j}b_{k}α^{j}β^{k} = ∑_{k=0}^{m} ∑_{j=0}^{n} b_{k}a_{j}β^{k}α^{j}

= ∑_{k=0}^{m} a_{k}β^{k} ∑_{j=0}^{n} a_{j}α^{j}.

Exercise 12.2.11

Setting t = 0 we have

0 ≠ det A = a_{0}

By the Cayley–Hamilton theorem

a_{0}I = ∑_{j=1}^{n} a_{j}A^{j}

so, multiplying by a_{0}^{−1}A^{−1}, we get

A^{−1} = −a_{0}^{−1} ∑_{j=1}^{n} a_{j}A^{j−1}.

Does not appear to be a good method. The computation of the a_{j} appears to require the evaluation of

det(tI − A)

which appears to require at least as many operations as inverting a matrix by Gaussian elimination.

Exercise 12.2.12

Let M be the smallest value of m with α^{m} = 0 We know that there exist a_{j} such that

α^{n} = ∑_{j=0}^{n−1} a_{j}α^{j}

and so either α^{n} = 0 or there exists a k with n − 1 ≥ k ≥ 0 such that

α^{n} = ∑_{j=k}^{n−1} a_{j}α^{j}

and a_{k} ≠ 0. But then applying α^{M−k−1} to both sides we get

0 = a_{k}α^{M−1}

so α^{M_{1}} = 0, contrary to our definition of M.

Exercise 12.3.1

(i) We have

det(tI − A_{1}) = det(tI − A_{2}) = t^{2}.

A_{1} and A_{2} are not similar since they have different ranks.

(ii) We have

det(tI − A_{1}) = det(tI − A_{2}) = det(tI − A_{3}) = t^{3}.

None of A_{1}, A_{2}, A_{3} are similar since they have different ranks. (iii) Set

A_{8} = (\begin{matrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) , A_{9} = (\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) .

and

A_{10} = (\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix})

Then

det(tI − A_{j}) = t^{4}

for 6 ≤ j ≤ 10. Now A_{6} has rank 1, A_{7} and A_{9} have rank 2, A_{8} has rank 0 and A_{9} has rank 3.

A_{7}^{2} = 0, A_{9}^{2} = (\begin{matrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})

so A_{7}^{2} has rank 0 and A_{9}^{2} rank 1. None of the A_{j} can be similar.

Exercise 12.3.3

(i) Since the characteristic polynomials have the form t^{n} so do the minimal polynomials. The minimal polynomial of A_{k} is m_{k}(t) = t^{r} where A_{k}^{r} = 0, A_{k}^{r−1} ≠ 0. Thus

m_{1}(t) = t, m_{2}(t) = t^{2}, m_{3}(t) = t, m_{4}(t) = t^{2}, m_{5}(t) = t^{3}m_{6}(t) = t^{2}, m_{7}(t) = t^{2}, m_{8}(t) = t, m_{9}(t) = t^{3}, m_{10}(t) = t^{4}

(ii) We have

det(tI − A) = (t − 1)^{2}(t − 2)^{2} = det(tI − B)

Now (A−I)(A−2I)^{2} ≠ 0 and (A−I)^{2}(A−I)^{2} ≠ 0 so A has minimal polynomial (t − 1)^{2}(t − 2)^{2}.

We also have (B − I)(B − 2I) ≠ 0 and (B − II)^{2} ≠ 0 but (B − I)^{2}(B − 2I) ≠ 0 so A has minimal polynomial (t − 1)^{2}(t − 2).

Exercise 12.3.6

(i) ∏_{i=1}^{r}(D − λ_{i}I) is a diagonal matrix with jth diagonal entry ∏_{i=1}^{r}(d_{jj} − λ_{i}) = 0 so ∏_{i=1}^{r}(D − λ_{i}I) = 0.

Now suppose, without loss of generality that d_{11} = λ_{1}. Then ∏_{i=2}^{r}(D− λ_{i}I) is a diagonal matrix with 1st diagonal entry ∏_{i=2}^{r}(d_{11} − λ_{i}) ≠ 0 so ∏_{i=2}^{r}(D − λ_{i}I) ≠ 0. Thus ∏_{i=1}^{r}(t − λ_{i}) is the minimal polynomial for D.

(ii) Choose a basis of eigenvectors. e_{1}, e_{2}, … e_{n}. With respect to this basis α has a diagonal matrix with j th diagonal entry the eigenvalue associated with e_{j}. Thus if λ is associated with k of the basis vectors det(tI − D) = det(tι − α) has (t − λ)^{k} as a factor but not (t − λ)^{k+1}

Now writing S for the set of eigenvalues which are not equal to λ

dim U_{λ} = dim(∏_{μ∈S}(μι−α)U) = dim span{∏_{μ∈S}(μι−α)U)e_{j} : 1 ≤ j ≤ n} = k

Exercise 12.3.7

(i) If P has a repeated root θ, then P (t) = (t − θ)^{k}Q(t) with Q a polynomial and k ≥ 2. We have

P′(t) = k(t − θ)^{k−1}Q(t) + (t − θ)^{k}Q′(t) = (t − θ)^{k−1}(kQ(t) + (t − θ)Q′(t)

so t − θ is a factor of both P (t) and P′(t).

The converse is true If P (t) = (t−θ)Q(t) with Q a polynomial having Q(θ) ≠ 0 then

P′(t) = Q(t) + (t − θ)^{k}Q′(t)

so P′(θ) ≠ 0.

(ii) Since A^{m} = I the minimal polynomial m_{A} of A must divide t^{m}−1. But t^{m} − 1 = 0 has no repeated roots (by part (i), or otherwise). Thus m_{a} has no repeated roots so A is diagonalisable.

If A is a real symmetric matrix then its eigenvalues are real so we have A = LDL^{−1} with L invertible (indeed we may take L orthogonal) and D real diagonal

D^{m} = I

so d_{jj}^{m} = 1 and d_{jj} is real so d_{jj} = ±1 so D^{2} + I so A^{2} = I.

Exercise 12.3.9

(i) Let ∂Q denote the degree of Q. If P ≠ {0} then the set

E = {∂P : P ∈ P \ {0}}

is a non-empty subset of the positive integers and so has a least member N.

Let Q_{0} ∈ P have degree N and leading coefficient a. If we set P_{0} = a^{−1}Q_{0}, then P_{0} is a monic polynomial of smallest degree in P.

If Q ∈ P then Q = SP + R with S, R polynomials and ∂R < N. Now R = Q − SP_{0} ∈ P so by the definition of N, R = 0 and Q = SP_{0}.

(ii) Observe that P satisfies the hypotheses of (i). If S divides each P_{j} then S divides each Q_{j}P_{j} and so divides P_{0}. (iii) Take P_{j}(t) = ∑_{j=1}^{r} Q_{j}(t) ∏_{i≠j}(t − λ_{i})^{m(i)} and apply (ii).

Exercise 12.3.11

Write α = α_{1} ⊕ α_{2} ⊕ … ⊕ α_{r} The expansion

u = u_{1} + u_{2} + … + u_{r}

with u_{j} ∈ U_{j} always exists a and is unique so

α(u) == α_{1}u_{1} + α_{2}u_{2} + … + α_{r}u_{r}

is well defined.

If u, v ∈ U and λ, μ then we can write

u = u_{1} + u_{2} + … + u_{r}
v = v_{1} + v_{2} + … + v_{r}

with u_{j}, v_{j} ∈ U_{j}. We then have

λu + μv = (λu_{1} + μv_{1}) + (λu_{2} + μv_{2}) + … + (λu_{r} + μv_{r}) and λu_{j} + μv_{j} ∈ U_{j} so

α(λu + μv) = α_{1}(λu_{1} + μv_{1}) + α_{2}(λu_{2} + μv_{2}) + … + α_{r}(λu_{r} + μv_{r})

= (λα_{1}u_{1} + μα_{1}v_{1}) + (λα_{2}u_{2} + μα_{2}v_{2}) + … + (λα_{r}u_{1} + μα_{r}v_{1}) = λ(α_{1}u_{1} + α_{2}u_{2} + … + α_{r}u_{r}) + μ(α_{1}v_{1} + α_{2}v_{2} + … + α_{r}v_{r}) = λαu + μαv

so α is linear.

Exercise 12.3.13

(i) We have

ι = ∑_{j=1}^{r} Q_{j}(α) ∏_{i≠j}(α − λ_{j}ι)^{m(j)},

so, if u ∈ U we have

u = ∑_{j=1}^{r} Q_{j}(α) ∏_{i≠j}(α − λ_{j}ι)^{m(j)}u = sum_{j=1}^{r}u_{j}

where

u_{j} = Q_{j}(α) ∏_{i≠j}(α − λ_{j}ι)^{m(j)}u.

Since

(α − λ_{j}ι)^{m(j)}u_{j} = (α − λ_{j}ι)^{m(j)}Q_{j}(α) ∏_{i≠j}(α − λ_{j}ι)^{m(j)}u

= Q_{j}(α) ∏_{j=1}^{r}(α − λ_{j}ι)^{m(j)}u

= Q_{j}(α)Q(α)u = Q_{j}(α)0 = 0

we have u_{j} ∈ U_{j}. Thus

U = U_{1} + U_{2} + … + U_{r}.

(ii) Next we observe that if v ∈ U_{k} with k ≠ j then

Q_{j}(α) ∏_{i≠j}(α − λ_{j}ι)^{m(j)}v = 0

Thus if v ∈ U_{k} we have

v = ∑_{j=1}^{r} Q_{j}(α) ∏_{i≠j}(α − λ_{i}ι)^{m(i)}v = Q_{k}(α) ∏_{i≠k}(α − λ_{i}ι)^{m(i)}v.

In particular ∏_{i≠k}(α − λ_{i}ι)^{m(i)}v ≠ 0 if v ≠ 0.

Now if v_{j} ∈ U_{j} and

∑_{j=1}^{r} v_{j} = 0

then

∏_{i≠k}(α − λ_{j}ι)^{m(i)}v_{k} = ∏_{i≠k}(α − λ_{j}ι)^{m(i)} ∑_{j=1}^{r} v_{j}0

so v_{k} = 0 for each k.

Thus

U = U_{1} ⊕ U_{2} ⊕ ⋯ ⊕ U_{r}

(iii) We have u ∈ U_{j} ⇒ (α−λ_{j})^{m(j)}u = 0 ⇒ (α−λ_{j})^{m(j)}αu = α(α−λ_{j})^{m(j)}u = 0 ⇒ αu ∈ U_{j}.

If u ∈ U then

u = ∑_{j=1}^{r} u_{j}

and

αu = α ∑_{j=1}^{r} u_{j} = ∑_{j=1}^{r} αu_{j} = ∑_{j=1}^{r} α_{j}u_{j} = α = α_{1} ⊕ α_{2} ⊕ ⋯ ⊕ α_{r}u

so

α = α_{1} ⊕ α_{2} ⊕ ⋯ ⊕ α_{r}.

(iiv) By the definition of U_{j}, (α − λ_{j}ι)^{m(j)}U_{j} = 0 so the minimal polynomial for α_{j} must have the form. (t − λ_{j})^{p(j)}. But If u ∈ U then u = ∑_{s=1}^{r} u_{s} with u_{s} ∈ U_{s}

∏_{j=1}^{r}(α − λ_{j})^{p(j)}u = ∑_{s=1}^{r} ∏_{j=1}^{r}(α − λ_{j})^{p(j)}u_{j} = ∑_{s=1}^{r} 0 = 0.

so the minimal polynomial must divide ∏_{j=1}^{r}(α−λ_{j})^{p(j)} so p(j) = m(j) for all J.

Exercise 12.3.14

Let e_{1}, e_{2}, …, e_{r} be a basis for U with respect to which α has matrix A and e_{r+1}, e_{r+2}, …, e_{r+s} be a basis for V with respect to which β has matrix B Then e_{1}, e_{2}, …, e_{r+s} is a basis for W with respect to which α ⊕ β has matrix

(\begin{matrix}A & 0 \\ 0 & B \end{matrix})

Thus

χ_{α⊕β}(t) = det (\begin{matrix}tI − A & 0 \\ 0 & tI − B \end{matrix}) = det(tI−A) det(tI−B) = χ_{α}(t)χβ(t).

If α has minimal polynomial m_{α}, β has minimal polynomial m_{β} and P is a polynomial

P (α ⊕ β) = 0 ⇔ P (α ⊕ β)(u + v) = 0∀u ∈ U, v ∈ V
⇔ P (α)u + P (β)v = 0∀u ∈ U, v ∈ V
⇔ P (α)u = P (β)v = 0∀u ∈ U, v ∈ V
⇔ m_{α} and m_{β} divide P

Thus the minimal polynomial of α ⊕ β is the lowest common multiple of (ie the monic polynomial of lowest degree dividing) m_{α} and m_{β}.

Exercise 12.3.15

(i) If (t − λ) is a factor of the characteristic polynomial then λ is an eigenvalue and (t − λ) must be a factor of the minimal polynomial. Thus we cannot choose S(t) = t + 1, Q(t) = t.

(ii) If m = 1 set A = 0. If m = n let A = (a_{ij}) be the n × n matrix with a_{i,i+1} = 1 a_{ij} = 0 otherwise.

If n > m > 1 let B = 0_{n−m} the n − m + ×n − m zero matrix and let C be the m × m matrix with c_{i,i+1} = 1 c_{ij} = 0 otherwise. Take

A = (\begin{matrix}B & 0 \\ 0 & C \end{matrix})

A − λI has characteristic equation (t − λ)^{n} and minimal polynomial (t − λ)^{m}

(iii) We can write P (t) = ∏ i = 1^{r}(t−λ_{i})^{m(i)} and Q(t) = ∏ i = 1^{r}(t− λ_{i})^{m(i)} with the λ_{j} distinct and m(j) ≥ p(j) ≥ 1. By (ii) we can find a m(j) × m(j) matrix B_{j} = A_{j} − λI_{m(j)} with characteristic polynomial (t − λ)^{m(j)} and minimal polynomial (t − λ)^{p(j)}. Take

A = (\begin{matrix}A_{1} & 0 & … & 0 & 0 \\ 0 & A_{2} & 0 & … & 0 & 0 \\ ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & 0 & … & A_{r−1} & 0 \\ 0 & 0 & 0 & … & 0 & A_{r} \end{matrix})

A has the required properties either by direct computation or using Exercise 12.3.14.

Exercise 12.4.3

Since αe_{j} = e_{j+1} for 1 ≤ j ≤ n − 1 and αe_{n} = 0 we have A = (a_{ij}) with a_{i,i+1} = 1 a_{ij} = 0 otherwise.

Exercise 12.4.4

If α^{n} ≠ 0, then e, αe, … α^{n}e are linearly independent a and a space of dimension n would contain n + 1 linearly independent vectors which is impossible.

Exercise 12.4.7

We have α^{k}e ≠ 0 for only finitely many values of k.

Exercise 12.4.11

Observe that (d_{jj}) is a 1 × 1 Jordan matrix.

Exercise 12.4.12

(i) A is the matrix of a linear map α : U → U

α = α_{1} ⊕ α_{2} … α_{r}

with U = U_{1} ⊕ U_{2} … U_{r} where α_{j} is a linear map on U_{j} with α_{j} = λ_{j}ι + β_{j} with β_{j} a nilpotent linear map on U_{j} with β_{j}^{k_{j}} = 0, β^{k_{j}−1} ≠ 0 dim U_{j} = k_{j} Thus

(λι − α)^{−k}0 = ⨁_{j=1}^{r}(λι − α_{j})^{−k}0

and

dim(λι − α)^{−k}0 = ∑_{j=1}^{r} dim(λι − α_{j})^{−k}0 = ∑_{λ=λ_{j}} min{k, k_{j}}.

Thus

{x ∈ ℂ^{n} : (λI − A)^{k}x = 0} = ∑_{λ=λ_{j}} min{k, k_{j}}

(ii) By (i)

∑_{λ=λ_{j}} min{k, k_{j}} = ∑_{λ=λ̃_{i}} min{k, k̃_{i}}

so r̃ = r and, possibly after renumbering, λ̃_{j} = λ_{j} and k̃_{j} = k_{j} for 1 ≤ j ≤ r.

Exercise 12.4.13

Observe by looking at the effect on z^{n} that T^{n+1} = 0 but T^{n} ≠ 0. Thus, since P_{n} has dimension n + 1, T has Jordan form

J_{n+1}(0) = (\begin{matrix}0 & 1 & 0 & 0 & … & 0 & 0 \\ 0 & 0 & 1 & 0 & … & 0 & 0 \\ 0 & 0 & 0 & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & … & 0 & 1 \\ 0 & 0 & 0 & 0 & … & 0 & 0 \end{matrix}) .

We observe that e_{r}(z) = z^{r} is an eigenvector with eigenvalue r [n ≥ r ≥ 0]. Thus the Jordan normal form is the diagonal matrix

(\begin{matrix}0 & 0 & 0 & 0 & … & 0 & 0 \\ 0 & 1 & 0 & 0 & … & 0 & 0 \\ 0 & 0 & 2 & 0 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & … & n − 1 & 0 \\ 0 & 0 & 0 & 0 & … & 0 & n \end{matrix}) .

Exercise 12.4.14

(i) If λ is a root of χ_{α} then we can find an associated eigenvector e. Since

e ∈ dim{u : (α − λι)(u) = 0}

we have m_{g}(λ) ≥ 1.

Now take a basis e_{1}, e_{2} …, e_{m_{g}(λ)} for

{u : (α − λι)(u) = 0}

and extend it to a basis e_{1}, e_{2} …, e_{n} for U. With respect to this basis α has matrix

A = (\begin{matrix}λI_{m_{g}(λ)} & 0 \\ R & S \end{matrix})

with I_{m_{g}(λ)} the m_{g}(λ) × (m_{g}(λ) identity matrix so

χ_{α}(t) = (t − λ)^{m_{g}(λ)}χ_{R}(t)

and m_{a}(λ) ≥ m_{g}(λ).

(ii) Suppose s ≥ r ≥ 1 Let B = (b_{ij}) be the r × r matrix with b_{i,i+1} = 1 and b_{ij} = 0 and A(r, s) the s × s matrix with

A(r, s) = (\begin{matrix}λB & 0 \\ 0 & 0 \end{matrix}) .

If α_{r,s} is the linear map associated with A(r, s) then, for this linear map,

m_{g}(λ) = \begin{cases}r & if λ = 0 \\ = 0 & otherwise \end{cases} , m_{a}(λ) = \begin{cases}s & if λ = 0 \\ = 0 & otherwise \end{cases}

If we now set β_{r,s} = μι − α_{r,s} then for this linear map

m_{g}(λ) = \begin{cases}r & if λ = μ \\ = 0 & otherwise \end{cases} , m_{a}(λ) = \begin{cases}s & if λ = μ \\ = 0 & otherwise \end{cases}

Now take a vector space

U = U_{1} ⊕ U_{2} ⊕ … ⊕ U_{r}

with dim U_{k} = n_{a}(λ_{k}). By the previous paragraph we can find γ_{k} : U_{k} → U_{k} linear such that for this linear map

m_{g}(λ) = \begin{cases}n_{g}(λ_{k})If & if λ = λ_{k} \\ = 0 & otherwise \end{cases} , n_{a}(λ) = \begin{cases}s & if λ = λ_{k} \\ = 0 & otherwise \end{cases}

If we set

α = γ_{1} ⊕ γ_{2} ⊕ … ⊕ γ_{r}

then the required properties can be read off.

