Page 301 - Advanced engineering mathematics

9.2 Diagonalization 281

Then P diagonalizes A:
⎛ ⎞
−300
−1 ⎝ 0
P AP = 1 0 ⎠ .
0 0 1
Again, we know this from Theorem 9.6, without explicitly computing the product P AP.
−1
If A has fewer than n linearly independent eigenvectors, then A is not diagonalizable.

We will now prove Theorem 9.6.
Proof Let the eigenvalues of A be λ 1 ,λ 2 ,··· ,λ n (not necessarily distinct). Suppose first that
these eigenvalues have corresponding linearly independent eigenvectors V 1 ,V 2 ,··· ,V n . These
form the columns of P, which we indicate by writing
⎛ ⎞
| | ··· |
.
P = V 1 V 2 ··· V n ⎠
⎝
| | ··· |
P is nonsingular because its columns are linearly independent.
Let D be the n ×n diagonal matrix having the eigenvalues of A, in the given order, down the
main diagonal. We want to prove that
−1
P AP = D.
We will prove this by showing by direct computation that
AP = PD.
First, recall that the product AP has as columns the product of A with the columns of P. Thus

column j of AP = A(column j of P)
= A(V j ) = λ j V j .
Now compute PD. As a convenience in understanding the computation, write
⎛ ⎞
v 1 j
⎜ ⎟
v 2 j
V j = ⎜ ⎟ .
⎝ ··· ⎠
v nj
Then
⎛ ⎞⎛ ⎞
··· 0 ··· 0
v 11 v 12 v 1n λ 1
··· 0 ··· 0
v 21 v 22 v 2n λ 2
⎜ ⎟⎜ ⎟
PD = ⎜ . . . . ⎟⎜ . . . ⎟
⎜
⎟⎜
⎝ . . . . . . . ⎠⎝ . . . . . . . ⎟
.
.
. ⎠
0 0
v n1 v n2 ··· v nn ··· λ n
⎛ ⎞
λ 1 v 11 λ 2 v 12 ··· λ n v 1n
λ 1 v 21 λ 2 v 22 λ n v 2n
⎜ ··· ⎟
⎜
= ⎜ . . . ⎟
⎝ . . . . . . . ⎟
.
. ⎠
λ 1 v n1 λ 2 v 2n ··· λ n v nn
⎛ ⎞
| | ··· |
= AP,
= λ 1 V 1 λ 2 V 2 ··· λ n V n ⎠
⎝
| | ··· |
since column j of this matrix is λ j V j .
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October 14, 2010 14:49 THM/NEIL Page-281 27410_09_ch09_p267-294