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MATA23H3 (19)
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# s11.pdf

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Department
Mathematics
Course
MATA23H3
Professor
Kathleen Smith
Semester
Winter

Description
University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT A23H 2013/2014 Solutions #11 1. (Fraleigh & Beauregard, pages 300 – 303) (18) T([x,y]) = [x − y,−x + y] so 1(e ) = [▯,−1],2T▯e ) = [−1,1] and the standard 1 −1 matrix represention is A = [1(e ),2(e )]−1 1 . The characteristic poly- ▯ ▯ ▯1 − λ −1 ▯ 2 2 nomial is p(λ) = det(A−λI) =▯ −1 1 − λ = (1−λ) −1 = λ −2λ = λ(λ−2) so the ▯igenva▯ues ▯re λ ▯ 0, λ = 2. We now ﬁnd the ▯s▯ociat▯d▯eigenvectors. λ = 0: 1 −1 ∼ 1 −1 . We put x = t giving x = t = t 1 , t ∈ R. −1 1 0 0 2 t 1 ▯ ▯ The eigenvectors associated to λ = 0 are of the form, t ▯= 0. ▯ ▯ ▯ ▯ 1▯ ▯ ▯ ▯ −1 −1 1 1 −t −1 λ = 2: ∼ . We put x2= t giving x = = t , t ∈ R. −1 −1 0 0 ▯ ▯ 1 −1 The eigenvectors associated to λ = 2 are of the form t t ▯= 0. 1 (20) T([1 ,2 ,3 ]) =1[x 24x + 3x 22x −3x ] so 1(e ) = [1,0,0]2 T(e ) = [0,4,2] and T(e 3 = [0,7,−1] ad the stndard matrix representation is A = ▯ ▯ 1 0 0 T(e ),T(e ),T(e ) = 0 4 7 . The characteristic polynomial is p(λ) = 1 2 3 ▯ 0 2 −1 ▯ ▯1 − λ 0 0 ▯ det(A−λI) = ▯ 0 4 − λ 7 ▯= (1−λ)[(4−λ)(1−λ)−14] = (1−λ)(λ −2 ▯ ▯ ▯ 0 2 −1 − λ▯ 3λ − 18) = (1 − λ)(λ − 6)(λ + 3) so the eigenvalues are λ = −3, λ = 1, λ = 6. We now ﬁd the ssocated egenvectors.     4 0 0 1 0 0 0 0 λ = −3: 0 7 7  ∼ 0 1 1  . We put 3 = t giving x =t  = t −1 , t ∈ 0 2 2 0 0 0 t 1   0 R. The eigenvectors associated to λ = −3 are of the form t t ▯= 0.      1   0 0 0 0 1 0 t 1 λ = 1: 0 3 7  ∼  0 0 1 . We put x = t giving x =0 = t 0 , t ∈ R. 1 0 2 2 0 0 0 0 0 MATA23H Solutions # 11 page 2   1   The eigenvectors associated to λ = 1 are of the form t t ▯= 0. 0       −5 0 0 1 0 0 0 λ = 6: 0 −2 7 ∼  0 −2 7  . We put x 3 = t giving x =  7t = 2   0 2 −7 0 0 0   t 0 0     t 2 , t ∈ R. The eigenvectors associated to λ = 6 are of the form t ▯= 0. 1 2 (23) (a) False; (b) False; (c) True; (d) True; (e) False; (f) True; (g) False; (h) True; (i) False; (j) True. Since general vector spaces have not been introduced, for the rest of this solution set, we will replace V by R . n (24) Let W = {v ∈ R | T(v) = λv}. W is nonempty since 0 ∈ W (T(0) = 0 = λ0). Let u, w ∈ W and let t ∈ R. Now T(u+w) = T(u)+T(w) = λu+λw = λ(u+ w) =⇒ u + w ∈ W. Also T(tu) = tT(u) = t(λu) = λ(tu) =⇒ tu ∈ W. Since the conditions are satisﬁed, W is a subspace of R . (25) Assume that A A has eigenvalue λ; i.e., (A A)v = λv for some nonzero vector T T v. Now (AA )Av = A(A A)v = A(λv) = λ(Av) so we see that, if λ is an eigenvalue of A A with eigenvector v, then λ is an eigenvalue of Awith eigenvector Av. On the other hand, if (AA )u = µu, then (A A)A u = A (AA )u = T T T T A (µu) = µ(A u). Therefore, if µ is an eigenvalue of AA with eigenvector u, then µ is also an eigenvalue of A A with eigenvector A u. (30) A square matrix A is invertible if and only if detA ▯= 0. Since detA = det(A − 0I), detA ▯= 0 if and only if det(A − 0I) ▯= 0. Now det(A − 0I) ▯= 0 if and only if 0 is not an eigenvalue of A. Hence A is invertible if and only if no eigenvalue of A is zero. (32) Assume that λ is an eigenvalue of A and that v is an associated eigenvector. Since (A + r I)v = Av + r I v = λv + r v = (λ + r)v, λ + r is an eigenvalue of A + r I and v is an associated