Page 846 - Advanced_Engineering_Mathematics o'neil

826 Answers to Selected Problems

2
13. p A (λ) = λ(λ − 8λ + 7),
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
14 6 0
0, ⎝ 7 ⎠ ;1, ⎝ 0 ⎠ ;7, ⎝ 0 ⎠
10 5 1
The Gerschgorin circles have radius 2 and center (1,0) and radius 5 and center (7,0).
2
15. p A (λ) = (λ − 1)(λ − 2)(λ + λ − 13),
−2 0
⎛ ⎞ ⎛ ⎞
⎠ ;2, ⎝ ⎠ ;
⎜−11⎟ ⎜0⎟
0 1
1, ⎝
1 0
⎛ √ ⎞ ⎛ √ ⎞
53 − 7 − 53 − 7
√ 0 √ 0
(−1 + ⎜ ⎟ ⎜ ⎟
⎠ ;(−1 −
0 53)/2, ⎝ 0 ⎠
53)/2, ⎝
2 2
The Gerschgorin circles have radius 2, center (−4,0) and radius 1 and center (3,0).
2
17. p A (λ) = λ − 5λ,

1 −2
0, ;5,
2 1
2
19. p A (λ) = λ − 10λ − 23,
√ √ √ √
2
2
5 + 2, 1 + ;5 − 2, 1 −
1 1
2
21. p A (λ) = (λ − 3)(λ + 2λ − 1),
⎛ ⎞ ⎛ √ ⎞ ⎛ √ ⎞
0 √ 1 + 2 √ 1 − 2
3, ⎝ 0 ⎠ ;−1 + 2, ⎝ 1 ⎠ ;−1 − 2, ⎝ 0 ⎠
1 0 4
23. If AE = λE,then
2
2
A E = A(AE) = A(λE) = λAE = λ E.
k
k
Repetition of this argument yields the general conclusion A E = λ E.
Section 9.2 Diagonalization
In Problems 1 through 9, P is given, or it is stated that the matrix is not diagonalizable.
√
√
−3 + 7i −3 − 7i
1.
8 8
3. Not diagonalizable (A does not have two linearly independent eigenvectors)
⎛ ⎞
0 5 0
5. ⎝ 1 1 −3 ⎠
0 0 2
7. A is not diagonalizable
⎛ ⎞
1 0 0 0
√ √
⎜ 0 1 (2 − 3 5)/41 (2 + 3 5)/41 ⎟
9. ⎜ √ √ ⎟
⎝ 0 0 (−1 + 5)/2 (−1 − 5)/2 ⎠
0 0 1 1
−1
−1
11. Since P AP = D,then A = PAP ,so
−1
−1
−1
k
A = (PDP )(PDP )···(PDP )
−1
k
= PD P .
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October 14, 2010 17:50 THM/NEIL Page-826 27410_25_Ans_p801-866