(iii) Suppose α has the associated Jordan form

A = (\begin{matrix}J_{k_{1}}(λ_{1}) \\ & J_{k_{2}}(λ_{2}) \\ & & J_{k_{3}}(λ_{3}) \\ & & & . & . & . \\ & & & & & & J_{k_{r−1}}(λ_{r−1}) \\ & & & & & & & J_{k_{r}}(λ_{r}) \end{matrix})

(using the notation of Theorem 12.4.10).

We observe that the characteristic polynomial

χ_{α}(t) = χ_{A}(t) = ∏_{j=1}^{r} χ_{J_{k_{j}}} (t) = ∏_{j=1}^{r}(t − λ_{j})^{k(j)}

so

m_{a}(λ) = ∑_{λ_{j}=λ} k(j).

Observe that the dimension of the space of solutions of

J_{k}x = λx

is 1 if λ = λ_{k} and zero otherwise we see that

m_{a}(λ) = ∑_{λ_{j}=λ} 1 = card{: λ_{j} = λ}

Exercise 12.5.1

Let U be a finite dimensional vector space over C. If α ∈ L(U, U) then the Jordan normal form theorem tells that we can write U = ⨁ j = 1^{r}U_{j} and find α_{j} ∈ L(U_{j}, U_{j}) such that α = ⨁ j = 1^{r}α_{j} and U_{j} has a basis with respect to which α_{j} has an n(j) × n(j) matrix

A_{j} = (\begin{matrix}λ & 1 & 0 & 0 & … & 0 & 0 \\ 0 & λ & 1 & 0 & … & 0 & 0 \\ 0 & 0 & λ & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & … & λ & 1 \\ 0 & 0 & 0 & 0 & … & 0 & λ \end{matrix}) .

Now

det(tι − α) = ∏_{j=1}^{r} det(tι − α_{j}) ∏_{j=1}^{r}(t − λ_{j})^{n(j)}

and

∏_{j=1}^{r}(α − λ_{j})^{n(j)} = ⨁ j = 1^{r} ∏_{i=1}^{r}(α_{j} − λ_{i})^{n(j)} = 0

(since (α_{j} − λ_{j})^{n(j)} = 0) so χ_{α}(α) = 0.

Lemma 12.2.1 tells us that there exists a non-zero polynomial P with P (α) = 0. This is all we need to show that there is a minimal polynomial. The proof of the Jordan form theorem only uses the minimal polynomial.

Exercise 12.5.2

Consider the s × s matrix

B = (\begin{matrix}λ & 1 & 0 & 0 & … & 0 & 0 \\ 0 & λ & 1 & 0 & … & 0 & 0 \\ 0 & 0 & λ & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & … & λ & 1 \\ 0 & 0 & 0 & 0 & … & 0 & λ \end{matrix}) .

If λ ≠ 0, B is invertible so rank B^{j} = s for all j ≥ 0. If λ = 0 then direct calculation shows that rank B^{j} = max{0, k − j} for all j ≥ 0.

If α = ∏_{k=1}^{r} α_{k} then rank α^{j} = ∑_{k=1}^{r} rank α_{k}^{j}

Thus if the Jordan normal form of an n × n A contains m nilpotent blocks

K_{p} = (\begin{matrix}0 & 1 & 0 & 0 & … & 0 & 0 \\ 0 & 0 & 1 & 0 & … & 0 & 0 \\ 0 & 0 & 0 & 1 & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & … & 0 & 1 \\ 0 & 0 & 0 & 0 & … & 0 & 0 \end{matrix})

with K_{p} an n(p) × n(p) matrix, then

rankA^{j} = (n − ∑_{p=1}^{m} n_{p}) + ∑_{p=1}^{m} max{0, n_{p} − j}

. (ii) Observe that if we set

r(j) = (n − ∑_{p=1}^{m} n_{p}) + ∑_{p=1}^{m} max{0, n_{p} − j}

then for some m = max_{p} n(p) =≤ n together with the condition

r_{0} − r_{1} ≥ r_{1} − r_{2} ≥ r_{2} − r_{3} ≥ … ≥ r_{m−1} − r_{m}.

Let q(j) = (s_{j} − s_{j−1}) − (s_{j+1} − s_{j}). Let A be the n × n matrix with qj nilpotent Jordan blocks of size j × j and s(m) 1 × 1 Jordan blocks (1). If α has matrix A with respect to some basis then α satisfies the required conditions.

Exercise 12.5.3

(i) We have

A_{1} = (\begin{matrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) , A_{2} = (\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) , A_{3} = (\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) ,

A_{4} = (\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}) , A_{5} = (\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix})

(ii) For A_{1} ẋ_{r} = 0 so x(t) = c a constant. For A_{2} ẋ_{1} = x_{2}, ẋ_{r} = 0 otherwise so x(t) = (c_{2}t + c_{1}, c_{2}, c_{3}, c_{4})^{T} with

c_{j} constants.

For A_{3} ẋ_{1} = x_{2}, ẋ_{2} = x_{3}, ẋ_{r} = 0 otherwise so x(t) = (c_{3}t^{2}/2 + c_{2}t + c_{1}, c_{2}t + c_{3}, c_{3}, c_{4})^{T} with c_{j} constants.

Similarly for A_{4}, x(t) = (c_{4}t^{3}/6 + c_{3}t^{2}/2 + c_{2}t + c_{1}, c_{4}t^{2}/2 + c_{3}t + c_{2}, c_{4}t + c_{3}, c_{4})^{T} with c_{j} constants.

Similarly for A_{5} x(t) = (c_{2}t + c_{1}, c_{2}, c_{4}t + c_{3}, c_{4})^{T} (iii) If we write x(t) = e^{λ}y(t) then

x′(t) = A_{j}x(t)

Thus the general solution of

y′(t) = (λI + A_{j})y(t)

is y(t) = e^{−λt}x(t) where x(t) is the general solution of x′(t) = Ax(t).

(iv) Suppose

B = (\begin{matrix}B_{1} & 0 & … & 0 & 0 \\ 0 & B_{2} & … & 0 & 0 \\ ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & … & B_{r−1} & 0 \\ 0 & 0 & … & 0 & B_{r} \end{matrix})

with B_{j} = λ_{j}I +C_{j} where C_{j} is an s(j)×s(j) nilpotent Jordan matrix. Then the general solution of

x′(t) = Bx(t)

is x = (y_{1}, y_{2}, …, y_{r})^{T} with

y_{j}(t) = e^{−λ_{j}t}(P_{j}^{(s(j))}(t), P_{j}^{(s(j)−1)}(t), …, P_{j}(t))^{T}

with P_{j} any polynomial of degree s(j) − 1.

(iv) Find M invertible and B a matrix in Jordan form such that

MAM^{−1} = B

then writing y = Mx we have

⋆ ẏ = By.

The general solution of

x′(t) = Ax(t).

is x = M^{−1}y where y is the general solution of ⋆.

Exercise 12.5.4

Observe that

(\begin{matrix}0 & 1 & 0 & … & 0 & 0 \\ 0 & 0 & 1 & … & 0 & 0 \\ & ⋮ & & & ⋮ \\ 0 & 0 & 0 & … & 1 & 0 \\ 0 & 0 & 0 & … & 0 & 1 \\ −a_{n−1} & −a_{n−2} & −a_{n−3} & … & −a_{1} & −a_{0} \end{matrix}) ( x_{0} ; x_{1} ; x_{2} ; ⋮ ; x_{n−2} ; x_{n−1}) = ( x_{0} ; x_{1} ; x_{2} ; ⋮ ; x_{n−2} ; x_{n−1})

⇔ ẋ_{j} = x_{j+1} [0 ≤ j ≤ n − 2], − ∑_{j=0}^{n−2} a_{j}x_{j} = x_{n−1}

⇔ x^{(n)} + ∑_{j=1}^{n−2} a_{j}x^{(n−j)} = 0

Exercise 12.5.5

Suppose U = ⨁_{j=1}^{r} U_{j}, U_{j} has dimension k_{j} β_{j} : U_{j} → U_{j} has β_{j}^{k(j)−1} ≠ 0, β_{j}^{k_{j}} = 0, α_{j} = λ_{j}ι + β_{j} and α = ⨁_{j=1}^{r} α_{j}.

Write s(λ) = max_{λ=λ_{j}} k_{j}. If P is a polynomial

P (α) = 0 ⇔ P (α_{j}) = 0 ∀j ↔ (t − λ)^{s}(λ)|P (t).

Thus the characteristic polynomial equals the minimal polynomial if and only if

∏_{j=1}^{r}(t − λ)^{k_{j}} = ∏_{s(λ)≠0}(t − λ)^{s(λ)}

that is to say, if and only if all the λ_{j} are distinct.

Exercise 12.5.6

Writing e_{j} for the vector with 1 in the jth place and zero everywhere else. By induction

A^{j}e_{n} = e_{n−j} + f_{n−j}

where

f_{n−j} ∈ span{e_{r} : n ≥ r ≥ n − j + 1}.

Thus the A^{j}e_{n} with n − 1 ≥ j ≥ 0 are linearly independent and

(∑_{j=0}^{n−1} b_{j}A^{j}) e = 0 ⇔ b_{j} = 0 ∀0 ≤ j ≤ n − 1.

Thus the minimal polynomial has degree at least n. Since the minimal polynomial divides the characteristic polynomial and the characteristic polynomial has degree n it follows that the two polynomials are equal. By Exercise 12.5.5, it follows that that there is a Jordan form associated with A in which all the blocks J_{k}(λ_{k}) have distinct λ_{k}.

Exercise 12.5.7

(i) If A is an n × n matrix consisting of a single Jordan block with associated eigenvalue λ and

ẋ = Ax

then writing y(t) = e^{−λt}x(t) we have

ẏ(t) = −λe^{−λt}x(t) + e^{−λt}ẋ(t) = −λe^{−λt}x(t) + e^{−λt}Ax(t) = Bẏ(t)

where B is a single Jordan block with associated eigenvalue 0. Thus

y^{(n)} = 0

and y(t) = ∑_{j=0}^{n−1} c_{j}t^{j} for some c_{j} ∈ ℝ.

Conversely, if y(t) = ∑_{j=0}^{n−1} c_{j}t^{j}, for some c_{j} ∈ ℝ, then

ẏ(t) = Bẏ(t).

Thus

x(t) = e^{λt} ∑_{j=0}^{n−1} c_{j}t^{j}

where the c_{j} ∈ ℝ are arbitrary.

(ii) Suppose that A is associated with a Jordan form B with blocks the n_{k} × n_{k} Jordan block J_{k}(λ_{k}) with all the lambda_{k} distinct [1 ≤ k ≤ r]. Writing A = MBM^{−1} with M non-singular we see that if y is a solution of ẏ = By if and only x = My is a solution of ẋ = Ax. Thus the only possible solutions of ⋆ are

x(t) = ∑_{k=1}^{r} e^{λ_{k}t}P_{k}(t).

with the P_{k} general polynomials of degree n_{k} − 1.

To show that every x of this form is a solution observe that (by linearity) it suffices to show that

e^{λ_{k}t}P_{k}(t)

is a solution with the P_{k} a general polynomials of degree n_{k} − 1. By considering e^{−λ_{k}t}x(t) we see that it suffices to show that if the characteristic equation of A has 0 as an p times repeated root the every polynomial of degree r − 1 satisfies ⋆. But if A has 0 as an p times repeated root then a_{0} = a_{1} = … = a_{r−1} = 0 so the previous sentence is automatic.

Exercise 12.5.8

(i) We have

\binom{r}{k − 1}+\binom{r}{k} = k \frac{r!}{k!(r + 1 − k)!}+(r+1−k)k \frac{r!}{k!(r + 1 − k)!} = (r+1)\frac{r!}{k!(r + 1 − k)!} = \binom{r + 1}{k}.

(ii) If k = 0 our system becomes

u_{r}(0) = u_{r−1}(0)

with the general solution u_{r}(0) = b_{0}.

If k = 1 we have u_{r}(0) = b_{0} as before and

u_{r}(1) − u_{r−1}(1) = b_{0}

If we set

v_{r}(1) = u_{r}(1) − b_{0}r

we obtain

v_{r}(1) − v_{r−1}(1) = 0

so as before v_{r}(1) = b_{1} for some freely chosen b_{1}. Thus u_{r}(1) = b_{0}r +b_{1}.

Suppose that the solution for a given k ≥ 0 is

u_{r}(k) = ∑_{j=0}^{k} b_{k−j}\binom{r}{j}.

Then

u_{r}(k + 1) − u_{r−1}(k + 1) = ∑_{j=0}^{k} b_{j}\binom{r}{j},

and setting

v_{r}(k + 1) = u_{r}(k + 1) − ∑_{j=0}^{k} b_{k−j}\binom{r}{j + 1},

we obtain

v_{r}(k + 1) − v_{r−1}(k + 1) = 0

so, as before, v_{r}(k + 1) = b_{k+1} for some freely chosen b_{k+1}. Thus

u_{r}(k + 1) = ∑_{j=0}^{k+1} b_{k+1−j}\binom{r}{j}.

Thus, by induction,

u_{r}(k) = ∑_{j=0}^{k} b_{k−j}\binom{r}{j}.

(ii) If we set u_{r}(k) = λ^{−r}v_{r}(k) then the u(k) satisfy the equations in (i). Thus

v_{r}(k) = λ^{r} ∑_{j=0}^{k} b_{k−j}\binom{r}{j}.

(iii) The equation

u_{r} + ∑_{j=0}^{n−1} a_{j}u_{j−n+r} = 0

may be rewritten as

u_{t+1} + Au_{t} = 0

where u_{t} is the column vector (u_{t}, u_{t+1}, …, u_{t+n−1}) and

A = (\begin{matrix}0 & 1 & 0 & … & 0 & 0 \\ 0 & 0 & 1 & … & 0 & 0 \\ & ⋮ & & & ⋮ \\ 0 & 0 & 0 & … & 1 & 0 \\ 0 & 0 & 0 & … & 0 & 1 \\ −a_{0} & −a_{1} & −a_{2} & … & −a_{n−2} & −a_{n−1} \end{matrix}) .

By Exercise 12.5.6 we know that we can associate A with a Jordan form B in which in which the q blocks J_{k}(λ_{k}) are of size s_{k} × s_{k} and have distinct λ_{k}.

Let B = M^{−1}AM with M invertible.

By (ii) the general solution of

v_{t+1} + Bv_{t} = 0

is

v_{t} ∑_{k=1}^{q} λ_{k}^{t}e_{k} ∑_{j=0}^{s_{k}} b_{k,s_{k}−j}\binom{r}{j}

so

u_{t} = ∑_{k=1}^{q} λ_{k}^{t}e_{k} ∑_{j=0}^{s_{k}} c_{k,s_{k}−j}\binom{r}{j}.

We have shown that u_{t} must be of the form just given. Direct computation shows that any u_{t} of the form just given is a solution.

Exercise 12.5.9

All roots characteristic equation the same.

(\begin{matrix}λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix}) , (\begin{matrix}λ & 1 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix}) , (\begin{matrix}λ & 1 & 0 \\ 0 & λ & 1 \\ 0 & 0 & λ \end{matrix}) .

Characteristic equation has two distinct roots

(\begin{matrix}λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & μ \end{matrix}) , (\begin{matrix}λ & 1 & 0 \\ 0 & λ & 0 \\ 0 & 0 & μ \end{matrix})

with λ ≠ μ. Characteristic equation has three distinct roots

(\begin{matrix}λ & 0 & 0 \\ 0 & μ & 0 \\ 0 & 0 & ν \end{matrix}) , (\begin{matrix}λ & 1 & 0 \\ 0 & λ & 0 \\ 0 & 0 & μ \end{matrix})

with λ, μ, ν distinct.

Exercise 12.5.10

We have A^{4} − A^{2} = 0 so the minimal polynomial m must divide t^{4} − t^{2} = t^{2}(t − 1)(t + 1)

Since (A − I)A = A^{2} − A ≠ 0 the minimal polynomial cannot be 1, t, t − 1 or t(t − 1)

(i) m(t) = t^{2}. The characteristic polynomial must be t^{5}. The Jordan blocks must be of size 2 × 2 or less. There are three possibilities

0, (\begin{matrix}J & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix}J & 0 & 0 \\ 0 & J & 0 \\ 0 & 0 & 0 \end{matrix}) with J = (\begin{matrix}J & 0 \\ 0 & 0 \end{matrix})

(ii) m(t) = t + 1. The Jordan normal form is −I giving one possibility.

(iii) m(t) = t^{2} − 1. The Jordan normal form is diagonal with r entries 1, 5 − r entries −1 [1 ≤ r ≤ 4] and the characteristic equation is (t − 1)^{r}(t + 1)^{s} There are four possibilities.

(iii) m(t) = t(t + 1). The Jordan normal form is diagonal with r entries −1, 5 − r entries −0 [1 ≤ r ≤ 4] and the characteristic equation is (t + 1)^{r}t^{s} There are four possibilities.

(iv) m(t) = t^{2}(t−e) with e = ±1. The Jordan normal form either has one block J and then r diagonal entries e and 3 − r diagonal entries 0 [1 ≤ r ≤ 3] giving a characteristic polynomial t^{5−r}(t−e)^{r} or two blocks J and a diagonal entry e giving a characteristic polynomial t^{4}(t − e) There are eight possibilities.

(v) m(t) = t(t − 1)(t + 1). (iii) m(t) = t^{2} − 1. The Jordan normal form is diagonal with r entries 1, s entries −1, u entries 0 [r, s, u ≥ 1, r + s + u = 5] and the characteristic equation is (t − 1)^{r}(t + 1)^{s}t^{u} There are six possibilities.

(vi) m(t) = t^{2}(t − 1)(t + 1). The Jordan normal form has one block J and r diagonal entries 1, s diagonal entries −1 [r, s ≥ 1, r + s ≤ 3] and 3 − r − s diagonal entries 0. The characteristic equation is (t − 1)^{r}(t + 1)^{s}t^{5−r−s}. There are three possibilities.

There are 29 possible Jordan forms.

Exercise 12.5.11

det(tI − M) = (t − 1) det (\begin{matrix}t − 1 & 0 & 0 \\ 1 & t − 2 & 0 \\ 0 & 0 & t − 2 \end{matrix}) = (t − 1)^{2}(t − 2)^{2}

Looking at the eigenvalue 1 we have

(\begin{matrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & −1 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix}) (\begin{matrix}x \\ y \\ z \\ w \end{matrix}) = (\begin{matrix}x \\ y \\ z \\ w \end{matrix})

implies

x + z = x
y = y
−y + 2z = z
2w = w

so

M(x, y, z, w)^{T} = (x, y, z, w)^{T} ⇔ (x, y, z, w) = (t, 0, 0, 0)

for general t.

Thus e_{2} = (1, 0, 0, 0)^{T} is an eigenvector eigenvalue 1. If (M −I)x = e then

z = 1
0 = 0
−y + z = 0
w = 0

Thus writing e_{1} = (0, 1, 1, 0)^{T} we have (M−I)e_{1} = e_{2} ≠ 0 (M−I)^{2}e_{1} = 0.