eigenvector. Hence the eigenvalues of A + r I are the eigenvalues of A increased by r while the corresponding eigenvector remains the same. T T T T T T (34) If A w = αw, then (A w) = (αw) or w A = αw ; hence the name “left” eigenvalue and “left” eigenvector. (35) Let Av = λv and let A w = αw. From #34, we have w A = αw and T (i) (w A)v = (αw )v = α(w v) = α(w ·v) T T T (ii) (w A)v = w (Av) = w (λv) = λ(w ·v) MATA23H Solutions # 11 page 3 Hence α(w ·v) = λ(w ·v) =⇒ (α − λ)w ·v = 0. Since α ▯= λ we must have w ·v = 0; i.e., w and v are orthogonal. (36) (a) Since detB = detB , we have det(A−λI) = det(A−λI) = det(A −λI). Hence the eigenvalues of A are the same as those of A . ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ 1 1 1 1 1 1 1 (b) Let 0 1 . Since 0 1 0 = 0 , 0 is an eigenvector of A associated ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ to λ = 1. However, A 1 = 1 0 1 = 1 ▯= t 1 for all t ∈ R. Hence 0 1 1 0 1 0 ▯ ▯ 1 is not an eigenvector of A . 0 −1 −1 (38) We start with the characteristic polynomial oAC : det(C AC − λI) = det(C−1AC−C −1(λI)C) = det(C−1(A−λI)C) = (detC −1)(det(A−λI))(detC) −1 −1 = (det(A−λI))(detC )(detC) = det(A−λI) det(C C) = det(A−λI) detI = det(A−λI). Hence C −1AC and A have the same characteristic polynomial and, consequently, have the same eigenvalues. ▯▯ ▯ ▯ ▯▯ ▯ ▯ 2 −1 1 0 2 − λ −1 (39) The characteristic polynomial is det − λ = det 1 3 0 1 1 3 − λ = (2 − λ)(3 − λ) + 1 = λ − 5λ + 7. We now compute A − 5A + 7I giving ▯ ▯2 ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ 2 −1 − 5 2 −1 + 7 1 0 = 3 −5 − 10 −5 + 7 0 = 0 0 . 1 3 1 3 0 1 5 8 5 15 0 7 0 0 Hence A has satisﬁed its own characteristic polynomial, illustrating the Cayley- Hamilton Theorem. (40) Let the characteristic polynomial of A be p(λ) = p λ + p λn−1 + ... + n n−1 p1λ + p0. Since A is invertible, λ = 0 is not an eigenvalue,0so p ▯= 0. Now Cayley-Hamilton gives p(A) =np A + n−1 An−1 + ... + p 1 + p 0 = O =⇒ n n−1 −1 pnA +p n−1A + ... +p 1 = −p I0 Multiplying both sides by A we have p An−1 +p An−2+ ... +p I = −p A −1 =⇒ A −1 = − 1 (p A n−1+ ··· + n n−1 1 0 p0 n p I). 1 ▯▯ ▯ −1 1 1 2 −1 Using the example from #39, we have A = − 7[A − 5I] = − 7 1 3 − ▯ ▯▯ ▯ ▯ ▯ 3 1▯ 5 0 1 −3 −1 7 7 0 5 = − 1 −2 = 1 2 . 7 − 7 7 2. (Fraleigh & Beauregard, page 315 – 317) ▯ ▯ ▯ ▯ −3 4 ▯−3 − λ 4 ▯ (1) A = . p(λ) = det(A − λI) ▯ ▯= (−3 − λ)(3 − λ) − 16 = 4 3 4 3 − λ λ − 25 = (λ + 5)(λ − 5). The eigenvalues are λ = −5, λ = 5. We now ﬁnd the associated eigenvectors. ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ 2 4 1 2 −2t −2 λ = −5: 4 8 ∼ 0 0 giving eigenvectors v = t = t 1 , t ▯= 0. ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ λ = 5: −8 4 ∼ −2 1 giving eigenvectors v =t = t 1 , t ▯= 0. 4 −2 0 0 2t 2 MATA23H Solutions # 11 page 4 ▯ ▯ ▯ ▯ ▯ ▯ −2 1 −2 1 −1 Choosing eigenvectors1 and 2 we get C = 1 2 . Then C AC = D = ▯ ▯ −5 0 . 0 5 ▯ ▯ ▯ ▯ 3 2 ▯3 − λ 2 ▯ (2) A = 1 4 . p(λ) = det(A − λI) =▯ 1 4 − λ = (3 − λ)(4 − λ) + 10 = (λ − 2)(λ − 5). The eigenvalues are λ = 2, λ = 5. We now ﬁnd the associated eigenv▯cto▯s. ▯ ▯ ▯ ▯ ▯ ▯ 1 2 1 2 −2t −2 λ = 2: 1 2 ∼ 0 0 giving eigenvectors v =t = t 1 , t ▯= 0. ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ λ = 5: −2 2 ∼ 1 −1 giving eigenvectors v == t 1 , t ▯= 0. 1 −1 0 0 t 1 ▯−2 ▯ ▯1▯ ▯−2 1 ▯ Choosing eigenvectors and we get C = . Then C1 AC = D = ▯ ▯ 1 1 1 1 2 0 . 0 5  ▯ ▯ 6 3 −3 6 − λ 3 −3 ▯  
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