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(math111)[2008](f)MATH111Q4~PPSpider^_10429.pdf
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MATH111 Quiz 4A Solutions
Page 1/4
November 27, 2008 by Daniel Zheng
..................................................40 mins allowed..................................................

Name: Student ID: Score:
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I. [6] Diagonalize matrix A = .10 .3 . You need to specify the diagonal 125
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matrix D and the invertible matrix P , such that A = P DP .1 .
Solutions.
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100 .2 .2 .3
D = 020 ,P = .110 . 002 101
MATH111 Quiz 4A Solutions Page 2/4
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II. Determine whether the following statements are true or not. If true, justify your
answer; if false, give a counterexample.
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[3] If x is an eigenvector of AB, then Bx is an eigenvector of BA.

2.
[3] Similar matrices always have exactly the same eigenvalues.


Solutions.
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1. False. It may occur that Bx = 0. Considering A = B = and 00
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x = . Then Bx =.
10

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2. True. Assume A = P BP .1, then
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det(A . I) = det(P BP .1 . I) = det(P (B . I)P .1) = det(B . I).
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MATH111 Quiz 4A Solutions Page 3/4
III. [4] Denote the eigenvalues of A as 1, ,p. Show that the matrices I A are invertible when |j| < 1, for all j =1, ,p.(Hint: Show det(I A) = 0.) .Proof. Since all eigenvalues are less than 1 in magnitude, 1 cant be eigenvalues of A. Thus, det(A I) .= 0. Hence det(I A) .= 0.
MATH111 Quiz 4A Solutions Page 4/4
IV. [4] Suppose A is diagonalizable and let p(t) be the characteristic polynomial of
A. De.ne p(A) to be the matrix formed by replacing each power of t by the corresponding power of A (tk Ak, with A0 = I). Show that p(A)= O, where
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O represents the zero matrix. (Hint: if A = P DP .1, then Ak = PDkP .1.) Proof. Assume A = P DP .1, and p(t)= a0 + a1t + + antn, then p(A)= a0I + a1A + + anAn = P (a0I + a1D + + anDn)P .1, using the conclusion Ak = PDkP .1 . Note that if we denote the eigenvalues as 1, ,n, counting the multiplicity, then
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p(1)
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Dn .
a0I + a1D + + an=. . (1) p(n)
Since p(t) is the characteristic polynomial, we have p(j) = 0 for all j. Therefore, matrix (1) is the zero matrix, so is p(A).