Looking at the eigenvalue 2 we have

(\begin{matrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & −1 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix}) (\begin{matrix}x \\ y \\ z \\ w \end{matrix}) = 2 (\begin{matrix}x \\ y \\ z \\ w \end{matrix})

implies

x + z = 2x
y = 2y
−y + 2z = 2z
2w = 2w

so

M(x, y, z, w)^{T} = (x, y, z, w)^{T} ⇔ (x, y, z, w) = (t, t, 0, s)

for general t and s. Thus the eigenspace has dimension 2 and basis (1, 1, 0, 0)^{T}, (0, 0, 0, 1)^{T}.

The characteristic polynomial is (t − 1)^{2}(t − 2)^{2} minimal polynomial is (t − 1)^{2}(t − 2). A Jordan form is given by

J = (\begin{matrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix})

An appropriate basis is (0, 1, 1, 0)^{T}, (1, 0, 0, 0)^{T} (1, 1, 0, 0)^{T}, (0, 0, 0, 1)^{T} and we can take M to be the matrix with these columns, that is to say

M = (\begin{matrix}0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) .

Exercise 13.2.2

(i) The tables are

+ 0 1 0 0 1 1 1 0  \begin{matrix}× & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{matrix}

By inspection of the addition table x + x = 0 for all x. (ii) x = −x ⇒ (1 + 1)x = 0 ⇒ x = 0.

Exercise 13.2.3

(i) If c ≠ 0 then

0 = c^{−1}(cd) = (c^{−1}c)d = d

(ii) If a^{2} = b^{2} then (a + b)(a − b) = a^{2} − b^{2} = 0 so a + b = 0 and/or b − a = 0 i.e. a = −b and or a = b.

(iii) By (ii) (and the fact that 1 + 1 ≠ 0, the equation x^{2} = c has no solutions or two solutions or c = 0 and we have x = 0. Thus every element y is a square root (just look at y^{2}) but every square has two distinct square roots or is 0. Thus k is odd, there are (k − 1)/2 non zero squares and k + 1)/2 squares in all.

(iv) If we work in Z_{2}, then 0^{2} = 0 and 1^{2} = 1. Every element is a square. (Note −1 = 1)

Exercise 13.2.4

Suppose x, y, z ∈ R λ, μ ∈ ℚ.

(i) (x ˙+y) ˙+z = (x + y) + z = x + (y + z) = (x ˙+y) ˙+z = x ˙+(y ˙+z). (ii) x ˙+y = x + y = y + x = y ˙+x. (iii) x ˙+0 = x + 0 = x. (iv) λ ˙×(x ˙+y) = λ(x + y) = λx + λy = λ ˙×x ˙+λ ˙×y.

(v) (λ + μ) ˙ ×x = (λ + μ)x = λx + μx = λ ˙×x ˙+μ ˙×y. (vi) (λμ) ˙×x = (λμ)x = λ(μx) = λ ˙×(μ ˙×x). (vii) 1 ˙ ×x = 1x = x and 0 ˙×x = 0x = 0.

Exercise 13.2.6

(i) If

MAM^{−1} = D with D = (\begin{matrix}λ & 0 \\ 0 & μ \end{matrix})

then

t^{2} − a = det(tI − A) = det(tI − D) = (t − λ)(t − μ)

so t^{2} = a has a solution.

Suppose c^{2} = a ≠ 0. The eigenvalues are ±c which are distinct so by our standard arguments A is diagonalisable.

(ii) Our standard argument shows that A is not diagonalisable. (iii) We know that 2 has no square root in Q so by (i), no M exists. (iv) Since some a do not have square roots the corresponding a are

not diagonalisable.

(v) If A = M^{−1}DM with D diagonal then I = A^{2} = M^{−1}D^{2}M so D^{2} = I. Thus (since we work in ℤ_{2}, D = I so A = I which is absurd.

If we set e_{1} = (1, 1)^{T} and e_{2} = (1, 0)^{T} we have a basis and Ae_{1} = e_{1}, Ae_{1} = e_{1} + e_{2}, Thus setting

M = (\begin{matrix}1 & 1 \\ 1 & 0 \end{matrix})

we have

MAM^{1} = (\begin{matrix}1 & 0 \\ 1 & 1 \end{matrix}) .

Exercise 13.2.7

The tables are

\begin{matrix}+ & (0, 0) & (1, 0) & (0, 1) & (1, 1) \\ (0, 0) & (0, 0) & (1, 0) & (0, 1) & (1, 1) \\ (1, 0) & (1, 0) & (0, 0) & (1, 1) & (0, 1) \\ (0, 1) & (0, 1) & (1, 1) & (0, 0) & (1, 0) \\ (1, 1) & (1, 1) & (0, 1) & (1, 0) & (0, 0) \end{matrix}

\begin{matrix}× & (0, 0) & (1, 0) & (0, 1) & (1, 1) \\ (0, 0) & (0, 0) & (0, 0) & (0, 0) & (0, 0) \\ (1, 0) & (0, 0) & (1, 0) & (1, 1) & (0, 1) \\ (0, 1) & (0, 0) & (0, 1) & (1, 1) & (1, 0) \\ (1, 1) & (1, 0) & (1, 1) & (1, 0) & (0, 1) \end{matrix}

We check the axioms for a field.

(i) (a, b) + (c, d) = (a + c, b + d) = (c, d) + (a, b). (ii) We have

((a, b) + (c, d)) + (e, f) = (a + c, b + d) + (e, f)
= ((a + c) + e, (b + d) + f)
= (a + (c + e), b + (d + f)).

(iii) (a, b) + (0, 0) = (a, b) (iv) (a, b) + (a, b) = (0, 0)

(v) We have

(a, b) × (c, d) = (ac + bd, ad + bc + bd)
= (ca + db, da + cb + db) = (c, d) × (a, b).

(vi) We have ((a, b) × (c, d)) × (e, f) = (ac + bd, ad + bc + bd) × (e, f) = ((ac + bd)e + (ad + bc + bd)f,

(ac + bd)f + (ad + bc + bd)e + (ad + bc + bd)f)

= (a(ce + df) + b(de + cf + df),

a(de + cf + df) + b(ce + df) + b(de + cf + df))

= (a, b) × ((c, d) × (e, f))

(vii) (a, b) × (1, 0) = (a, b). (viii) By inspection of the multiplication table we see that if (a, b) ≠

(0, 0), we can find (c, d) with a × (a, b) × (c, d) = (1, 0).

(ix) We have

(a, b) × ((c, d) + (e, f)) = (a, b) × (c + e, d + f)

= (a(c + e) + b(c + e), a(d + f) + b(c + e) + b(d + f)) = (ac + bd, ad + bd + bc) + (ae + bf, af + bf + be) = (a, b) × (c, d) + (a, b) × (e, f)

(x) By inspection of the multiplication table, all the elements are squares.

Exercise 13.2.8

(i) Suppose 2 ≠ 0.

If A = B + C with B^{T} = B and C^{T} = −C then

2B = B + C + B − C = (B + C) + (B + C)^{T} = A + A^{T}

so B = 2^{−1}(A + A^{T}) and C = A − B = 2^{−1}(A − A^{T}).

Conversely if B = 2^{−1}(A + A^{T}) and C = A − B = 2^{−1}(A − A^{T}) then

B^{T} = 2^{−1}(A^{T} + A^{T T}) = B
C^{T} = 2^{−1}(A^{T} − A^{T T}) = −C
B + C = 2^{−1}(A + A^{T}) + 2^{−1}(A − A^{T}) = A

(ii) Suppose 2 = 0. Then a + a = (1 + 1)a = 2a = 0a = 0 so a = −a and (a_{ij}) = (−a_{ij}) so A = −A. Thus

A = A^{T} ⇔ A = −A^{T} .

We work over Z_{2}. If A = B+C with B^{T} = B, C^{T} = C then A^{T} = A. Thus

(\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix})

cannot be written as the sum of a symmetric and an antisymmetric matrix.

On the other hand the zero matrix 0 and the identity matrix I are symmetric so antisymmetric and

0 = 0 + 0 = I + I.

Exercise 13.2.9

(i) Observe that 0^{2} + 0 = 0 + 0 and 1^{2} + 1 = 1 + 1 = 0 so x^{2} = x = 0 for all x ∈ ℤ_{2}.

(ii) If 𝔾 is a finite field there are only a finite number of distinct functions f : 𝔾 → 𝔾 but there are infinitely many polynomials. Thus there must exist distinct polynomials P and Q which give rise to the same function. Consider P − Q.

Exercise 13.3.1

If U is a vector space over 𝔾 then W ⊆ U is a subspace if and only if.

(i) 0 ∈ W. (ii) x, y ∈ W → x = y ∈ W. (iii) λ ∈ 𝔾, x ∈ W ⇒ λx ∈ W. If 𝔾 = ℤ_{2} then, if λ ∈ 𝔾, x ∈ W either λ = 0 and λx = 0 so, if

condition (ii) is true λx ∈ W, or λ = 1 and λx = x ∈ W. Thus we can drop condition (iii) in this case.

Exercise 13.3.2

If U is infinite then all three statements are false. If U is finite then (since U spans itself) it has a finite spanning set

and so a basis e_{1}, e_{2}, …, e_{m}. The elements of U are the vectors

∑_{j=1}^{m} x_{j}e_{j}

with x_{j} ∈ ℤ_{2}, each choice of the x_{j} giving a distinct vector so U has 2^{m} elements.

If U is isomorphic to Z_{2}^{q} they must have the same number of elements so q = m.

The map θ : Z_{2}^{m} → U given by

θ(x_{1}, x_{2}, …, x_{m}) = ∑_{j=1}^{m} x_{j}e_{j}.

Thus the three given statements are equivalent.

Exercise 13.3.3

We have

α(λe+μf) = ∑_{j=1}^{n} p_{j}(λe_{j} +μf_{j}) = λ ∑_{j=1}^{n} p_{j}e_{j} +μ ∑_{j=1}^{n} p_{j}f_{j} = λα(e)+μα(f)

so α is linear.

Let e(j) be the vector with 1 in the jth place and 0 elsewhere. If α is linear, then, setting p_{j} = αe(j) we have

α(e) = α (∑_{j=1}^{n} e_{j}e(j)) = ∑_{j=1}^{n} e_{j}αe(j) = ∑_{j=1}^{n} p_{j}e_{j}.

Exercise 13.3.4

Codeword (0, 0, 0, 1, 1, 1, 1). Error in fourth place gives (0, 0, 0, 0, 1, 1, 1). z_{1} = 0, z_{2} = 0, z_{3} = 1 tells us error in 4th place. Recover (0, 0, 0, 1, 1, 1, 1).

Exercise 13.3.5

The tape will be accepted if each line contains an even number of errors. Since the probability of errors is small (and the number of bits on each line is small) the probability of one error is much greater than the probability of an odd number of errors greater than 1. Thus

Pr(odd number of errors in one line) ≈ Pr(exactly one error in one line)

= 8 × 10^{−4} × (1 − 10^{−4})^{7} ≈ 8 × 10^{−4}.

Since the the probability λ of an odd number of errors in one line is very small but there are a large number N of lines, we may use the Poisson approximation to get

1 − Pr(odd number of errors in some line) ≈ e^{−λN} ≈ e^{−8} ≈ 0.00034

and conclude that the probability of acceptance is less than .04%.

If we use the Hamming scheme, then, instead of having 7 freely chosen bits (plus a check bit) on each line, we only have 4 freely chosen bits (plus three check bits plus an unused bit) per line so we need approximately

\frac{1}{4} × 7 × 10^{4} = 1.75 × 10^{4}

lines.

If a line contains at most one error, it will be correctly decoded. A line will fail to be correctly decoded if it contains exactly two errors and it may fail to be correctly decoded if it contains more than two errors. Since the probability of errors is small (and the number of bits on each line is small), the probability of two errors is much greater than the probability of more than two. Thus

Pr(decoding failure for one line) ≈ Pr(exactly two errors in one line)

= \binom{7}{2} × (10^{−4})^{2} × (1 − 10^{−4})^{5} ≈ 21 × 10^{−8}.

Since the the probability of a decoding error in one line is very small but there are a large number of lines, we may use the Poisson approximation to get

Pr(decoding error for some line) = 1 − Pr(no decoding error in any line)
≈ 1 − e^{−21×10^{−8}×17500} ≈ 1 − e^{−.003675} ≈ .9963

and conclude that the probability of a correct decode is greater than 99.6%.

Exercise 13.3.6

(i) Observe that

(\begin{matrix}1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 \end{matrix}) (\begin{matrix}c_{1} \\ c_{2} \\ c_{3} \\ c_{4} \\ c_{5} \\ c_{6} \\ c_{7} \end{matrix}) = (\begin{matrix}0 \\ 0 \\ 0 \end{matrix})

can be rewritten as

c_{1} + c_{3} + c_{5} + c_{7} ≡ 0
c_{2} + c_{3} + c_{6} + c_{7} ≡ 0
c_{4} + c_{5} + c_{6} + c_{7} ≡ 0.

(ii) Observe that Ae_{j}^{T} is the j row of A and that the jth row of A is the binary digits of j written in reverse order.

By linearity A(c + e_{j})^{T} = Ac^{T} + Ae_{j}^{T} = Ae_{j}^{T} = (a_{1}, a_{2}, a_{3})^{T} so the statement follows.

Exercise 13.3.7

Ae_{i}^{T} ≠ Ae_{j}^{T} for i ≠ j since otherwise an error in the jth place would have the same effect as an error in the ith place. Since

x + y = 0 ⇔ x = y

we have

A(e_{i}^{T} + e_{j}^{T}) ≠ 0^{T}

Thus by linearity

A(c + e_{i})^{T} + e_{j})^{T}) = Ac + Ac^{T} + Ae_{j}^{T}
= A(e_{i}^{T} + e_{j}^{T}) ≠ 0

and c + e_{i})^{T} + e_{j} is not a codeword.

(ii) Ae_{j} ∈ ℤ^{3} \ {0}. Since the Ae_{j} are distinct for distinct j and {j : 1 ≤ j ≤ 7} has the same number of elements as ℤ^{3} \ {0}^{T} we have bijection and so since

A(e_{i}^{T} + e_{j}^{T}) ≠ 0

there must be a k with

A(e_{i}^{T} + Ae_{j}^{T}) = A(e_{k}^{T}).

If k = j we would have Ae_{i}^{T} = 0^{T} which is impossible. Thus j ≠ k and similarly j ≠ i.

If we have an error in the i and jth place the Hamming system will assume an error in the kth place.

Exercise 13.3.9

Observe that if A = (a_{ij})

c ∈ C ⇔ Ac^{T} = 0^{T}

⇔ ∑_{j=1}^{n} a_{ij}c_{j} = 0 for 1 ≤ j ≤ r

⇔ α_{j}c = 0 for 1 ≤ j ≤ r

where α_{j} ∈ (Z_{2}^{n})′ is given by

α_{j}c = ∑_{j=1}^{n} a_{ij}c_{j}.

Exercise 13.3.10

The null space of a linear map is a subspace.

Exercise 13.3.12

Let α be the linear map associated with A for the standard bases. Since A has rank r so does α and

dim C = dim α^{−1}(0) = n − rank(α) = n − r

so C has 2^{n−r} elements.

Exercise 13.3.13

(i) Suppose C has basis g(j) [1 ≤ r ≤ n]. Consider an r×n matrix U with jth row g(j). We can use elementary elementary row operations so that (after reordering coordinates if necessary) we get V = (I B). Since elementary row operations leave the space spanned by the rows of a matrix unchanged, C has basis the rows e(j) of V. This gives the required result.

(ii) The rank nullity theorem tells us that the Hamming code has dimension 4. If we do not interchange columns then considering the effect of choosing one of the c_{3}, c_{5}, c_{6}, c_{7} to be 1 and the rest 0 we obtain 4 independent vectors.

(1, 1, 1, 0, 0, 0, 0), (1, 0, 0, 1, 1, 0, 0)
(0, 1, 0, 1, 0, 1, 0), (1, 1, 0, 1, 0, 0, 1)

which therefore form a basis.

(iii) α is linear and surjective so (since the spaces have the same dimension) an isomorphism.

(λx + μy, β(λx + μy)) = α(λx + μy)
= λαx + μαy
= λ(x, βx) + μ(y, βy)
= (λx + μy, λβx + μβy)

so

β(λx + μy) = λβx + μβy

and β is linear.

(iv) Neither is true. Take n = 2r b(j) = 0.

Exercise 13.3.14

The probability of exactly 2 errors in a line (ie codeword), in which case the correction is erroneous, is

\binom{7}{2} (\frac{1}{10})^{2} (\frac{9}{10})^{5} ≈ 0.124

The probability of no such problem in 17 500 lines is less than (9/10)^{17500} which is negligible.

Exercise 13.3.15

If p = 1/2 then, whatever the message sent, the received sequence is a series of independent random variables each taking the value 0 with probability 1/2 and the value 1 with probability 1/2.

If p > 1/2, then replacing each received 0 by 1 and each received 1 by 0 gives a system for the probability of an error in one bit is 1 − p.

Exercise 13.3.16

By definition

A_{1} = (1), A_{2} = (\begin{matrix}1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix}) , A_{3} = (\begin{matrix}1 & 0 & 1 & 0 & 1 & ) & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 \end{matrix})

and

A_{4} = (\begin{matrix}1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{matrix})

(ii) The 2^{r}th column has a 1 in the rth place with all other entries zero, so these columns are linearly independent and A has rank n.

(iii) Let us write j = ∑_{i=1}^{n+1} a_{ij}2^{i−1} with a_{ij} ∈ {0, 1}. Then, if 0 ≤ j < 2^{n}, a_{ij} = a_{i(j+2^{n})} for 1 ≤ i ≤ n − 1 and a_{n,j} = 0, a_{n,(j+2^{n})} = 1. We have a_{i2^{n−1}} = δ_{in}.

Exercise 13.3.17

Let e_{k} be a row vector of length 2^{n} − 1 with 1 in the kth place an zero elsewhere. Write k = ∑_{i=1}^{n−1} b_{i}2^{i−1} with b_{i} ∈ {0, 1}. Then, if c is a code word,

A(c + e_{k})^{T} = Ae_{k}^{T} = b,

so we can recover from a single mistake by the rules:-

(1) If x received and Ax^{T} = 0^{T}, assume message correct.

(2) If x received and Ax^{T} = b^{T}, assume mistake in ∑_{i=1}^{n} b_{i}2^{i−1}th place.

Since A has full rank (see previous question), the rank-nullity theorem tells us that C has 2^{n−r} elements.

Exercise 13.3.18

Since the error rate is very low we can use the Poisson approximation. (But exact calculation is easy using a calculator.) We have a rate (5 × 10^{7}) × 10^{−7} = 10 so the probability that no error occurs is roughly e^{−5} which is very small.

By the Poisson approximation (or directly by bounding terms) the probability of more than one error in a line is roughly

p = \frac{1}{2!}λ^{2}e^{−λ}

with λ = 63 × 10^{−7}. Thus

p ≈ 2 × 10^{−10}

There are roughly 10^{5} lines so the probability of an error in the decoded message is (using the Poisson approximation) less than about exp(−2× 10^{−5}) which is pretty small.

