Page 303 - Advanced engineering mathematics

9.2 Diagonalization 283

Notice that, if P diagonalized A, then P would have to have eigenvectors of B as columns.
Then P would have to have the form

α β
P =
0 0
for some nonzero α and β. But this matrix is singular, with no inverse, because |P|= 0.

The key to diagonalizing A is the existence of n linearly independent eigenvectors.
By Theorem 9.2, one circumstance in which this always happens is that A has n distinct
eigenvalues.

COROLLARY 9.1

An n × n matrix with n distinct eigenvalues must be diagonalizable.

EXAMPLE 9.12
Let
⎛ ⎞
−200 5
⎜ 1 3 0 0 ⎟
A = ⎜ ⎟ .
⎝ 0 4 4 0 ⎠
2 0 0 −3
√ √
A has eigenvalues 3,4,(−5 + 41)/2 and (−5 − 41)/2. Because these are distinct, A has 4
linearly independent eigenvectors and therefore is diagonalizable. There is a matrix Psuch that
⎛ ⎞
3 0 0 0
⎜0 4 0 0
−1 √ ⎟
P AP = ⎜ ⎟ .
⎝ 0 0 (−5 + 41)/2 0 ⎠
√
0 0 0 (−5 − 41)/2
We do not have to actually write down P (this would require finding eigenvectors) or compute
P −1 to draw this conclusion.

SECTION 9.2 PROBLEMS

⎛ ⎞
In each of Problems 1 through 10, produce a matrix P that 5 0 0
diagonalizes the given matrix, or show that the matrix is 5. ⎝ 1 0 3 ⎠
−1
not diagonalizable. Determine P AP. Hint: Keep in mind 0 0 −2
that it is not necessary to compute P to know this product
⎛ ⎞
matrix. 0 0 0
6. ⎝ 1 0 2 ⎠
0 1 3
0 −1
1.
4 3
⎛ ⎞
−2 0 1

5 3
2. 7. ⎝ 1 1 0 ⎠
1 3
0 0 −2

1 0
3. ⎛ ⎞
−4 1 2 0 0
8. ⎝ 0 2
1 ⎠
−5 3
4. 0 −1 2
0 9
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October 14, 2010 14:49 THM/NEIL Page-283 27410_09_ch09_p267-294