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Mathematics

MATA23H3

Kathleen Smith

Winter

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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] ad the stndard 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 ﬁd the ssocated egenvectors.
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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