Exercise 14.1.3

C([a, b]) is a subset of the vector space ℝ^{[a,b]} of functions f : [a, b] → ℝ. We observe that 0 ∈ C([a, b]) and that

f, g ∈ C([a, b]) ⇒ λf + μg ∈ C([a, b])

so C([a, b]) is a subspace of ℝ^{[a,b]} and so a vector space.

We have

⟨f, f⟩ = ∫_{a}^{b} f(x)^{2} dx ≥ 0

and since the integral of a positive continuous function is zero if and only if the function is zero

⟨f, f⟩ = 0 ⇒ ∫_{a}^{b} f(x)^{2} dx = 0 ⇒ f × f = 0 ⇒ f = 0.

Also

⟨f, g⟩ = ∫_{a}^{b} f(x)g(x) dx = ∫_{a}^{b} g(x)f(x) dx = ⟨g, f⟩,

⟨f, g + h⟩ = ∫_{a}^{b} f(x)(g(x) + h(x) dx
= ∫_{a}^{b} f(x)g(x) + f(x)h(x) dx
= ⟨f, g⟩ + ⟨f, g⟩

and

⟨λf, g⟩ = ∫_{a}^{b} λf(x)g(x) dx = λ ∫_{a}^{b} f(x)g(x) dx = λ⟨f, g⟩.

Exercise 14.1.4

(i) If x = 0 or y = 0, we have equality. If λ is a real number then we have

0 ≤ ∥λx + y∥^{2}

= ⟨λx + y, λx + y⟩

= λ^{2}∥x∥^{2} + 2λ⟨x, y⟩ + ∥y∥^{2}

= (λ∥x∥ + \frac{⟨x, y⟩}{∥x∥})^{2} + ∥y∥^{2} − (\frac{⟨x, y⟩}{∥x∥})^{2} .

If we set

λ = −\frac{⟨x, y⟩}{∥x∥^{2}},

we obtain

0 ≤ (λx + y) · (λx + y) = ∥y∥^{2} − (\frac{x · y}{∥x∥})^{2} .

Thus

⋆  ∥y∥^{2} − (\frac{⟨x, y⟩}{∥x∥})^{2} ≥ 0

with equality only if

0 = ∥λx + y∥

so only if

λx + y = 0.

Rearranging the terms in ⋆ we obtain

⟨x, y⟩^{2} ≤ ∥x∥^{2}∥y∥^{2}

and so

|⟨x, y⟩| ≤ ∥x∥∥y∥

with equality only if

λx + y = 0.

We observe that, if λ′, μ′ ∈ ℝ are not both zero and λ′x = μ′y, then |⟨x, y⟩| = ∥x∥∥y∥.

Thus |⟨x, y⟩| ≤ ∥x∥∥y with equality if and only if x and y are linearly dependent.

(ii) By definition ∥x∥ ≥ 0. Further

∥x∥ ≥ 0 ⇔ ⟨x, x⟩ = 0 ⇔ x = 0.

Since ⟨λx, λx⟩ = λ^{2}⟨x, x⟩, we have ∥λx∥ = |λ|∥x∥.

Finally

∥x + y∥^{2} = ∥x∥^{2} + 2⟨x, y + ∥y∥^{2}
≤ ∥x∥^{2} + 2∥x∥∥y∥ + ∥y∥^{2}
= (∥x∥ + ∥y∥)^{2},

so ∥x + y∥ ≤ ∥x∥ + ∥y∥.

(iii) If we use the inner product of Exercise 14.1.3, the Cauchy– Schwarz inequality takes the form

(∫_{a}^{b} f(t)g(t) dt)^{2} ≤ ∫_{a}^{b} f(t)^{2} dt ∫_{a}^{b} g(t)^{2} dt.

Exercise 14.1.5

(i) We have p_{0} = 1, so the result is true for n = 0. If p_{r} is polynomial of degree r for r ≤ n − 1, then

∑_{j=0}^{n−1} \frac{⟨q_{n}, p_{j}⟩}{∥p_{j}∥^{2}}

is a polynomial of degree at most n − 1, so

p_{n} = q_{n} − ∑_{j=0}^{n−1} \frac{⟨q_{n}, p_{j}⟩}{∥p_{j}∥^{2}}p_{j}

is a monic polynomial of degree n.

p_{1}(x) = x − \frac{⟨q_{1}, p_{0}⟩}{∥p_{0}∥^{2}}p_{0}(x) = x.

p_{1}(x) = x^{2} − ∑_{j=0}^{1} \frac{⟨q_{n}, p_{j}⟩}{∥p_{j}∥^{2}}p_{j}(x)

= x^{2} − (∫_{−1}^{1} t^{2} dt) (∫_{−1}^{1} 1^{2} dt) = x^{2} − \frac{1}{3}.

(ii) 0 is a polynomial of degree less than n. If P and Q are polynomials of degree at most n so is λP = μQ. Writing q_{r}(x) = x^{r}, we have the trivial equality

∑_{r=0}^{n} a_{r}x^{r} = (∑_{r=0}^{n} a_{r}q_{r}) (x)

so the q_{j} with 0 ≤ j ≤ n span P_{n}. Since a non-zero polynomial of degree at most n can have at most n zeros

∑_{r=0}^{n} a_{r}q_{r} = 0 ⇒ a_{r} = 0 [0 ≤ r ≤ n]

so the q_{r} form a basis and P_{n} has dimension n + 1.

The p_{j}∥p_{j}∥^{−1} form a collection of n + 1 orthonormal vectors so form an orthonormal basis.

(iii) If P is monic of degree n + 1, then

P = p_{n+1} + Q

with Q a polynomial of degree at most n Now

P ⊥ P_{n} ⇒ (P − p_{n+1}) ⊥ P_{n} ⇒ Q ⊥ P_{n}

⇒ Q ⊥ Q ⇒ ⟨Q, Q⟩ = 0 ⇒ Q = 0 ⇒ P = p_{n+1}.

By definition p_{n+1} is monic and

p_{n+1} ⊥ span{p_{0}, p_{1}, …, p_{n}}

and so, since P_{n} span{p_{0}, p_{1}, …, p_{n}}, we have p_{n+1} ⊥ P_{n}.

Exercise 14.1.11

(i) Observe that

∑_{j=1}^{n} |λ_{j}| ≥ ∑_{j=1}^{n} λ_{j} = ∫_{−1}^{1} 1 dt = 2.

Similarly ∑_{j=n+1}^{2n} |λ_{j}| ≥ 2.

(ii) We have

∑_{j=1}^{2n} μ_{j}P (t_{j}) = (ϵ^{−1} + 1) ∑_{j=1}^{n} λ_{j}P (t_{j}) − ϵ^{−1} ∑_{j=n+1}^{2n} λ_{j}P (t_{j}) = ∫_{−1}^{1} P (t) dt

for all polynomials P of degree n or less. Let f be the simplest piecewise linear functional with

f(t_{j} = \begin{cases}ϵ & if 1 ≤ j ≤ n and λ_{j} ≥ 0 \\ −ϵ & if 1 ≤ j ≤ n and λ_{j} < 0 \\ ϵ & if n + 1 ≤ j ≤ 2n and λ_{j} < 0 \\ −ϵ & if n + 1 ≤ j ≤ 2n and λ_{j} ≥ 0 \end{cases}

with

f(−1) = 0  if −1 ∉ {t_{j} : 1 ≤ j ≤ 2n}
f(1) = 0  if 1 ∉ {t_{j} : 1 ≤ j ≤ 2n}.

Then |f(t)| ≤ ϵ for all t ∈ [−1, 1], but

∑_{j=1}^{2n} μ_{j}P (t_{j}) = ∑_{j=1}^{2n} |μ_{j}| ≥ 2(1 + ϵ^{−1}) + 2ϵ^{−1}.

Exercise 14.1.12

(i) Observe that

⟨f, f⟩ = ∫_{a}^{b} f(x)^{2}r(x) dx ≥ 0,

and, since x ↦ f(x)^{2}r(x) is continuous and positive

⟨f, f⟩ = 0 ⇒ ∫_{a}^{b} f(x)^{2}r(x) dx = 0
⇒ f(x)^{2}r(x) = 0 ∀x
⇒ f(x) = 0 ∀x

Automatically

⟨f, g⟩ = ∫_{a}^{b} f(x)g(x)r(x) dx = ∫_{a}^{b} g(x)f(x)r(x) dx = ⟨g, f⟩

and

⟨λ_{1}f_{1} + λ_{2}f_{2}, g⟩ = ∫_{a}^{b}(λ_{1}f_{1}(x) + λ_{2}f_{2}(x))g(x)r(x) dx
= ∫_{a}^{b} λ_{1}f_{1}(x)g(x) + λ_{2}f_{2}g(x)r(x) dx
= λ_{1}⟨f_{1}, g⟩ + λ_{2}⟨f_{2}, g⟩.

(ii) This is just the Cauchy–Schwartz inequality

⟨f, f⟩⟨g, g⟩ ≥ ⟨f, g⟩^{2}.

(iii) We define P_{n} inductively. Set P_{0} = 1. Suppose that P_{j} is a monic polynomial of degree j with the P_{j} orthogonal with respect to our inner product [0 ≤ j ≤ n − 1]. Then, if Q_{n}(x) = x^{n},

P_{n} = Q_{n} − ∑_{j=0}^{n−1} \frac{⟨Q_{n}, P_{j}⟩}{⟨P_{j}, P_{j}⟩}

is a monic polynomial of degree n orthogonal to all the P_{j} with j ≤ n − 1.

Since the subspace P_{n−1} of polynomials of degree at most n − 1 has dimension n and the P_{j} are non-zero and mutually orthogonal the P_{j} with j ≤ n−1 form a basis for P_{n−1} and ⟨P_{n}, Q⟩ = 0 for all polynomials Q of degree n − 1 or less.

(iii) Let x_{1}, x_{2}, … x_{k} be the real roots of P_{n} which have odd order and lie in (a, b). Setting Q(x) = ∏_{j=1}^{k}(x − x_{k}), we see that P_{n}Q is single signed so

⟨P_{n}, Q⟩ = ∫_{a}^{b} P_{n}(x)Q(x)r(x) dx ≠ 0

and k ≥ n. Thus k = n and P_{n} has n distinct real roots all lying in (a, b).

If we set e_{v}(x) = ∏_{j≠v,1≤j≤n}(x − x_{j})(x_{v} − x_{j})^{−1} and

λ_{v} = ∫_{a}^{b} e_{v}(x)r(x) dx

then (since f − ∑_{v=1}^{n} f_{(}x_{v})e_{v} is a polynomial of degree at most n − 1 vanishing at n points) we have f = ∑_{v=1}^{n} f(x_{v})e_{v} and

∫_{a}^{b} f(x)r(x) dx = ∑_{v=1}^{n} λ_{v}f_{(}x_{v})

for all f polynomials of degree n − 1 or less.

If f is polynomials of degree 2n − 1 or less, then f = gP_{n} + h with g and h polynomials of degree n − 1 or less, so

∫_{a}^{b} f(x)r(x) dx = ∫_{a}^{b} g(x)P_{n}r(x) dx + ∫_{a}^{b} h(x)r(x) dx

= 0 + ∑_{v=1}^{n} λ_{v}g_{(}x_{v})

= ∑_{v=1}^{n} λ_{v}g_{(}x_{v}) + ∑_{v=1}^{n} λ_{v}0

= ∑_{v=1}^{n} λ_{v}g_{(}x_{v}) + ∑_{v=1}^{n} λ_{v}P_{n}(x_{v})h(x_{v})

= ∑_{v=1}^{n} λ_{v}f(x_{v}).

Exercise 14.1.14

We need to show that ⟨P_{n}, qP_{n−1}⟩ > 0. Now qP_{n−1} and P_{n} are both monic polynomials of degree n, so

qP_{n−1} − P_{n} has degree at most n − 1. Thus

⟨P_{n}, qP_{n−1} − P_{n}⟩ = 0

and

⟨P_{n}, qP_{n−1}⟩ = ⟨P_{n}, P_{n}⟩ > 0.

Exercise 14.1.16

Observe that

0 ≤ ∫_{1}^{1} f_{n}(x)^{2} dx = (2^{n+1} − 1) ∫_{−2^{−8n}}^{2^{−8n}} f_{n}(x)^{2} dx
≤ 2^{n+1}2^{−8n+1}2^{2n} = 2^{−5n+2} → 0

as n → ∞ so ∥f_{n}∥ → 0.

However, if u is an integer with |u| < 2^{m} − 1, then f(u2^{−m}) = 2^{n} when n ≥ m and so f_{n}(u2^{−m}) → ∞ as n → ∞.

Exercise 14.2.1

If dim U = 0, there is nothing to prove.

Otherwise, we construct the e_{j} inductively. Since dim U > 0 we can find a a ≠ 0. Set e_{1} = ∥a∥^{−1}a. If e_{1}, e_{2}, …, e_{r} have been found as orthonormal vectors either r = n and we have an orthonormal basis or we can find

b ∉ span{e_{1}, e_{2}, …, e_{r}}

The Gram–Schmidt method now produces an e_{r+1} with e_{1}, e_{2}, …, e_{r+1} orthonormal.

The process terminates in an orthonormal basis.

Since the e_{j} are linearly independent and span the map

θ (∑_{j=1}^{n} x_{j}e_{j}) = (x_{1}, x_{2}, …, x_{n})^{T}

is a well defined bijection. Since

θ (λ ∑_{j=1}^{n} x_{j}e_{j} + μ ∑_{j=1}^{n} y_{j}e_{j}) = θ (∑_{j=1}^{n}(λx_{j} + μy_{j})e_{j})

= (λx_{1} + μy_{1}, λx_{2} + μy_{2}, …, λx_{n} + μy_{n})^{T}

= λ(x_{1}, x_{2}, …, x_{n})^{T} + μ(y_{1}, y_{2}, …, y_{n})^{T}

= λθ (λ ∑_{j=1}^{n} x_{j}e_{j}) + μθ (∑_{j=1}^{n} y_{j}e_{j})

θ is linear. Finally,

θ (∑_{j=1}^{n} x_{j}e_{j}))·θ (∑_{j=1}^{n} y_{j}e_{j}) = ∑_{j=1}^{n} x_{j}y_{j} = ⟨θ (∑_{j=1}^{n} x_{j}e_{j}) , θ (∑_{j=1}^{n} y_{j}e_{j})⟩ .

Exercise 14.2.3

Choose an orthonormal basis

e_{1}, e_{2}, …, e_{r}

for V. Then by Bessel’s inequality

∥a − ∑_{j=1}^{r} λ_{j}e_{j}∥

has a unique minimum when λ_{j} = ⟨a, e_{j}⟩.

If x ∈ V

⟨a − x, v⟩ = 0 ∀v ∈ V ⇔ ⟨a − x, e_{j}⟩ = 0 ∀1 ≤ j ≤ r

⇔ ⟨x, e_{j}⟩ = ⟨a, e_{j}⟩ ∀1 ≤ j ≤ r

⇔ x = ∑_{j=1}^{r}⟨a, e_{j}⟩e_{j}

so we are done.

Exercise 14.2.5⋆

Exercise 14.2.6

Observe that

⟨u, a⟩ = ⟨∑_{j=1}^{n} α(e_{j})e_{j}, u⟩

= ∑_{j=1}^{n} α(e_{j})⟨e_{j}, a⟩

= α (∑_{j=1}^{n}⟨e_{j}, a⟩e_{j})

= αu.

Exercise 14.2.7

We have a map α : U → ℝ so we only need check linearity. But, if u, v ∈ U and λ, μ ∈ ℝ we have α(λu + μv) = ⟨λu + μv, a⟩ = λ⟨u, a⟩ + μ⟨v, a⟩ = λαu + μαv,

so we are done.

Exercise 14.2.10

Observe that

⟨u, Φ(λα + μβ)v⟩ = ⟨(λα + μβ)u, v⟩

= ⟨λα(u) + μβ(u), v⟩ = λ⟨αu, v⟩ + μ⟨βu, v⟩

= λ⟨u, α^{∗}v⟩ + μ⟨u, β^{∗}v⟩ = ⟨u, (λα^{∗} + μβ^{∗})v⟩

= ⟨u, (λΦα + μΦβ)v⟩

for all u ∈ U, so

Φ(λα + μβ)v = (λΦα + μΦβ)v

for all v ∈ U so

Φ(λα + μβ) = λΦα + μΦβ

for all α, β ∈ L(U, U) and λ, μ ∈ 𝔽, so Φ is linear.

Exercise 14.2.12

D is a subset of the inner product space C([0, 1]). 0 ∈ D and

λ, μ ∈ ℝ, f, g ∈ D ⇒ λf + μg ∈ D.

Thus D is a subspace of the inner product space C([0, 1]) and so an inner product space.

Further,

α(λf + μg) = (p(λf + μg)′)′ = (λpf′ + μgf′)′ = λαf + μαg. so α is linear.

By integration by parts,

⟨αf, g⟩ = ∫_{0}^{1}(f′p)′(t)g(t) dt
= [f′(t)p(t)g(t)]_{0}^{1} − ∫_{0}^{1} f′(t)p(t)g′(t) dt
= − ∫_{0}^{1} f′(t)p(t)g′(t) dt
= ⟨αg, f⟩ = ⟨f, αg⟩,

so α^{∗} = α.

If we write (βf)(t) = tf(t) then β is a self-adjoint linear functional.

Exercise 14.3.2

We have

⟨z, λw⟩_{P} = ⟨λw, z⟩_{P}^{∗} = (λ^{∗}⟨w, z⟩_{P} )^{∗} = λ⟨w, z⟩_{P}^{∗} = λ⟨z, w⟩_{P} .

Let ⟨ , ⟩_{M} be defined as in the question. (i) Since ⟨z, z⟩_{P} is real and positive, so is ⟨z, z⟩_{M} = ⟨z, z⟩_{P}^{∗}.

(ii) ⟨z, z⟩_{M} = 0 ⇒ ⟨z, z⟩_{P} = 0 ⇒ z = 0.

(iii) ⟨λz, w⟩_{M} = ⟨λz, w⟩_{P}^{∗} = (λ^{∗}⟨z, w⟩_{P} )^{∗} = λ⟨z, w⟩_{P}^{∗} = λ⟨z, w⟩_{M}.

(iv) We have

⟨z + u, w⟩_{M} = ⟨z + u, w⟩_{P}^{∗}
= (⟨z, w⟩_{P} + ⟨u, w⟩_{P} )^{∗}
= ⟨z, w⟩_{P}^{∗} + ⟨u, w⟩_{P}^{∗}
= ⟨z, w⟩_{M} + ⟨u, w⟩_{M}

(v) ⟨w, z⟩_{M} = ⟨w, z⟩_{P}^{∗} = ⟨z, w⟩_{P}^{∗∗} = ⟨z, w⟩_{M}^{∗}.

Essentially same arguments show that if ⟨ , ⟩_{M} is a mathematician’s inner product then

⟨z, w⟩_{P} = ⟨z, w⟩_{M}^{∗}

defines a physicist’s inner product.

Exercise 14.3.3

C([a, b]) is a subset of the vector space ℂ^{[a,b]} of functions f : [a, b] → ℂ. We observe that 0 ∈ C([a, b]) and that

f, g ∈ C([a, b]) ⇒ λf + μg ∈ C([a, b])

so C([a, b]) is a subspace of ℝ^{[a,b]} and so a vector space.

