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Unformatted text preview: Solutions to quiz # 12 (December 2)
1. First, we ﬁnd the eigenvalues of A. The sum of the eigenvalues is 3 + 4 = 7, the
product is 4 · 3 − 2 · 1 = 10, so the eigenvalues are 2 and 5. Next, we ﬁnd eigenvectors. An
eigenvector with eigenvalue 2 is found by solving the system
21
21
so 0
21
−→
0
00 −1
is an eigenvector. An eigenvector with eigenvalue 5 is found by solving the system
2
−1 1
2 −2 so 0
,
0 1
1 
 0
1 −1
−→
0
00 0
,
0 is an eigenvector. Next, we write
4
−1
1
=x
+y
,
13
2
1 that is, x −1
2 y
1
1 1
 4 −→
0
 13 −1
3  −4
1
−→
 21
0 0
1 3
,
7 so x = 3 and y = 7. Finally,
A k v = 3A k Answer: Ak v = 1
−1
7 · 5k − 3 · 2k
1
−1
.
=
+ 7 · 5k
= 3 · 2k
+ 7A k
1
7 · 5k + 3 · 2k+1
2
1
2
7 · 5k − 3 · 2k
.
7 · 5k + 3 · 2k+1 2. First, we ﬁnd the eigenvalues from the equation 1−λ
a
0
0 = det 0
1−λ
b = (1 − λ)2 (2 − λ),
0
0
2−λ
so the eigenvalues are λ = 1 and λ = 2. To diagonalize the matrix, we need three linearly
independent eigenvectors. We get exactly one of these eigenvectors from the eigenvalue
λ = 2, so we need two linearly independent eigenvectors with eigenvalue λ = 1. That is,
we need 0a0
dim ker 0 0 b = 2,
001
1 that is, we need 0
1 = rank 0
0 a0
0a0
0 b = rank 0 0 0 .
01
001 In other words, we need a = 0.
Answer: The matrix is diagonalizable if and only if a = 0 (the value of b doesn’t matter). 2 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.
 Fall '08
 Conger
 Linear Algebra, Algebra, Eigenvectors, Vectors

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