Finite-Dimensional Linear Algebra Errata for the first printing

Transcription

1 Finite-Dimensionl Liner Algebr Errt for the first printing Mrk S. Gockenbch Jnury 6, 011 The following corrections will be mde in the second printing of the text, expected in 011. Pge 41: Exercise 5: S is subset S is subspce. Pge 65: Exercise 14: belongs in Section.7. (Not chnged in the second printing.) Pge 65: Exercise 16: should red (cf. Exercise.3.1), not (cf. Exercise..1). Pge 71: Exercise 9 (b): Z 5 4 Z 4 5. Pge 7: Exercise 11: over V over F. Pge 7: Exercise 15: i = 1,,..., k j = 1,,..., k (twice). Pge 79: Exercise 1: x 3 = x = 3. Pge 8: Exercise 14(): Ech A i nd B i hs degree n + 1 should red A i, B i P n+1 for ll i = 0, 1,..., n. Pge 100, Exercise 11: K : C[, b] C[, b] K : C[c, d] C[, b] Pge 114, Line 9: L : F n R m L : F n F m. Pge 115: Exercise 8: S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} X = {(1, 1, 1), (0, 1, 1), (0, 0, 1)}. S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, X = {(1, 1, 1), (0, 1, 1), (0, 0, 1)}. Pge 116, Exercise 17(b): F mn F mn. Pge 11, Exercise 3: T : R 4 R 3 T : R 4 R 4. Pge 14, Exercise 15: T : X/ker(L) R(U) T : X/ker(L) R(L). Pge 14, Exercise 15: T ([x]) = T (x) for ll [x] X/ker(L) T ([x]) = L(x) for ll [x] X/ker(L). 1

2 Pge 19, Exercise 4(b): Period is missing t the end of the sentence. Pge 130, Exercise 8: L : Z 3 3 Z 3 3 should red L : Z 3 5 Z 3 5 Pge 130, Exercise 13(b): T defines... S defines.... Pge 131, Exercise 15: K : C[, b] C[c, d] C[, b] K : C[c, d] C[, b]. Pge 131, Exercise 15: In the first displyed eqution, s t should be s b. Pge 138, Exercise 7(b): define defined. Pge 139, Exercise 8(b): define defined. Pge 139, Exercise 1: In the lst line, sp{x 1, x,..., x n } sp{x 1, x,..., x k }. Pge 139, Exercise 1: The proposed pln for the proof is not vlid. Insted, the instructions should red: Choose vectors x 1,..., x k X such tht {T (x 1 ),..., T (x k )} is bsis for R(T ), nd choose bsis {y 1,..., y l } for ker(t ). Prove tht {x 1,..., x k, y 1,..., y l } is bsis for X. (Hint: First show tht ker(t ) sp{x 1,..., x k } is trivil.) Pge 140, Exercise 15: In the displyed eqution, A ii A ii. Pge 168: Definition 13 defines the djcency mtrix of grph, not the incidence mtrix (which is something different). The correct term (djcency mtrix) is used throughout the rest of the section. (Chnge incidence to djcency in three plces: the title of Section , Pge 168 line -, Pge 169 line 1.) Pge 199, Eqution (3.41d): x 1, x 0 x 1, x 0. Pge 04, Exercise 10: α 1,..., α k R α 1,..., α k 0. Also, C should not be boldfce in the displyed formul. Pge 1, Exercise 9: m > n m < n. Pge 4, Corollry 194: i = 1,,..., m. Pge 51, Exercise 18(e): for ech i = 1,,..., t for ech w = w = [ 0 v [ 0 v (Tht is, the comm period.) Pge 56, Exercise 13: First line should red Let X be finite-dimensionl vector spce over C with bsis.... References in prt (b) to F n n, F k k, F k l, F l l replced with C n n, etc. Also, in prt (b), Prove tht [T ] X replced with Prove tht [T ] X,X. Pge 64, Exercise 3: Add Assume {p, q} is linerly independent. Pge 71, Exercise 3:... we introduced the incidence mtrix... should be... we introduced the djcency mtrix.... Pge 8, Exercise 6: S = sp{(1, 3, 3, ), (3, 7, 11, 4)} S = sp{(1, 4, 1, 3), (4, 7, 19, 3)}. Pge 8, Exercise 7(b): N (A) col(a) N (A) col(a) = {0}. ], ].

3 Pge 83, Exercise 1: Lemm 5.1. Lemm 9. Pge 306, Exmple 5: B = {p 0, D(p 0 ), D (p 0 )} = {x, x, } should be B = {D (p 0 ), D(p 0 ), p 0 } = {, x, x }. Also, [T ] B,B [D] B,B (twice). Similrly, A defined s {, 1 + x, 1 x + x } nd [T ] A,A [D] A,A. Pge 308, Exercise 3: Suppose X is vector spce... Suppose X is finite-dimensionl vector spce.... Pge 311, Line 7: corresponding to λ corresponding to λ i. Pge 316, Exercise 6(f): Should end with ; insted of.. Pge 317, Exercise 15: ker((t λi) ) = ker(a λi) ker((t λi) ) = ker(t λi). Pge 3, displyed eqution (5.1): The lst line should red v r = λv r. Pge 35, Exercise 9: If U(t 0 ) is singulr, sy U(t)c = 0 for some c C n, c 0 If U(t 0 ) is singulr, sy U(t 0 )c = 0 for some c C n, c 0. Pge 331, Line 16:... is t lest t is t lest s + 1. Pge 337, Line -7: Pge 339, Line 13: [ n ] 1/p x p = x p i for ll x R n. i=1 [ n ] 1/p x p = x i p for ll x R n. i=1 [ ] 1/p b f p = f(x) p for ll x C[, b]. [ ] 1/p b f p = f(x) p for ll x C[, b]. Pge 339, Line -1: Replce The spce L p (, b) is defined to be the spce of ll (Lebesgue... with Roughly speking, the spce L p (, b) is the spce of ll (Lebesgue... Pge 340, Line : b b f(x) p dx <, f(x) p dx <, Pge 340, Line 3: Replce Lebesgue mesure nd integrtion, nd the resulting L p spces,... with Lebesgue mesure nd integrtion, nd the precise definition of the L p spces,... 3