We have

⟨f, f⟩ = ∫_{a}^{b} |f(x)|^{2} dx ≥ 0

and since the integral of a positive continuous function is zero if and only if the function is zero

⟨f, f⟩ = 0 ⇒ ∫_{a}^{b} |f(x)|^{2} dx = 0 ⇒ f × f = 0 ⇒ f = 0.

Also

⟨f, g⟩ = ∫_{a}^{b} f(x)g(x)^{∗} dx = ∫_{a}^{b}(g(x)f(x)^{∗})^{∗} dx
= (∫_{a}^{b} g(x)f(x)^{∗} dx)^{∗} = ⟨g, f⟩,

and

⟨f, g + h⟩ = ∫_{a}^{b} f(x)(g(x) + h(x)^{∗} dx
= ∫_{a}^{b} f(x)g(x)^{∗} + f(x)h(x)^{∗} dx = ⟨f, g⟩ + ⟨f, g⟩,

whilst

⟨λf, g⟩ = ∫_{a}^{b} λf(x)g(x)^{∗} dx = λ ∫_{a}^{b} f(x)g(x)^{∗} dx = λ⟨f, g⟩.

Exercise 14.3.4

(i) We have

⟨f, e_{j}⟩ = ⟨∑_{r=1}^{n} a_{r}e_{r}, e_{j}⟩

= ∑_{r=1}^{n} a_{r}⟨e_{r}, e_{j}⟩ = ∑_{r=1}^{n} a_{r}δ_{rj} = a_{j}

Note that

⟨e_{j}, f⟩ = ⟨f, e_{j}⟩^{∗} = a_{j}^{∗}.

(ii) We have ⟨e_{r}, e_{s}⟩ = ∫_{0}^{1} exp(2πirt) exp(−2πist) dt = ∫_{0}^{1} exp (2πi(r−s)t) dt = δ_{rs}.

Exercise 14.3.5

(i) Observe that

⟨v, e_{r}⟩ = ⟨x, e_{r}⟩ − ∑_{j=1}^{k}⟨x, e_{r}⟩⟨e_{j}, e_{r}⟩ = ⟨x, e_{r}⟩ − ⟨x, e_{r}⟩ = 0

for all 1 ≤ r ≤ k.

(ii) If v = 0, then

x = ∑_{j=1}^{k}⟨x, e_{j}⟩e_{j} ∈ span{e_{1}, e_{2}, …, e_{k}}.

If v ≠ 0, then ∥v∥ ≠ 0 and

x = v + ∑_{j=1}^{k}⟨x, e_{j}⟩e_{j}

= ∥v∥e_{k+1} + ∑_{j=1}^{k}⟨x, e_{j}⟩e_{j} ∈ span{e_{1}, e_{2}, …, e_{k+1}}.

(iii) If e_{1}, e_{2}, …, e_{k} do not form a basis for U, then we can find

x ∈ U \ span{e_{1}, e_{2}, …, e_{k}}.

Defining v as in part (i), we see that v ∈ U and so the vector e_{k+1} defined in (ii) lies in U. Thus we have found orthonormal vectors e_{1}, e_{2}, …, e_{k+1} in U. If they form a basis for U we stop. If not, we repeat the process. Since no set of n + 1 vectors in U can be orthonormal (because no set of n + 1 vectors in U can be linearly independent), the process must terminate with an orthonormal basis for U of the required form.

Exercise 14.3.7

Let V have an orthonormal basis e_{1}, e_{2}, …, e_{n−1}. By using the Gram–Schmidt process, we can find e_{n} such that e_{1}, e_{2}, …, e_{n} are orthonormal and so a basis for U. Setting b = e_{n}, we have the result.

Exercise 14.3.8

Uniqueness. If αu = ⟨u, a_{1}⟩ = ⟨u, a_{2}⟩ for all u ∈ U, then

⟨u, a_{1} − a_{2}⟩ = 0

for all u ∈ U and, choosing u = a_{1} − a_{2}, we conclude, in the usual way, that a_{1} − a_{2} = 0.

Existence. If α = 0, then we set a = 0. If not, then α has rank 1 (since α(U) = ℝ) and, by the rank-nullity theorem, α has nullity n − 1. In other words,

α^{−1}(0) = {u : αu = 0}

has dimension n − 1. By Lemma 14.2.2, we can find a b ≠ 0 such that

α^{−1}(0) = {x ∈ U : ⟨x, b⟩ = 0},

that is to say,

α(x) = 0 ⇔ ⟨x, b⟩ = 0

If we now set

a = ∥b∥^{−2}α(b)b,

we have

αx = 0 ⇔ ⟨x, a⟩ = 0

and

αa = ∥b∥^{−2}α(b)^{2} = ⟨a, a⟩.

Now suppose that u ∈ U. If we set

x = u − \frac{αu}{αa}a,

then αx = 0 so ⟨x, a⟩ = 0, that is to say,

0 = ⟨u − \frac{αu}{αa}a, a⟩ = ⟨u, a⟩ − \frac{αa}{⟨a, a⟩}αu = ⟨u, a⟩

and we are done.

Exercise 14.3.9

Since

⟨λu + μv, a⟩ = λ⟨u, a⟩ + μ⟨u, a⟩

θ(a ∈ U′. The previous exercise tells us that θ is surjective.

θ(a) = θ(b) ⇒ ⟨b − a, b − a⟩ ⇒ a = b

so θ is injective.

Finally,

θ(λa + μb)(u) = ⟨u, λa + μb⟩
= λ^{∗}⟨u, a⟩ + μ^{∗}⟨u, b⟩
= λ^{∗}θ(a)u + μ^{∗}θ(b)u
= (λ^{∗}θ(a) + μ^{∗}θ(b))u

for all u ∈ U. Thus

θ(λa + μb) = λ^{∗}θ(a) + μ^{∗}θ(b)

as required.

Exercise 14.3.11

Observe that, if v ∈ U, the map

θ(u) = ⟨αu, v⟩

lies in U′, so the Riesz representation theorem tells us that there is a unique α^{∗}v ∈ U such that

⟨αu, v⟩ = ⟨u, α^{∗}v⟩

for all v ∈ U.

Since

⟨u, α^{∗}(λv + μw)⟩ = ⟨αu, (λv + μw)⟩
= λ^{∗}⟨αu, v⟩ + μ^{∗}⟨αu, w⟩
= λ^{∗}⟨u, α^{∗}v⟩ + μ^{∗}⟨u, α^{∗}w⟩
= ⟨u, λα^{∗}v + μα^{∗}w⟩

for all u ∈ U, we have

α^{∗}(λv + μw) = λα^{∗}v + μα^{∗}w

so α^{∗} ∈ L(U, U).

Again,

⟨u, (λα + μβ)^{∗}v⟩ = ⟨(λα + μβ)u, v⟩
= λ⟨αu, v⟩ + μ⟨βu, v⟩
= λ⟨u, α^{∗}v⟩ + μ⟨u, β^{∗}v⟩
= ⟨u, λ^{∗}α^{∗}v + μ^{∗}β^{∗}v⟩
= ⟨u, (λ^{∗}α^{∗} + μ^{∗}β^{∗})v⟩

for all u ∈ U, so

(λα + μβ)^{∗}v = (λ^{∗}α^{∗} + μ^{∗}β^{∗})v

for all v ∈ U, so

(λα + μβ)^{∗} = λ^{∗}α^{∗} + μ^{∗}β^{∗}

and he map Ψ : L(U, U) → L(U, U) given by Ψα = α^{∗} satisfies

Ψ(λα + μβ) = λ^{∗}Ψα + μ^{∗}Ψβ.

Finally,

⟨α^{∗∗}u, v⟩ = ⟨v, α^{∗∗}u⟩ = ⟨α^{∗}v, u⟩^{∗}
= ⟨u, α^{∗}v⟩ = ⟨αu, v⟩

for all v ∈ U, so

α^{∗∗}u = u

for all u ∈ U, so α^{∗∗} = α and Ψ is its own inverse, so bijective so an anti-isomorphism.

Exercise 14.3.12

If α has matrix A = (a_{ij}) and α^{∗} has matrix B = (b_{ij}) with respect to the given basis, then

b_{rj} = ⟨∑_{i=1}^{n} b_{ij}e_{i}, e_{r}⟩

= ⟨α^{∗}e_{j}, e_{r}⟩ = ⟨e_{r}α^{∗}e_{j}⟩^{∗} = ⟨αe_{r}, e_{j}⟩

= ⟨∑_{k=1}^{n} a_{kr}e_{k}, e_{j}⟩^{∗}

= a_{jr}^{∗},

so B = A^{∗}.

Exercise 14.3.13

(i) Since V and V^{⊥} are complementary,

dim V + dim V^{⊥} = dim U.

For the same reason, dim V^{⊥} +dim V^{⊥⊥} = dim U, so dim V^{⊥⊥} +dim V.

Since

x ∈ V ⇒ ⟨x, y⟩ = ⟨y, x⟩ = 0 ∀y ∈ V^{⊥}

we have V ⊆ V^{⊥⊥} so V = V^{⊥⊥}.

(ii) The statement is equivalent to saying that, if u ∈ U, there are unique a ∈ V and b ∈ V^{⊥} such that u = a + b. and this is the same as saying that V and V^{⊥} are complementary.

(iii) We have

λ_{1}u_{1} + λ_{2}u_{2} = (λ_{1}πu_{1} + λ_{2}πu_{2}) + (λ_{1}(ι − π)u_{1} + λ_{2}(ι − π)u_{2}) and λ_{1}πu_{1} + λ_{2}πu_{2} ∈ V, λ_{1}(ι − π)u_{1} + λ_{2}(ι − π)u_{2} ∈ V^{⊥} so, by uniqueness,

π(λ_{1}u_{1} + λ_{2}u_{2}) = λ_{1}πu_{1} + λ_{2}πu_{2}.

Thus π and so ι − π are linear.

Since πu ∈ V we have, by uniqueness π^{2}U = π(π(u) = πu for all u ∈ U so π = π^{2}.

If x, y ∈ U, then

⟨πx, y⟩ = ⟨πx, πy + (ι − π)y⟩
= ⟨πx, πy⟩ + ⟨πx, (ι − π)y⟩
= ⟨πx, πy⟩
= ⟨πx, πy⟩ + ⟨(ι − π)x, πy⟩
= ⟨x, πy⟩

so π = π^{∗}

Exercise 14.3.14

(i)⇒(ii)

Let U = α(V) and W = (ι − α)(V). Since

v = αv + (ι − α)v

U + W = V. If u ∈ U then u = αv for some v ∈ V so

αu = α^{2}v = αv = u = ι_{U}u.

If w ∈ W then u = (ι − α)v for some v ∈ V so

αw = (α − α^{2})v = 0.

Finally, if u ∈ U, w ∈ W the results of the previous paragraph give

⟨u, w⟩ = ⟨αu, (ι − α)w⟩ = ⟨u, α(ι − α)w⟩ = ⟨u, 0⟩ = 0. (ii)⇒(iii) Immediate. (iii)⇒(iv) Take an orthonormal basis for U and an orthonormal ba-

sis for W. Together they form an orthonormal basis for V with the required properties.

(iv)⇒(i) Let A be the matrix of α with respect to the specified basis. Then A^{∗} = A and A^{2} = A, so the results for α follow.

An orthogonal projection automatically obeys the condition α^{2} = α. Give ℂ^{2} the standard inner product. Let

α(z, w) = (\frac{1}{3}z + \frac{2}{3}w, \frac{1}{3}z + \frac{2}{3}w).

By inspection, α is linear and α^{2} = α. By observing that the matrix A of α with respect to the standard basis is real, but not symmetric, we see that A^{∗} ≠ A so α^{∗} ≠ α. Thus α is a projection but not an orthogonal projection.

If α and β are orthogonal projections and αβ = βα, then

(αβ)^{2} = (αβ)(αβ) = α(βα)β = α(αβ)β = α^{2}β^{2} = αβ and (αβ)^{∗} = (βα)^{∗} = α^{∗}β^{∗} = αβ.

Exercise 14.3.15

We have

ρ_{W}^{2} = 4π_{W}^{2} − 4π_{w} + ι = ι.

∥ρ_{W} x∥^{2} = ⟨(2π_{W} − ι)x, (2π_{W} − ι)x⟩ = ⟨(2π_{W} − ι)x, (2π_{W} − ι)x⟩

= ⟨x, x⟩ = ∥x∥^{2}

so ρ_{W} is an isometry.

ρ_{W} ρ_{W^{⊥}} = (2π_{W} − ι)(2π_{W^{⊥}} − ι)

= 4π_{W} π_{W^{⊥}} − 2π_{W} − 2π_{W^{⊥}} + ι = 0 − 2ι + ι = −ι.

Exercise 15.1.1

d_{1}(A, B) = max_{i,j} |a_{ij} − b_{ij}| = ∥A − B∥_{1}.

(i) |a_{ij}| ≥ 0 ∀i, j so ∥A∥_{1} ≥ 0. (ii) ∥A∥_{1} = 0 ⇒ |a_{ij}| = 0 ∀i, j ⇒ a_{ij} = 0 ∀i, j ⇒ A = 0. (iii) ∥λA∥_{1} = max_{i,j} |λa_{ij}| = max_{i,j} |λ||a_{ij}| = |λ| max_{i,j} |a_{ij}| = |λ∥A∥_{1}. (iv) |a_{ij} − b_{ij}| ≤ |a_{ij}| + |b_{ij}| ∀i, j so

max_{i,j} |a_{ij} − b_{ij}| ≤ max_{i,j} |a_{ij}| + max_{i,j} |b_{ij}|

and ∥A + B∥_{1} ≤ ∥A∥_{1} + ∥B∥_{1}. d_{2}(A, B) = ∑_{i,j} |a_{ij} − b_{ij}| = ∥A − B∥_{2}.

(i) |a_{ij}| ≥ 0 ∀i, j so

∥A∥_{2} = ∑_{i,j} |a_{ij}| ≥ 0.

(ii) ∥A∥_{2} = 0 ⇒ |a_{ij}| = 0 ∀i, j ⇒ a_{ij} = 0 ∀i, j ⇒ A = 0. (iii) We have

∥λA∥_{2} = ∑_{i,j} |λa_{ij}| = ∑_{i,j} |λ||a_{ij}|

= |λ| ∑_{i,j} |a_{ij}| = |λ|∥A∥_{2}

(iv) We have

∥A + B∥_{2} = ∑_{i,j} |a_{ij} + b_{ij}| ≤ ∑_{i,j} |a_{ij}| + |b_{ij}|

= ∑_{i,j} |a_{ij}| + ∑_{i,j} |b_{ij}| = ∥A∥_{2} + ∥B∥_{2}.

d_{3}(A, B) = (∑_{i,j} |a_{ij} − b_{ij}|^{2})^{1/2} = ∥A − B∥_{3}. ∥ ∥_{3} is just the standard norm for an inner product space of dimension

n^{2}.

Exercise 15.1.3

Observe that, using the Cauchy–Schwarz inequality,

∥αx∥ = (∑_{i=1}^{n} |∑_{j=1}^{n} a_{ij}x_{j}|^{2})^{1/2} ≤ (∑_{i=1}^{n} ∑_{j=1}^{n} |a_{ij}|^{2} ∑_{j=1}^{n} |x_{j}|^{2})^{1/2}

= (∑_{i=1}^{n} ∑_{j=1}^{n} |a_{ij}|^{2}∥x∥^{2})^{1/2} ≤ (∑_{i=1}^{n} ∑_{j=1}^{n} |a_{ij}|^{2})^{1/2} ∥x∥.

Exercise 15.1.7

Since U = (ι−α)^{−1}(0) and W = α^{−1}(0) are orthogonal complements, we can write any v ∈ V in the form

v = u + w

with u ∈ U and w ∈ W. Since u ⊥ w,

∥v∥^{2} = ∥u∥^{2} + ∥w∥^{2}

and

∥πv∥ = ∥u∥ ≤ ∥v

so ∥π∥ ≤ 1.

Since π ≠ 0, we can find x with u = πx ≠ 0 so ∥u∥ ≠ 0, Since π(u = u, we have ∥π∥ = 1.

We use row vectors. Let K > 0. If α(x, y) = (x + Ky, 0), then α is linear and α^{2}(x, y) = (x + Ky, 0) = α(x, y), so α is a projection. Since ∥(0, 1)∥ = 1 and ∥α(0, 1)∥ = K, ∥α∥ ≥ K.

Exercise 15.1.9

If e_{1}, e_{2}, …, e_{n} is an orthonormal basis for U, then the map

θ (∑_{j=1}^{n} x_{j}e_{j}) = (x_{1}, x_{2}, …, x_{n})^{T}

is a linear inner product preserving (and so norm preserving) isomorphism θ : U → 𝔽^{n} where 𝔽^{n} is equipped with the standard dot product.

Exercise 15.1.10

Observe that

∥Ix∥ = ∥x∥

so ∥I∥ = 1.

(i) ∥I + (−I)∥ = ∥0∥ = 0, ∥I∥ = ∥ − I∥ = 1. (ii) ∥I + I)∥ = ∥2I∥ = 2, ∥I∥ = 1. (iv) ∥II∥ = ∥I∥ = 1. We now attack (iii). If

J = (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) ,

then

∥J(x, y)^{T} ∥ = ∥(y, 0)^{T} ∥ = |y| ≤ ∥(x, y)^{T} ∥

so ∥J∥| ≤ 1. However ∥(1, 0)^{T} ∥ = ∥(0, 1)^{T} ∥ = 1 and J(0, 1)^{T} = (1, 0)^{T} so ∥J∥ = 1 but

∥JJ∥ = ∥0∥ = 0.

Exercise 15.1.11

det(tI − A) = (t − 1)^{2} − μ^{2} = t^{2} − 2t + η

so the eigenvalues are 1 ± (1 − η^{2})^{1/2}

The eigenvectors corresponding to 1 + (1 − η^{2})^{1/2} are given by

x + μy = (1 + (1 − η^{2})^{1/2})x
μx + y = (1 + (1 − η^{2})^{1/2})y

i.e x = y (as we could spot directly). The eigenvectors corresponding to 1 − (1 − η^{2})^{1/2} are given by

x + μy = (1 − (1 − η^{2})^{1/2})x
μx + y = (1 − (1 − η^{2})^{1/2})y

i.e x = −y (as we could spot directly).

Thus we have a basis of orthonormal eigenvectors e_{1} = 2^{−1/2}(1, 1)^{T} eigenvalue λ_{1} = 1 + (1 − η^{2})^{1/2} and e_{2} = 2^{−1/2}(1, −1)^{T} eigenvalue λ_{2} = 1 − (1 − η^{2})^{1/2}.

We have

∥T (x_{1}e_{1} + x_{2}e_{2}∥ = ∥λ_{1}x_{1}e_{1} − λ_{2}x_{2}e_{2}∥
= ((λ_{1}x_{1})^{2} + (λ_{2}x_{2}))^{1/2}
≤ λ_{1}(x_{1}^{2} + x_{2}^{2})^{1/2}  ≤ λ_{1}∥(x_{1}e_{1} + x_{2}e_{2}∥

so ∥A∥ ≤ λ_{1} (in fact, since Ae_{1} = λ_{1}e_{1} we have ∥A∥ = λ_{1}) and small changes in x produce small changes in y.

