Page 300 - Advanced engineering mathematics

280 CHAPTER 9 Eigenvalues, Diagonalization, and Special Matrices

EXAMPLE 9.9
Let
⎛ ⎞
−11 3
A = ⎝ 2 1 4 ⎠ .
1 0 −2
√ √
The eigenvalues are −1,(−1 + 29)/2 and (−1 − 29)/2, and corresponding eigenvectors are
√ √
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 3 + 29 3 − 29
√ √
, 10 + 2 29 , 10 − 2 29 .
⎝ −3 ⎠ ⎝ ⎠ ⎝ ⎠
1 2 2
These are linearly independent because the eigenvalues are distinct. Use these eigenvectors as
columns of P to form
√ √
⎛ ⎞
1 3 + 29 3 − 29
√ √
P = −3 10 + 2 29 10 − 2 29 .
⎝
⎠
1 2 2
We find that
√ √ √
√ ⎛ 232/ 29 −116/ 29 232/ 29 ⎞
29 √ √ √
−1
P = ⎝ 16 − 2 29 −1 + 29 −19 + 5 29 ⎠
812 √ √ √
−16 − 2 29 1 + 29 19 + 5 29
and
⎛ ⎞
−1 0 0
√
−1 ⎝ 0
P AP = (−1 + 29)/2 0 ⎠ ,
√
0 0 (−1 − 29)/2
with the eigenvalues down the main diagonal in the order of the eigenvalues listed for
columns of P.
In this example, P −1 is an unpleasant matrix. One of the values of Theorem 9.6 is that it
−1
tells us what P AP looks like, without actually having to determine P −1 and carry out this
product.
Although n distinct eigenvalues guarantee that A is diagonalizable, an n × n matrix with
fewer than n distinct eigenvalues may still be diagonalizable. This will occur if we are able to
find n linearly independent eigenvectors.

EXAMPLE 9.10
Let
⎛ ⎞
5 −4 4
A = 12 −11 12 ⎠
⎝
4 −4 5
as in Example 9.5. We found the eigenvalues −3,1,1, with a repeated eigenvalue. Nevertheless,
we were able to find three linearly independent eigenvectors. Use these as columns to form
⎛ ⎞
1 1 0
P = 3 0 1 ⎠ .
⎝
1 −1 1

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October 14, 2010 14:49 THM/NEIL Page-280 27410_09_ch09_p267-294