4 Pge 356, Exercise 9:... such tht {x 1, x, x 3, x 4 }.... such tht {x 1, x, x 3, x 4 } is n orthogonl bsis for R 4. Pge 356, Exercise 13:... be linerly independent subset of V... be n orthogonl subset of V Pge 356, Exercise 14:... be linerly independent subset of V... be n orthogonl subset of V Pges : Miscellneous exercises 1 1 numbered. Pge 365, Exercise 6 ( 7):... under the L (0, 1) norm should be... under the L (0, 1) inner product. Pge 383, Line 1: col(t ) col(a) nd col(t ) col(a). Pge 383, Exercise 3:... bsis for R 4... bsis for R 3. Pge 384, Exercise 6: bsis bses. Pge 385, Exercise 14: Exercise Exercise Tht exercise lso Exercise Pge 385, Exercise 15: See Exercise 6.4 See Exercise Pge 400, Exercise 4: ( f(x) = f + b ) (t + 1). ( f(x) = f + b ) (x + 1). Pge 410, Exercise 1: The problem should specify l = 1, k(x) = x + 1, f(x) = 4x 1. Pge 411, Exercise 6: u(l) = 0. u(l) = 0 (i.e. there should not be period fter 0). Pge 44, Exercise 1:... prove (1)... prove (6.50). Pge 43, Exercise 9: G 1/ is the inverse of G 1/ G 1/ is the inverse of G 1/. Pge 433, Exercise 16:... so we will try to estimte the vlues u(x 1 ), u(x ),..., u(x n )... so we will try to estimte the vlues u(x 1 ), u(x ),..., u(x n 1 ). Pge 438, Exercise 3:... define T : R n F n... define T : F n F n. Pge 448, Exercise 8: In the formul for f, 00x 1x 00x 1x. Also, ( 1., 1) (1, 1). Pge 453, Exercise 6: Add: Assume g(x(0)) hs full rnk. Pge 46, Exercise 6: Add: λ 1 λ. Pge 475, Exercise 10: A = GH A = GQ. Pge 476, Exercise 15(): m n A F = A ij for ll A Cm n. i=1 j=1 4

5 m n A F = A ij for ll A C m n. i=1 j=1 Pge 476, Exercise 15: No need to define C F gin. Pge 501, lst prgrph: The text fils to define k l (mod p) for generl k, l Z. The following text dded: In generl, for k, l Z, we sy tht k l (mod p) if p divides k l, tht is, if there exists m Z with k = l + mp. It is esy to show tht, if r is the congruence clss of k Z, then p divides k r, nd hence this is consistent with the erlier definition. Moreover, it is strightforwrd exercise to show tht k l (mod p) if nd only if k nd l hve the sme congruence clss modulo p. Pge 511, Theorem 381: A (k) ij = A ij M (k) ij = A ij. Pge 516, Exercise 8: A (1), A (),..., A (n 1) M (1), M (),..., M (n 1). Pge 516, Exercise 10: n / n/ n n. Pge 53, Exercise 6(b):... the columns of AP re the columns of AP T re.... Pge 535, Theorem 401: A 1 A (twice). Pge 536, Exercise 1:... be ny mtrix norm be ny induced mtrix norm.... Pge 554, Exercise 4:... b is consider the dt b is considered the dt.... Pge 554, Exercise 5:... b is consider the dt b is considered the dt.... Pge 563, Exercise 7: Let v R m be given nd define α = ± x, let {u 1, u,..., u m } be n orthonorml bsis for R m, where u 1 = x/ x... Let v R m be given, define α = ± v, x = αe 1 v, u 1 = x/ x, nd let {u 1, u,..., u m } be n orthonorml bsis for R m,.... Pge 571, Exercise 3: Prove tht the ngle between A k v 0 nd x 1 converges to zero s k Prove tht the ngle between A k v 0 nd sp{x 1 } = E A (λ 1 ) converges to zero s k. Pge 575, line 15: 3n n 3n + n 5. Pge 575, line 16: n squre roots n 1 squre roots. Pge 580, Exercise 3:... requires 3n n rithmetic opertions, plus the clcultion of n squre roots, requires 3n +n 5 rithmetic opertions, plus the clcultion of n 1 squre roots,.... Pge 585, line 19: originl subsequence originl sequence. Pge 585, line 0: originl subsequence originl sequence. Pge 604, Exercise 4: Theorem 4 Theorem 451. Pge 608, line 18:... exists rel number exists s rel number.... 5