Observe that taking, y = e_{2}, we have x = A^{−1} = λ_{2}^{−1} which is very large.

(ii) Observe that looking at A^{−1} we have a basis of orthonormal eigenvectors e_{1} = 2^{−1/2}(1, 1)^{T} eigenvalue λ_{1} = 1 + (1 − η^{2})^{−1/2} and e_{2} = 2^{−1/2}(1, −1)^{T} eigenvalue λ_{2} = 1 − (1 − η^{2})^{−1/2}. Thus arguing as in (i), ∥A^{−1}∥ = λ_{2}^{−1}.

Now det B = det A det A^{−1} = det AA^{−1} = det I = 1 and

B^{−1} = (\begin{matrix}A^{−1} & 0 \\ 0 & A \end{matrix}) .

By looking at vectors of the form (0, 0, x, y) we see that ∥B∥ ≥ ∥A^{−1}∥ and by looking at vectors of the form (x, y, 0, 0) we see that ∥B∥ ≥ ∥A^{−1}∥. Thus if η is very small, both ∥B∥ and ∥B^{−1}∥ are very large.

Exercise 15.1.12

(i) We have c(A) = ∥A∥∥A^{−1}| ≥ ∥AA^{−1}∥ = ∥I∥ = 1. (ii) We have

c(λA) = ∥λA∥∥(λA)^{−1}∥ = ∥λA∥∥λ^{−1}A^{−1}| = |λ||λ^{−1}|∥A∥∥A^{−1}∥ = c(A).

Exercise 15.1.13

Let us use an orthonormal basis e_{1}, e_{2}, …, e_{n}. Observe that

n^{−2} ∑_{i,j} |a_{ij}| ≤ n^{−2} ∑_{i,j} max_{r,s} |a_{rs}| = max_{r,s} |a_{rs}|.

and that

|a_{rs}| ≤ (∑_{j=1}^{n} |a_{js}|^{2})^{1/2}

= ∥∑_{j=1}^{n} a_{js}e_{j}∥

= ∥αe_{s}∥ ≤ ∥α∥∥e_{s}∥ = ∥α∥.

If x = ∑_{j=1}^{n} x_{j}e_{j} then |x_{j}| = |⟨x, e_{j}⟩| ≤ ∥x∥ so

∥αx∥ = ∥∑_{r=1}^{n} x_{r}α(e_{r})∥ ≤ ∑_{r=1}^{n} |x_{r}|∥α(e_{r})∥

≤ ∥x∥ ∑_{r=1}^{n} ∥α(e_{r})∥ = ∥x∥ ∑_{r=1}^{n} ∥∑_{j=1}^{n} a_{jr}e_{j}∥

≤ ∥x∥ ∑_{r=1}^{n} ∑_{j=1}^{n} |a_{jr}|∥e_{j}∥ ≤ ∥x∥ ∑_{r=1}^{n} ∑_{j=1}^{n} |a_{jr}|,

so

∥α∥ ≤ ∑_{i,j} |a_{ij}|.

Finally,

∑_{i,j} |a_{ij}| ≤ ∑_{i,j} max_{r,s} |a_{rs}| = n^{2} max_{r,s} |a_{rs}|.

Since

∥αx∥ ≤ (∑_{i=1}^{n} ∑_{j=1}^{n} |a_{ij}|^{2})^{1/2}

for all ∥x∥ ≤ 1, we have

∥αy∥ ≤ (∑_{i=1}^{n} ∑_{j=1}^{n} |a_{ij}|^{2})^{1/2} ∥y∥

for all y and

∥α∥ ≤ (∑_{i,j} a_{ij}^{2})^{1/2} .

Exercise 15.2.2

A triangular matrix A has characteristic polynomial

χ_{A}(t) = ∏_{j=1}^{n}(t − a_{ii})

so, if A is real, its characteristic equation has all its roots real. Thus, if α is triangularisable, its characteristic equation has all its roots real. Consider the matrix

B = (\begin{matrix}C & 0 \\ 0 & I \end{matrix})

with I the n − 2 × n − 2 identity matrix and

C = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix})

B has characteristic polynomial (t−1)^{n−2}(t^{2} +1) so not all its roots are real. If β is the linear map corresponding to B with respect to some basis then β is not triangularisable.

If dim V = 1 a 1 × 1 matrix is triangular so all endomorphisms are triangularisable.

The fact that A = QR with R upper triangular and Q orthogonal does not imply that there is an invertible matrix P and an upper triangular matrix T with A = P AP^{−1}.

Exercise 15.2.4

Let A be the matrix of an endomorphism α with respect to some orthonormal basis. We can find α(n) ∈ L(U, U) with ∥α(n) − α∥ → 0 as n → ∞ such that the characteristic equations of the α(n) have no repeated roots. Let A(n) be the matrix of α(n) with respect to the given basis. Then

max_{i,j} |a_{ij}(n) − a_{ij}| ≤ ∥α(n) − α∥ → 0

as n → ∞.

Exercise 15.3.2

We have

det(tI − A) = (t − 1/2)(t − 1/4),

so A has eigenvalues 1/2 and 1/4

By induction on n,

A^{n}(0, 1)^{T} = (2^{−n+1}K, 0)^{T}

for all n ≥ 1. Thus, if K = 2^{N}L, we have ∥A^{N}(0, 1)^{T} ∥ > L.

Exercise 15.3.3

We consider column vectors. Observe that, if x, y ∈ ℝ^{n} and z = x + iy, then

∥z∥^{2} = ∑_{j=1}^{n} |z_{j}|^{2}

= ∑_{j=1}^{n}(|x_{j}|^{2} + |y_{j}|^{2})

= ∥x∥^{2} + ∥y∥^{2}

Using the notation introduced in the first paragraph,

∥A^{n}z∥^{2} = ∥A^{n}(x + iy)∥^{2}
= ∥A^{n}x + iA^{n}y)∥^{2}
= ∥A^{n}x∥^{2} + ∥A^{n}y)∥^{2},

so

∥A^{n}x∥, ∥A^{n}y)∥ → 0 ⇒ ∥A^{n}z)∥ → 0.

The result follows.

Exercise 15.3.7

(i) We have

c_{ij} = ∑_{k=1}^{n} a_{ik}b_{kj} = ∑_{k=1}^{j+s} a_{ik}b_{kj} = ∑_{j+s+1≤k≤i−r+1} a_{ik}b_{kj} = 0

if i − r ≤ j + s + 1. (ii) If D = (d_{ij}) is diagonal, then ∥D∥ = max |d_{jj}| = ρ(D).

Exercise 15.3.8

The computations by which we obtained the result for r, s ≥ 0 remain valid and the result remains true.

Exercise 15.3.11

Let A be an n × n matrix with non-zero diagonal entries and b ∈ 𝔽^{n} a column vector. Suppose that the equation

Ax = b

has the solution x^{∗}. Let us write D for the n × n diagonal matrix with diagonal entries the same as those of A and set B = A−D. If x_{0} ∈ ℝ^{n} and

x_{j+1} = D^{−1}(b − Bx_{j}),

then

∥x_{j} − x^{∗}∥ ≤ ∥D^{−1}B∥^{j}

and ∥x_{j} − x^{∗}∥ → 0 whenever ρ(D^{−1}B) < 1. If ρ(D^{−1}) = 1. we can find an x_{0} such that ∥x_{j} − x^{∗}∥ ↛ 0.

To prove this, observe that −D^{−1}Bx^{∗} = D^{−1}b − x^{∗} and so

x_{j+1} − x^{∗} = D^{−1}(b − Bx_{j}) − x^{∗} = −D^{−1}B(x_{j} − x^{∗})

Thus

∥x_{n} − x^{∗}∥ ≤ ∥(D^{−1}B)^{n}∥∥x_{0} − x^{∗}∥

If ρ(D^{−1}B) < 1 ∥(D^{−1}B)^{n}∥ → 0 the result follows.

If ρ(D^{−1}B) ≥ 1 we can find an eigenvector e with eigenvalue having absolute value at least 1. If we set x_{0} = x^{∗} + e_{0} convergence fails.

To prove the last part, suppose, if possible, that we can find an eigenvector y of D^{−1}B with eigenvalue λ such that |λ| ≥ 1. Then D^{−1}By = λy and so By = λDy. Thus

∑_{j≠i}^{n} a_{ij}y_{j} = λa_{ii}y_{i}

and so

|a_{ii}||y_{i}| ≤ ∑_{j≠i} |a_{ij}||y_{j}|

for each i and (since y ≠ 0)

|a_{ii}||y_{i}| < ∑_{j≠i} |a_{ij}||y_{j}|

for at least one value of i.

Summing over all i and interchanging the order of summation, we get

∑_{i=1}^{n} |a_{ii}||y_{i}| ≤ ∑_{i=1}^{n} ∑_{j≠i} |a_{ij}||y_{j}| = ∑_{j=1}^{n} ∑_{i≠j} |a_{ij}||y_{j}|

= ∑_{j=1}^{n} |y_{j}| ∑_{i≠j} |a_{ij}| < ∑_{j=1}^{n} |y_{j}||a_{jj}| = ∑_{i=1}^{n} |a_{ii}||y_{i}|

which is absurd.

Thus all the eigenvalues of D^{−1}B have absolute value less than 1 and we may apply the first part of the question.

Exercise 15.4.3

If α has matrix A with respect to given orthonormal basis, α^{∗} has matrix A^{∗}, αα^{∗} has matrix AA^{∗} and α^{∗}α has matrix A^{∗}A.

Thus αα^{∗} = α^{∗}α ⇔ A^{∗}A = AA^{∗}.

Exercise 15.4.6

If α is diagonalisable with distinct eigenvalues λ_{1}, λ_{2}, …, λ_{m}, then U is the direct sum of the spaces

E_{j} = {e : αe = λ_{j}e}.

Moreover E_{j} ⊥ E_{k}, that is to say

e_{j} ∈ E_{j}, e_{k} ∈ E_{k} ⇒ ⟨e_{j}, e_{k}⟩ = 0.

when j ≠ k. It follows that if π_{j} is the (unique) orthogonal projection with π_{j}U = E_{j} we have π_{j}π_{k} = 0 for all j ≠ k If u ∈ U, we can write u uniquely as

⋆  u = e_{1} + e_{2} + … + e_{m}

with e_{j} ∈ E_{j}. Since π_{j}π_{k} = 0 for all j ≠ k we have π_{j}e_{k} = π_{j}π_{k}e_{k} = 0. Thus, applying π_{j} to both sides of ⋆, π_{j}u = e_{j}. Thus

ιu = ∑_{j=1}^{m} π_{j}u

and

αu = ∑_{j=1}^{m} αe_{j} = ∑_{j=1}^{m} λ_{j}e_{j} = ∑_{j=1}^{m} λ_{j}π_{j}u.

for all u ∈ U. It follows that

ι = π_{1} + π_{2} + … + π_{m} and α = λ_{1}π_{1} + λ_{2}π_{2} + … + λ_{m}π_{m}.

Conversely, if the stated conditions hold and we write E_{j} = π_{j}U, we see that E_{j} is a subspace and π_{i}E_{j} = π_{i}π_{j}E_{j} = 0 for i ≠ j. Since iota = π_{1} + π_{2} + … + π_{m}, we have

u = ιu = ∑_{j=1}^{m} π_{j}u = ∑_{j=1}^{m} u_{j}

with u_{j} = π_{j}u ∈ E_{j}. On the other hand, if e_{j} ∈ E_{j} and

∑_{j=1}^{m} e_{j} = 0

then applying π_{i} to both sides we get e_{i} = 0 for all i. Thus

U = E_{1} ⊕ E_{2} ⊕ … ⊕ E_{m}.

Moreover, since π_{j}π_{k} = 0 we have E_{j} ⊥ E_{k} for k ≠ j.

Let E_{j} be an orthonormal basis for E_{j}. The set E = ⋃_{j=1}^{m} E_{j} is an orthonormal basis for U. Since α has a diagonal matrix with respect to this basis, α is normal.

α is self-adjoint if and only if there exist distinct λ_{j} ∈ ℝ and othogonal projections π_{j} such that π_{k}π_{j} = 0 when k ≠ j,

ι = π_{1} + π_{2} + … + π_{m} textand α = λ_{1}π_{1} + λ_{2}π_{2} + … + λ_{m}π_{m}.

Exercise 15.4.7

If α is unitary, then α = α^{−1} so

αα^{∗} = ι = αα^{∗}

and α is normal.

β unitary ⇔ ββ^{∗} = ι

⇔ α^{−1}α^{∗}(α^{−1}α^{∗})^{∗} = ι ⇔ α^{−1}α^{∗}α^{∗∗}(α^{∗})^{−1} = ι ⇔ α^{−1}α^{∗}α(α^{∗})^{−1} = ι ⇔ α^{∗}α = αα^{∗} ⇔ α normal

Exercise 15.4.9

(i) Choose an orthonormal basis such that α is represented by a diagonal matrix D with d_{jj} = e^{iθ_{j}} with θ_{j} real. Let f(t) be represented by a diagonal matrix D with d_{jj}(t) = e^{iθ_{j}t}. Then

∥f(t) − f(s)∥ ≤ max_{j} |e^{iθ_{j}t} − e^{iθ_{j}s}| ≤ max |θ_{j}||t − s|

so f is continuous, f(t) is unitary, f(0) = ι, f(1) = α.

Recall that the unitary maps form a group. Let α = β^{−1}γ. Since α is unitary we can find f as in the first paragraph. Set g(t) = βf(t).

(ii) By considering matrix representations, we know that det : L(U, U) → ℝ is continuous. Thus, if f is as stated, det f(t) is a continuous function of t. But det f(1) = −1, det f(0) = 1 so, by the intermediate value theorem, we can find an s ∈ [0, 1] such that det f(s) = 0 and so f(s) is not invertible.

Exercise 16.1.5

If U is finite dimensional and

α(x, y) = 0 for all x ∈ U ⇒ y = 0,

then θ_{R} : U → U′ is an isomorphism. Proof. We observe that the stated condition tells us that θ_{R} is injective since

θ_{R}(y) = 0 ⇒ θ_{R}(y)(x) = 0 for all x ∈ U
⇒ α(x, y) = 0 for all x ∈ U
⇒ y = 0.

Since dim U = dim U′, it follows that θ_{R} is an isomorphism. □

Exercise 16.1.9

Observe, that if x ≠ 0, then at least one of the following two things must occur:-

(A) x_{1} ≠ 0, so β((x_{1}, x_{2})^{T} , (0, 1)^{T} ) = x_{1} ≠ 0.

(B) x_{2} ≠ 0, so β((x_{1}, x_{2})^{T} , (1, 0)^{T} ) = −x_{2} ≠ 0.

Thus β is non-degenerate, but

β(x, x) = x_{1}x_{2} − x_{2}x_{1} = 0

for all x.

We cannot find an α of the type required. If α is degenerate, then there exists an x ≠ 0 with α(x, y) = 0 for all y so, in particular, α(x, x) = 0.

Exercise 16.1.14

Observe that, with the notation of Definition 16.1.12,

q(u + v) + q(u − v) = α(u + v, u + v) + α(u − v, u − v)

= α(u, u + v) + α(v, u + v)

+ α(u, u − v) + α(v, u − v) = α(u, u) + α(u, v) + α(v, u) + α(v, v)

+ α(u, u) − α(u, v) − α(v, u) + α(v, v) = 2α(u, u) + 2α(u, u) = 2(q(u) + q(v)).

Exercise 16.1.16

Observe that

∑_{i=1}^{n} ∑_{j=1}^{n} x_{i}b_{ij}x_{j} = ∑_{j=1}^{n} ∑_{i=1}^{n} x_{j}b_{ij}x_{i},

so

∑_{i=1}^{n} ∑_{j=1}^{n} x_{i}b_{ij}x_{j} = ∑_{i=1}^{n} ∑_{j=1}^{n} x_{i} \frac{b_{ij} + b_{ji}}{2} x_{j}

and q is a quadratic form with associated symmetric matrix A = \frac{1}{2}(A + A^{T}).

Exercise 16.1.17

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 1}

is a sphere.

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 0}

is a point.

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = −1} = ∅.

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} − x_{3}^{2} = 1}

is a one sheeted hyperboloid of revolution obtained by revolving the hyperbola

{x ∈ ℝ^{3} : x_{1}^{2} − x_{3}^{2} = 1, x_{2} = 0}

about its semi-minor axis Ox_{3}.

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} − x_{3}^{2} = 0}

is a circular cone.

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} − x_{3}^{2} = −1}

is a two sheeted hyperboloid of revolution obtained by revolving the hyperbola

{x ∈ ℝ^{3} : x_{1}^{2} − x_{3}^{2} = −1, x_{2} = 0}

about its semi-major axis Ox_{3}. (The two sheeted hyperboloid has two connected components. The one sheeted hyperboloid has one.)

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} = 1}

is a circular cylinder.

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} = 0}

is the straight line give along the axis Ox_{3}.

{x ∈ ℝ^{3} : x_{1}^{2} + x_{2}^{2} = −1} = ∅.

{x ∈ ℝ^{3} : x_{1}^{2} − x_{2}^{2} = 1} and x ∈ ℝ^{3} : x_{1}^{2} − x_{2}^{2} = −1}

are hyperbolic cylinders.

{x ∈ ℝ^{3} : x_{1}^{2} − x_{2}^{2} = 0}

is a pair of planes intersecting at right angles.

{x ∈ ℝ^{3} : x_{1}^{2} = 1}

is a pair of parallel planes.

{x ∈ ℝ^{3} : x_{1}^{2} = 0}

is a plane.

{x ∈ ℝ^{3} : x_{1}^{2} = −1} = ∅.
{x ∈ ℝ^{3} : 0 = 1} = {x ∈ ℝ^{3} : 0 = −1} = ∅

and

{x ∈ ℝ^{3} : 0 = 0} = ℝ^{3}.

Exercise 16.1.23

By a rotation (which changes neither volume nor determinant) we may suppose that our ellipsoid is

∑_{i=1}^{n} d_{i}y_{i}^{2} ≤ L.

By successive scale changes w_{i} = d_{i}^{1/2}y_{i} which multiply volume by d_{i}^{1/2} we can transform our ellipsoid to

∑_{i=1}^{n} w_{i}^{2} ≤ L

Now make a scale change in every direction u_{i} = L^{−1/2}w_{i}, to obtain the unit sphere of volume V_{n}

Our original volume must have been L^{n/2} ∏_{j=1}^{n} d_{j}^{−1/2}V_{n} i.e,

(det A)^{−1/2}L^{n/2}V_{n}.

Exercise 16.1.25

We know that

f_{Y}(y) = K exp (−\frac{1}{2} ∑_{i=1}^{n} ∑_{j=1}^{n} d_{j}y_{j}^{2})

is a density function and so

∫_{ℝ^{n}} f_{Y}(y) dV (y) = 1.

