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9.1 Eigenvalues and Eigenvectors 269

has a nontrivial solution E. The condition for this is that the coefficient matrix be singular
(determinant zero), hence that
|λI n − A|= 0.

If expanded, the determinant on the left is a polynomial of degree n in the unknown λ, and
is called the characteristic polynomial of A. Thus
p A (λ) =|λI n − A|.

This polynomial has n roots for λ (perhaps some repeated, perhaps some or all complex).
These n numbers, counting multiplicities, are all of the eigenvalues of A. Corresponding to each
eigenvalue λ, a nontrivial solution of
(λI n − A)X = O

is an eigenvector.
We can summarize this discussion as follows.

THEOREM 9.1 Eigenvalues and Eigenvectors of A

Let A be an n × n matrix of numbers. Then

1. λ is an eigenvalue of A if and only if λ is a root of the characteristic polynomial of A.
This occurs exactly when

p A (λ) =|λI n − A|= 0.
Since p A (λ) has degree n, A has n eigenvalues, counting each eigenvalue as many times
as it appears as a root of p A (λ).
2. If λ is an eigenvalue of A, then any nontrivial solution E of
(λI n − A)X = O
is an eigenvector of A associated with λ.
3. If E is an eigenvector associated with the eigenvalue λ, then so is cE for any nonzero
number c.

EXAMPLE 9.3
Let
⎛ ⎞
1 −1 0
A = 0 1 1 ⎠ ,
⎝
0 0 −1
as in Example 9.2. The characteristic polynomial is

λ − 1 1 0

2
p A (λ) =|λI 3 − A|= 0 λ − 1 −1 = (λ − 1) (λ + 1).

0 0 λ + 1
This polynomial has roots 1,1,−1 and these are the eigenvalues of A. The root 1 has multiplicity
2 and must be listed twice as an eigenvalue of A. A has three eigenvalues.

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October 14, 2010 14:49 THM/NEIL Page-269 27410_09_ch09_p267-294