Thus

1 = K ∫_{−∞}^{∞} ∫_{−∞}^{∞} … ∫_{−∞}^{∞} exp (−\frac{1}{2} ∑_{j=1}^{n} d_{j}y_{j}^{2}) dy_{1}dy_{2} … dy_{n}

= K ∫_{−∞}^{∞} ∫_{−∞}^{∞} … ∫_{−∞}^{∞} ∏_{j=1}^{n} exp(−d_{j}y_{j}^{2}/2) dy_{1}dy_{2} … dy_{n}

= K ∏_{j=1}^{n} ∫_{−∞}^{∞} exp(−d_{j}y_{j}^{2}/2) dy_{j}

= K ∏_{j=1}^{n} (d_{j}^{−1/2} ∫_{−∞}^{∞} exp(−t_{j}^{2}/2) dt_{j})

= K(2π)^{n/2} ∏_{j=1}^{n} d_{j}^{−1/2}.

Thus

K = (2π)^{−n/2} (∏_{j=1}^{n} d_{j})^{1/2} .

Exercise 16.1.26

(i) Suppose α sesquilinear, β Hermitian, γ skew-Hermitian and

α = β + γ.

Then

α(u, v) = β(u, v) + γ(u, v)

and

α(v, u)^{∗} = (β(v, u) + γ(v, u))^{∗} = β(u, v) − γ(v, u).

Thus

β(u, v) = \frac{1}{2}(α(u, v) + α(v, u)^{∗})
γ(u, v) = \frac{1}{2}(α(u, v) − α(v, u)^{∗})

Conversely, if α is sesquilinear, the definitions

β(u, v) = \frac{1}{2}(α(u, v) + α(v, u)^{∗})
γ(u, v) = \frac{1}{2}(α(u, v) − α(v, u)^{∗})

give sesquilinear forms (the point at issue is that if (u, v) ↦ α(u, v) is sesquilinear, so is the map (u, v) ↦ α(v, u)^{∗}) with

β(v, u) = \frac{1}{2}(α(v, u) + α(u, v)^{∗}) = β(u, v)^{∗}
γ(v, u) = \frac{1}{2}(α(v, u)^{∗} − α(v, u)) = −γ(v, u)^{∗}

(ii) Since α(u, v) = α(v, u)^{∗} for all u, v, we have α(u, u) = α(u, u)^{∗}, so α(u, u) is real for all u ∈ U.

(iii) Set β = iα. Then α skew-Hermitian ⇔ α(v, u) = −α(u, v)^{∗} ∀v, u ⇔ β(u, v)^{∗} = (iα(u, v))^{∗} = −iα(u, v)^{∗} = iα(v, u) = β(v, u) ∀v, u ⇔ β Hermitian.

(iv) Observe that

α(u + v, u + v) = α(u, u) + α(u, v) + α(v, u) + α(u, u)
= α(u, u) + α(u, v) + α(u, v)^{∗} + α(v, v)
= α(u, u) + 2ℜα(u, v) + α(v, v).

Thus

α(u − v, u − v) = α(u, u) + 2ℜα(u, v) + α(v, v)

and

α(u + v, u + v) − α(u − v, u − v) = 4ℜα(u, v).

It follows that α(u+iv, u+iv)−α(u−iv, u−iv) = 4ℜα(u, iv) = −4ℜ(iα(u, v) = ℑ(u, v). The result follows.

(v) Choose any orthonormal basis f_{1}, f_{2}, …, f_{n} for U. Define a matrix A = (a_{pq}) by the formula

a_{pq} = α(f_{p}, f_{q}).

By sesquilinearity

α (∑_{p=1}^{n} z_{p}f_{p}, ∑_{q=1}^{n} w_{q}f_{q}) = ∑_{p=1}^{n} ∑_{q=1}^{n} z_{p}a_{pq}w_{q}^{∗}

for all z_{p}, w_{q} ∈ ℂ. Since α is Hermitian

a_{qp} = α(f_{q}, f_{p}) = α(f_{p}, f_{q})^{∗} = a_{pq}^{∗}

Thus A is Hermitian and we can find a unitary matrix M such that M^{∗}AM = D where D is a real diagonal matrix whose entries are the eigenvalues of A appearing with the appropriate multiplicities. If we set e_{q} = ∑_{p=1}^{n} m_{pq}f_{p} then (since M is unitary) the e_{1}, e_{2}, …, e_{n} are an orthonormal basis and by direct calculation

α (∑_{r=1}^{n} z_{r}e_{r}, ∑_{s=1}^{n} w_{s}e_{s}) = ∑_{t=1}^{n} d_{t}z_{t}w_{t}^{∗}

for all z_{r}, w_{s} ∈ ℂ.

Exercise 16.1.27

Suppose that U is vector space over ℂ and α : U × U → ℂ is a bilinear form. Let us set

θ_{R}(w)z = α(z, w)

for z, w ∈ U.

We observe that

θ_{R}(w)(λ_{1}z_{1} + λ_{2}z_{2}) = α(λ_{1}z_{1} + λ_{2}z_{2}, w)
= λ_{1}α(z_{1}, w) + λ_{2}α(z_{2}, w)
= λ_{1}θ_{R}(w)z_{1} + λ_{2}θ_{R}(w)z_{2}

for all λ_{1}, λ_{2} ∈ ℂ and z_{1}, z_{2} ∈ U. Thus θ_{R}(w) ∈ U′ for all w ∈ U.

Now

θ_{R}(λ_{1}w_{1} + λ_{2}w_{2})z) = α(z, λ_{1}w_{1} + λ_{2}w_{2})
= λ_{1}^{∗}α(z, w_{1}) + λ_{2}^{∗}α(z, w_{2})
= λ_{1}^{∗}θ_{R}(w_{1})z + λ_{2}^{∗}θ_{R}(w_{2})z
= (λ_{1}^{∗}θ_{R}(w_{1}) + λ_{2}^{∗}θ_{R}(w_{2}))z

for all z ∈ U. Thus

θ_{R}(λ_{1}w_{1} + λ_{2}w_{2}) = λ_{1}^{∗}θ_{R}(w_{1}) + λ_{2}^{∗}θ_{R}(w_{2})

and θ_{R} : U → U′ is an anti-linear map.

Suppose in addition we know that U is finite dimensional and that

α(z, w) = 0 for all z ∈ U ⇒ w = 0.

Then θ_{R} is injective since

θ_{R}(w) = 0 ⇒ θ_{R}(w)(z) = 0 for all z ∈ U
⇒ α(z, w) = 0 for all w ∈ U
⇒ w = 0.

Since dim U = dim U′, it follows that θ_{R} is an isomorphism. Alternatively, observe that if e_{1}, e_{2}, …, e_{n} form a basis for U

∑_{j=1}^{n} λ_{j}θ_{R}e_{j} = 0 ⇒ θ_{R} (∑_{j=1}^{n} λ_{j}^{∗}e_{j}) = 0

⇒ ∑_{j=1}^{n} λ_{j}^{∗}e_{j} = 0

⇒ λ_{j}^{∗} = 0 ∀j ⇒ λ_{j} = 0 ∀j

so θ_{R}e_{1}, θ_{R}e_{2}, …, θ_{R}e_{n} are linearly independent so a basis for U′ so span U′. Thus θ_{R} is surjective.

Exercise 16.2.4

Let f_{1}, f_{2}, …, f_{n} be the basis associated with S. Define

α (∑_{i=1}^{n} x_{i}f_{i}, ∑_{j=1}^{n} y_{j}f_{j}) = ∑_{k=1}^{n} x_{k}y_{k}

for all x_{i}, y_{j} ∈ ℝ. Then, by inspection, α is a symmetric form on ℝ^{n}.

By Theorem 16.2.1 we can find a basis e_{1}, e_{2}, …, e_{n} and positive integers p and m with p + m ≤ n such that

α (∑_{i=1}^{n} x_{i}e_{i}, ∑_{j=1}^{n} y_{j}e_{j}) = ∑_{k=1}^{p} x_{k}y_{k} − ∑_{k=p+1}^{p+m} x_{k}y_{k}

for all x_{i}, y_{j} ∈ ℝ. We have

q (∑_{i=1}^{n} y_{i}e_{i}) = α (∑_{i=1}^{n} y_{i}e_{i}, ∑_{j=1}^{n} y_{j}e_{j}) = ∑_{k=1}^{p} y_{k}^{2} − ∑_{k=p+1}^{p+m} y_{k}^{2}

as required.

Exercise 16.2.8

A is two quadrants, B two lines at right angles. (1, 0), (−1, 1) ∈ A, but (0, 1) = (1, 0) + (−1, 1) ∉ A. Thus A is not

a subspace.

(1, 0), (−1, 1) ∈ B, but (0, 1) = (1, 0) + (−1, 1) ∉ B. Thus A is not a subspace.

Exercise 16.2.9

(i) Choose M orthogonal so that M^{T} AM = D with D diagonal. Since matrix rank is unchanged by multiplication on the left or right by invertible matrices, D has the same matrix rank as A. But the signature rank of A and the matrix rank of D are both the number of non-zero entries of D. Thus the signature and matrix ranks are identical.

(ii) We continue with the notation of (i). The rank of q is the number of non-zero characteristic roots (multiple roots counted multiply). Now

signature q = (no. strictly pos. entries D)−(no. strictly neg. entries D),

so the signature of D is the number of strictly positive characteristic roots minus the number of strictly negative characteristic roots of A (multiple roots counted multiply).

Exercise 16.2.10

(i) Observe that

γ(P z) = (P z^{T})A(P z) = (z^{T} P^{T})A(P z) = z^{T} (P^{T} AP)z.

(ii) We cannot use results on the diagonalisation of real symmetric matrices or Hermitian matrices, but we can use completing the square.

(α ) Suppose that A = (a_{ij})_{1≤i,j≤n} is a symmetric n × n matrix with a_{11} ≠ 0, and we set b_{ij} = a_{11}^{−1}(a_{11}a_{ij} − a_{1i}a_{1j}). Then B = (b_{ij})_{2≤i,j≤n} is a symmetric matrix. Further, if z ∈ ℂ^{n} and (after choosing one square root a_{11}^{1/2} of a_{11}) we set

w_{1} = a_{11}^{1/2} (z_{1} + ∑_{j=2}^{n} a_{1j}a_{11}^{−1/2}z_{j})

and w_{j} = z_{j} otherwise, we have

∑_{i=1}^{n} ∑_{j=1}^{n} z_{i}a_{ij}z_{j} = w_{1}^{2} + ∑_{i=2}^{n} ∑_{j=2}^{n} w_{i}b_{ij}w_{j}.

(β) Suppose that A = (a_{ij})_{1≤i,j≤n} is a symmetric n × n matrix. Suppose further that σ : {1, 2, …, n} → {1, 2, …, n} is a permutation (that is to say σ is a bijection) and we set b_{ij} = a_{σi σj}. Then C = (c_{ij})_{1≤i,j≤n} is a symmetric matrix. Further, if z ∈ ℂ^{n} and we set w_{j} = x_{σ(j)}, we have

∑_{i=1}^{n} ∑_{j=1}^{n} z_{i}a_{ij}z_{j} = ∑_{i=1}^{n} ∑_{j=1}^{n} w_{i}c_{ij}w_{j}.

(γ) Suppose that n ≥ 2, A = (a_{ij})_{1≤i,j≤n} is a symmetric n×n matrix and a_{11} = a_{22} = 0, but a_{12} ≠ 0. Then there exists a symmetric n × n matrix C with c_{11} ≠ 0 such that, if z ∈ ℂ^{n} and we set w_{1} = (z_{1} +z_{2})/2, w_{2} = (z_{1} − z_{2})/2, w_{j} = z_{j} for j ≥ 3, we have

∑_{i=1}^{n} ∑_{j=1}^{n} z_{i}a_{ij}z_{j} = ∑_{i=1}^{n} ∑_{j=1}^{n} w_{i}c_{ij}w_{j}.

Combining (α), (β) and (γ) we have the following.

(δ) Suppose that A = (a_{ij})_{1≤i,j≤n} is a non-zero symmetric n × n matrix. Then we can find an n × n invertible matrix M = (m_{ij})_{1≤i,j≤n} and a symmetric (n − 1) × (n − 1) matrix B = (b_{ij})_{2≤i,j≤n} such that, if z ∈ ℂ^{n} and we set w_{i} = ∑_{j=1}^{n} m_{ij}z_{j}, then

∑_{i=1}^{n} ∑_{j=1}^{n} z_{i}a_{ij}z_{j} = w_{1}^{2} + ∑_{i=2}^{n} ∑_{j=2}^{n} w_{i}b_{ij}w_{j}.

Repeated use of (δ) gives the desired result.

(iii) Observe that if P is invertible so is P^{T}, so (using matrix rank)

rank(P^{T} AP) = rank A.

Thus, if

γ(P z) = ∑_{u=1}^{r} z_{u}^{2},

we have r = rank A.

(iv) Observe that γ(iz) = −γ(z).

(v) By (i) and (ii) we can choose a basis e_{j} for ℂ^{n} so that

γ (∑_{j=1}^{n} z_{j}e_{j}) = ∑_{u=1}^{r} z_{u}^{2}.

Let s be the integer part of r/2. If F is the subspace spanned by the vectors e_{2k−1} − ie_{2k} with 1 ≤ k ≤ s and the vectors e_{l} with r + 1 ≤ l ≤ n. Then F has dimension at least m and any subspace E of F with dimension m will have desired properties.

(vi) Let E_{0} = ℂ^{m}. Using (v) and proceeding inductively we can find subspaces E_{j} such that E_{j} is a subspace of E_{j−1}, dim E_{j} ≥ [2^{−j}n] (using the standard integer part notation) and γ_{j}|E_{j} = 0 [1 ≤ j ≤ k]. We have dim E_{k} ≥ 1, so we may choose a non-zero z ∈ E_{k} to obtain the required result.

Exercise 16.2.12

If a = 0, we have Example 16.2.11. The rank is 3 and the signature −1.

If a ≠ 0, then, setting y_{1} = (x_{1} − 2^{−1}a^{−1}x_{2} − 2^{−1}a^{−1}), y_{2} = x_{2}, y_{3} = x_{3}, we have x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}+ax_{1}^{2} = ay_{1}^{2}+4^{−1}a^{−1}y_{2}^{2}+4^{−1}a^{−1}y_{2}^{2}+(1−2^{−1}a^{−1})y_{2}y_{3} Setting u_{1} = y_{1}, u_{2} = y_{2} + 2a(1 − 2^{−1}a^{−1})y_{3} u_{3} = y_{3}, we have

x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1} + ax_{1}^{2} = au_{1}^{2} + 4^{−1}a^{−1}u_{2}^{2} − a(1 − 2^{−1}a^{−1})^{2}u_{3}. If a = 1/2, we see that q has rank and signature 2. If a > 0 and a ≠ 1/2, q has rank 3 and signature 1. If a < 0, q has rank 3 and signature −1.

Exercise 16.2.13

(i) We know that we can find a unitary matrix U such that C = U^{∗}AU is diagonal with real entries. Let D be the diagonal matrix with d_{ii} = 1 if c_{ii} = 0 and d_{ii} = c_{ii}^{−1/2} (positive square root) otherwise. Setting P = DU we see that P^{∗}AP is diagonal with diagonal entries taking the values 1, −1 or 0.

(ii) We have

(∑_{r=1}^{n} ∑_{s=1}^{n} z_{r}a_{rs}z_{s}^{∗})^{∗} = ∑_{r=1}^{n} ∑_{s=1}^{n} z_{r}^{∗}a_{rs}^{∗}z_{s}

= ∑_{r=1}^{n} ∑_{s=1}^{n} z_{r}^{∗}a_{sr}z_{s}

= ∑_{r=1}^{n} ∑_{s=1}^{n} z_{r}a_{rs}z_{s}^{∗}

so ∑_{r=1}^{n} ∑_{s=1}^{n} z_{r}a_{rs}z_{s}^{∗} is real.

(iii) If P_{1}^{∗}AP_{1} has k entries 1, then if E is the subspace of ℂ^{n} spanned by P_{1}e with e a column vector with ith entry 1 and all other entries 0 when the ith diagonal entry of P_{1}^{∗}AP_{1} is 1, we have

dim E = k, z^{∗}Az > 0 ∀z ∈ E \ {0}.

If P_{2}^{∗}AP_{2} has k′ entries 1 then if F is the subspace of C^{n} spanned by P_{2}e with e a column vector with ith entry 1 and all other entries 0 when the ith diagonal entry of P_{1}^{∗}AP_{1} is −1 or 0, we have

dim F = n − k′, z^{∗}Az ≤ 0 ∀z ∈ F.

Thus E ∩F = {0}, dim E +dim F ≤ n and n−k′ +k ≤ n. It follows that k ≤ k′. The same argument shows that k′ ≤ k so k = k′.

In the same way (or by considering −A) we see that P_{1}^{∗}AP_{1} and P_{2}^{∗}AP_{2} have the same number of entries −1 and so since they have the same number of entries 1 and −1 must have the same number of entries 0 along the diagonal.

Exercise 16.3.2

(i) Let U be a vector space over ℝ. A quadratic form q : U → ℝ is said to be negative semi-definite if

q(u) ≤ 0 for all u ∈ U

and strictly negative definite if

q(u) < 0 for all u ∈ U with u ≠ 0.

By inspection q : U → ℝ is strictly negative definite (respectively negative semi-definite) if and only if −q is strictly positive definite (respectively positive semi-definite).

(ii) If q is a quadratic form over a finite dimensional real vector space, then we can find a basis e_{1}, e_{2}, …, e_{n} such that

q (∑_{j=1}^{n} x_{j}e_{j}) = ∑_{j=1}^{r} x_{j}^{2} − ∑_{j=r+1}^{r+s} x_{j}^{2}.

Setting

q_{1} (∑_{j=1}^{n} x_{j}e_{j}) = ∑_{j=1}^{r} x_{j}^{2} and q_{2} (∑_{j=1}^{n} x_{j}e_{j}) = − ∑_{j=r+1}^{r+s} x_{j}^{2}

we see that q = q_{1} + q_{2} with q_{1} a positive semi-definite quadratic form and q_{2} a negative semi-definite quadratic form.

The observation

0 = 0 + 0 = x^{2} − x^{2}

shows that the decomposition is not unique even for a space of dimension 1.

Exercise 16.3.3

We have

∑_{i=1}^{n} ∑_{j=1}^{n} c_{i}(𝔼X_{i}X_{j})c_{j} = 𝔼 (∑_{j=1}^{n} c_{j}X_{j})^{2} ≥ 0

with equality if and only if

Pr (∑_{j=1}^{n} c_{j}X_{j} = 0) = 1.

Exercise 16.3.4

We know that we can find a basis e_{j} such that

α (∑_{j=1}^{n} x_{j}e_{j}) = ∑_{j=1}^{u} x_{j}^{2} − ∑_{j=u+1}^{v} x_{j}^{2}.

If v < n, then q(e_{n}) = 0 and q is not strictly positive definite. If v ≥ u + 1, q(e_{u+1}) = −1 so q is not positive semi-definite.

By inspection, u = v = n implies q strictly positive definite so q is strictly positive definite if and only if n = u = V i.e. q has rank and signature n.

By inspection, u = v implies q positive semi-definite so q is positive semi-definite if and only if u = v, i.e. q has rank and signature equal.

Finally q is negative semi-definite if and only if −q is positive semidefinite, i.e. q has signature equal to minus its rank.

Exercise 16.3.5

The conditions

(i) α(u, v) = α(v, u) (ii) α(λu, v) = λα(u, v) (iii) α(u + w, v) = α(u, v) + α(w, v)

(for all u, v, w ∈ U and λ ∈ ℝ) say that α is a quadratic form. The additional conditions (iv) α(u, u) ≥ 0 ∀u ∈ U

(v) α(u, u) = 0 ⇒ u = 0 say that α gives rise to a positive definite quadratic form.

Conditions (i) to (v) say that α is an inner product.

Exercise 16.3.6

By translation we may take a = 0 and by subtracting a constant we may take f(0) = 0.

Since f is smooth we know by the local Taylor’s theorem that

f(h) = \frac{∂f}{∂x_{i}} (0)h_{i} + \frac{∂f}{∂x_{i}∂x_{j}} (0)h_{i}h_{j} + ϵ(h)∥h∥^{2}

with ϵ(h) → 0 as ∥h∥ → 0. (We use the summation convention.)

(i) Suppose we have a minimum at 0. If e_{i} is the vector with 1 in the ith place and 0 elsewhere

\frac{∂f}{∂x_{i}} (0) = lim_{h_{i}→0+} \frac{f(he_{i})}{h} ≥ 0

and

\frac{∂f}{∂x_{i}} (0) = lim_{h_{i}→0−} \frac{f(he_{i})}{h} ≤ 0

so \frac{∂f}{∂x_{i}} = 0.

We now have

f(h) = H_{ij}h_{i}h_{j} + ϵ(h)∥h∥^{2}

with H the Hessian. Let u be any non-zero vector We have

0 ≤ f(hu) = h^{2}(H_{ij}u_{i}u_{j} + ϵ(hu)∥u∥^{2}),

so

0 ≤ H_{ij}u_{i}u_{j} + ϵ(hu)∥u∥^{2}

and, allowing h → 0,

0 ≤ H_{ij}u_{i}u_{j}.

Thus H is positive semi-definite.

(ii) We have

f(h) = H_{ij}h_{i}h_{j} + ϵ(h)∥h∥^{2}

with H the Hessian. Since u ↦ Hu is continuous and the surface S of the unit ball is compact, we know that H takes a minimum on S which is non-zero so there exists an η > 0 with

H(u) ≥ η

for all ∥u∥ = 1.

Thus

f(h) ≥ ∥h∥^{2}((η − ϵ(h)).

There exists a δ > 0 such that |ϵ(h)| < η/2 for all ∥h∥ < δ.

If ∥h∥ < δ, then f(h) ≥ (η/2)∥h∥^{2}.

Thus f has a strict minimum at 0.

(ii) Define f : ℝ → ℝ by f(x) = x^{4}. We have a strict minimum at 0, but the Hessian H = (f′′(0)) = (0) is not strictly positive definite.

Exercise 16.3.7

Since

(\frac{∂f}{∂x}(x, y), \frac{∂f}{∂y} (x, y)) = (cos x sin y, sin x cos y)

the stationary points are

(x, y) = (nπ, mπ) and(x, y) = ((n + \frac{1}{2})π, (m + \frac{1}{2})π).

The Hessian

H = (\begin{matrix}− sin x sin y & cos x cos y \\ cos x cos y & − sin x sin y \end{matrix})

so if (x, y) = (nπ, mπ)

H = (\begin{matrix}0 & (−1)^{n+m} \\ (−1)^{n+m} & 0 \end{matrix})

which is neither positive semi-definite nor negative semi-definite (note for example that 4xy = (x + y)^{2} − (x − y)^{2}) and we have a saddle, whilst if (x, y) = ((n + \frac{1}{2})π, (m + \frac{1}{2})π)

H = (\begin{matrix}(−1)^{n+m} & 0 \\ 0 & (−1)^{n+m} \end{matrix})

which is strictly positive definite if n+m is even (so we have a minimum) and strictly negative definite if n + m is even (so we have a maximum).

Exercise 16.3.11

We have

Ã^{T} = (L̃L̃^{T})^{T} = L̃^{T T} L̃^{T} = L̃L̃^{T} = Ã,

so Ã is symmetric.

Observe that

x^{T} Ãx = x^{T} L̃L̃^{T} x = ∥L̃^{T} x∥^{2} ≥ 0

so Ã is symmetric positive semi-definite. The case L = 0 shows that Ã need not be strictly positive definite.

If L is lower triangular with non-zero diagonal entries, then det L ≠ 0, so L is non-singular and

x^{T} Ax = 0 ⇒ x^{T} LL^{T} x = 0
⇒ ∥L^{T} x∥^{2} = 0
⇒ ∥L^{T} x∥ = 0
⇒ L^{T} x = 0
⇒ x = 0,

so A = LL^{T} is a symmetric strictly positive definite matrix.

Exercise 16.3.12

If, at the first stage a_{11} < 0, then A is not positive definite.

If a_{11} > 0, then either the Schur matrix B is positive definite, in which case, A is or the Schur matrix B is not positive definite, in which case, A is not. (Consider vectors whose first term is 0.)

If at each stage, the upper corner entry is strictly positive then our proof shows that A is positive definite. If at some stage it is not, we know that A is not positive definite.

Exercise 16.3.14

We need to look at l_{1} = (2, −3, 1)^{T}. Now

A − l_{1}l_{1}^{T} = (\begin{matrix}4 & −6 & 2 \\ −6 & 8 & −5 \\ 2 & −5 & 14 \end{matrix}) − (\begin{matrix}4 & −6 & 2 \\ −6 & 9 & −3 \\ 2 & −3 & 1 \end{matrix}) = (\begin{matrix}0 & 0 & 0 \\ 0 & −1 & −2 \\ 0 & −2 & 13 \end{matrix}) .

The Schur matrix

B = (\begin{matrix}−1 & −2 \\ −2 & 13 \end{matrix})

has strictly negative upper corner entry, so A is not positive semidefinite.

Exercise 16.3.15

(i) Look at the first formula of the proof of part (ii) of Theorem16.3.10.

(ii) Looking at the proof of of part (i) of Theorem16.3.10 we see that the computation of l involves roughly n operations and the computation of B involves roughly n^{2} operations. Thus the reduction of our problem from n × n matrices to (n − 1) × (n − 1) matrices involves roughly 2n^{2} operations and the total number of operations required is roughly

2 ∑_{r=1}^{n} r^{2} ≈ \frac{2}{3}n^{3}

operations.

[If the reader’s definition of an operation agrees with mine, this argument shows that we can certainly take K = 1 (and, if we really needed, as close to 2/3 as we wished) for n large.]

Exercise 16.3.16

If we take l_{1} = (1, \frac{1}{2}, \frac{1}{3})^{T}, then

H_{3} − l_{1}l_{1}^{T} = (\begin{matrix}1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{matrix} ) − (\begin{matrix}1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{3} & \frac{1}{6} & \frac{1}{9} \end{matrix} )

= (\begin{matrix}0 & 0 & 0 \\ 0 & \frac{1}{12} & \frac{1}{12} \\ 0 & \frac{1}{12} & \frac{4}{45} \end{matrix} )

If we take l_{2} = ((\frac{1}{12})^{1/2}, (\frac{1}{12})^{1/2})^{T}, then

(\begin{matrix}\frac{1}{12} & \frac{1}{12} \\ \frac{1}{12} & \frac{4}{45} \end{matrix}) − l_{2}l_{2}^{T} = (\begin{matrix}0 & 0 \\ 0 & \frac{1}{180} \end{matrix})

Thus H_{3} = LL^{T} with

L = (\begin{matrix}1 & 0 & 0 \\ 2^{−1} & 12^{−1/2} & 0 \\ 3^{−1} & 12^{−1/2} & 180^{−1/2} \end{matrix})

Second part If we take l_{1} = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4})^{T}, then

H_{4} − l_{1}l_{1}^{T} = (\begin{matrix}1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & λ \end{matrix}) − (\begin{matrix}1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{6} & \frac{1}{8} \\ \frac{1}{3} & \frac{1}{6} & \frac{1}{9} & \frac{1}{12} \\ \frac{1}{4} & \frac{1}{8} & \frac{1}{12} & \frac{1}{16} \end{matrix} )

= (\begin{matrix}0 & 0 & 0 & 0 \\ 0 & \frac{1}{12} & \frac{1}{12} & \frac{3}{40} \\ 0 & \frac{1}{12} & \frac{4}{45} & \frac{1}{12} \\ 0 & \frac{3}{40} & \frac{1}{12} & λ − \frac{1}{16} \end{matrix} )

If we take l_{2} = ((12)^{−1/2}, (12)^{−1/2}, (12)^{1/2} \frac{3}{40})^{T} then

(\begin{matrix}\frac{1}{12} & \frac{1}{12} & \frac{3}{40} \\ \frac{1}{12} & \frac{4}{45} & \frac{1}{12} \\ \frac{3}{40} & \frac{1}{12} & λ − \frac{1}{16} \end{matrix} ) − l_{2}l_{2}^{T} = (\begin{matrix}\frac{1}{12} & \frac{1}{12} & \frac{3}{40} \\ \frac{1}{12} & \frac{4}{45} & \frac{1}{12} \\ \frac{3}{40} & \frac{1}{12} & λ − \frac{1}{16} \end{matrix} ) − (\begin{matrix}\frac{1}{12} & \frac{1}{12} & \frac{3}{40} \\ \frac{1}{12} & \frac{1}{12} & \frac{3}{40} \\ \frac{3}{40} & \frac{3}{40} & \frac{27}{400} \end{matrix} )

= (\begin{matrix}0 & 0 & 0 \\ 0 & 180^{−1} & 120^{−1} \\ 0 & 120^{−1} & λ − \frac{13}{100} \end{matrix} )

If we take l_{3} = ((180)^{−1/2}, (180)^{1/2}120^{−1})^{T})

(\begin{matrix}180^{−1} & 120^{−1} \\ 120^{−1} & λ − \frac{13}{100} \end{matrix}) − l_{3}l_{3}^{T} = (\begin{matrix}0 & 0 \\ 0 & λ − \frac{57}{400} \end{matrix})

Thus λ_{0} = 57/400. We observe that

λ_{0} − \frac{1}{7} = \frac{1}{2800}.

(The ‘Hilbert matrices’ H_{n} are ‘only just positive definite’ and ‘only just invertible’ and are often used to test computational methods.)

If λ ≥ λ_{0} then H_{4} = LL^{T} with

L = (\begin{matrix}1 & 0 & 0 & 0 \\ 2^{−1} & 12^{−1/2} & 0 & 0 \\ 3^{−1} & 12^{−1/2} & 180^{−1/2} & 0 \\ 4^{−1} & (12)^{1/2} \frac{3}{40} & (180)^{1/2}120^{−1} & (λ − \frac{57}{400})^{1/2} \end{matrix})

Exercise 16.3.17

(i) The result is true when n = 1, since then the only root is a_{0} which is positive by hypothesis.

Suppose the result is true for m ≥ n ≥ 1 and P is a polynomial of the given form of degree m + 1. We observe that P′ is a polynomial of the given form of degree m and so all its real zeros are strictly positive. Thus, if m + 1 is even, P′(t) < 0 for t ≤ 0, so P is decreasing as t runs from −∞ to 0. But P (0) = a_{0} > 0 so P (t) > 0 for t ≤ 0 and P has no non-positive real zeros. If m + 1 is odd, P′(t) > 0 for t ≤ 0, so P is increasing as t runs from −∞ to 0. But P (0) = −a_{0} < 0, so P (t) < 0 for t ≤ 0 and P has no non-positive real zeros.

The result follows by induction.

(ii) The only if part is immediate. To obtain the if part apply part (i) to

P (t) = det(tI − A) = ∏_{j=1}^{n}(t − λ_{j}) = t^{n} + ∑_{j=1}^{n}(−1)^{n−j}a_{j}t^{j},

observing that

a_{n−1} = ∑_{j=1}^{n} λ_{j}, a_{n−2} = ∑_{j≠i} λ_{i}λ_{j} > 0, a_{n−3} = ∑_{i,j,k distinct} λ_{i}λ_{j}λ_{k}. . . . .

(iii) Let

A = (\begin{matrix}1 & −3 \\ 3 & 1 \end{matrix}) .

Then

det(tI − A) = (t − 1)^{2} + 9 = t^{2} − 2t + 8

is a polynomial with no real roots (since (t − 1)^{2} + 9 > 0 for all real t).

(iv) Observe that

det(tI − A) = t^{3} − b_{2}t^{2} + b_{1}t − b_{0}

with

b_{2} = a_{11} + a_{22} + a_{33}
b_{1} = a_{11}a_{22} + a_{22}a_{33} + a_{33}a_{22} − a_{12}a_{21} − a_{23}a_{32} − a_{31}a_{13} > 0

b_{0} = det A

and apply the results above.

Exercise 16.3.19

Suppose P is invertible with P^{T} AP and P^{T} BP diagonal. Since A is not semi-positive definite P^{T} AP is not and, since A is not semi-negative definite, P^{T} AP is not. Thus

P^{T} AP = (\begin{matrix}a & 0 \\ 0 & b \end{matrix})

with a and b non-zero of opposite sides. Observing that

(\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix}c & 0 \\ 0 & d \end{matrix}) (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) = (\begin{matrix}c & 0 \\ 0 & d \end{matrix})

and

(\begin{matrix}λ & 0 \\ 0 & μ \end{matrix}) (\begin{matrix}c & 0 \\ 0 & d \end{matrix}) (\begin{matrix}λ & 0 \\ 0 & μ \end{matrix}) = (\begin{matrix}λ^{2}c & 0 \\ 0 & μ^{2}d \end{matrix})

we see that there is a non-singular matrix Q with

Q^{T} (\begin{matrix}a & 0 \\ 0 & b \end{matrix}) Q = A

and Q^{T} CQ diagonal whenever C is diagonal.

Setting M = P Q we see that M is invertible M^{T} AM = A and M^{T} BM is diagonal.

Let

M = (\begin{matrix}a & b \\ c & d \end{matrix}) .

Since M^{T} AM = A we have

(\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix}) = A = M^{T} AM

= (\begin{matrix}a & c \\ b & d \end{matrix}) (\begin{matrix}a & b \\ −c & −d \end{matrix})

= (\begin{matrix}a^{2} − c^{2} & ab − cd \\ ba − cd & b^{2} − d^{2} \end{matrix})

Thus ab = cd and d > 0, a > 0

On the other hand

M^{T} BM = (\begin{matrix}a & c \\ b & d \end{matrix}) (\begin{matrix}c & d \\ a & b \end{matrix}) = (\begin{matrix}2ac & ad + bc \\ bc + ad & 2bd \end{matrix})

so if M^{T} BM is diagonal bc = −ad.

We now get

c^{2}d = abc = −a^{2}d

so, since d ≠ 0, we have c = a = 0 which contradicts the condition a^{2} − c^{2} = 1.

(ii) Since det A, det B ≠ 0, A and B have rank 2. A has signature 0 by inspection. B has signature 0 since xy = \frac{1}{4}((x + y)^{2} − (x − y)^{2}).

The transformation x ↦ Mx leaves lines x = y and x = −y unchanged (or interchanges them) since these are the asymptotes to x^{2} − y^{2} = K for all K. It therefore gives dilation along the directions x = y x = −y (and possibly interchanges these two axes) and will therefore not transform xy = k in the desired manner.

(iii) C is strictly negative definite and the same argument which gave simultaneous diagonalisation when one form is strictly positive definite will work if one form is strictly negative definite.

Exercise 16.4.2

We have

q (∑_{i=1}^{n} x_{i}e_{i}) = α (∑_{i=1}^{n} x_{i}e_{i}, ∑_{j=1}^{n} x_{j}e_{j})

= ∑_{i=1}^{n} ∑_{j=1}^{n} x_{i}a_{ij}x_{j}

= \frac{1}{2} ∑_{i=1}^{n} ∑_{j=1}^{n}(x_{i}a_{ij}x_{j} + x_{i}a_{ji}x_{j}) = \frac{1}{2} ∑_{i=1}^{n} ∑_{j=1}^{n} 0 = 0.

Exercise 16.4.3

Observe that

(iA)^{∗} = −iA^{∗} = −iA^{T} = iA.

Thus iA is Hermitian and so has real eigenvalues. If −λ is an eigenvalue of iA then iλ = (−i)(−λ) is an eigenvalue of A = (−i)(iA). Thus the eigenvalues of A are purely imaginary.

Observe that

(M^{T} BM)^{T} = M^{T} B^{T} M^{T T} = −M^{T} BM

so the eigenvalues of M^{T} BM are purely imaginary. But M^{T} BM is a real diagonal matrix whose diagonal entries are its eigenvalues which are thus real and so must be zero. Thus M^{T} BM = 0 and so B = 0.

Exercise 16.4.6

(i) Let u_{j} [1 ≤ j ≤ n] form a basis for an n-dimensional complex vector space U. Let α be the antisymmetric form associated with A for this basis.

Let e_{j} [1 ≤ j ≤ n] be the basis found in Theorem 16.4.5. If M is the appropriate basis change matrix M^{T} AM has the required form.

(ii) Observe that 2m = rank M^{T} AM = rank A.

Exercise 16.4.7

Observe that, by Theorem 16.4.5, if α is non-singular, 2m = n, so n is even.

Exercise 16.4.8

(i) True. Since we can find a non-singular M such that M^{T} AM has m copies of

(\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix})

along the diagonal and all other entries 0

rank A = rank M^{T} AM = 2m.

(ii) False. Take

A = (\begin{matrix}0 & 2 \\ −2 & 0 \end{matrix}) .

(iii) True. The non-real roots of a real polynomial occur in conjugate pairs. P_{A} is a real polynomial with roots the eigenvalues (over ℂ) and these are purely imaginary (see Exercise 16.4.3). Thus

P_{A}(t) = t^{n−2m} ∏_{j=1}^{m}(t + id_{r})(t − id_{r}) = ∏_{j=1}^{m}(t^{2} + d_{r}^{2})

with d_{r} real.

(iv) False. Take

A = (\begin{matrix}0 & 4 \\ −1 & 0 \end{matrix}) .

(v) True. Let A be the n × n matrix with the m matrices

(\begin{matrix}0 & d_{r} \\ −d_{r} & 0 \end{matrix})

along the diagonal and all other entries 0.

Exercise 16.4.9

We observe that

A is skew-Hermitian ⇔ A = iB where B is Hermitian.

(i) If A is skew-Hermitian, then, since iA is Hermitian, we can find unitary P such that iP^{∗}AP = P^{∗}(iA)P is a real diagonal matrix D and so P^{∗}AP = −iD is diagonal with diagonal entries purely imaginary.

(ii) If A is skew-Hermitian, then, since iA is Hermitian, Exercise 16.2.13 tells us that we can an invertible matrix P such that P^{∗}iAP such that P^{∗}iAP is diagonal with diagonal entries taking the values 1, −1 or 0 and so P^{∗}AP is diagonal with diagonal entries taking the values i, −i or 0.

(iii) Suppose that A is a skew-Hermitian matrix and P_{1}, P_{2} are invertible matrices such that P_{1}^{∗}AP_{1} and P_{2}^{∗}AP_{2} are diagonal with diagonal entries taking the values i, −i or 0. Then the number of entries of each type is the same for both diagonal matrices.

This follows at once from Exercise 16.2.13 (iii) by considering iA.