256 CHAPTER 12. SPECTRAL THEORY

Here is another more interesting example of a defective matrix.

Example 12.1.14 Let

A =

 2 −2 −1−2 −1 −214 25 14

 .

Find the eigenvectors and eigenvalues.

In this case the eigenvalues are 3,6,6 where we have listed 6 twice because it is a zeroof algebraic multiplicity two, the characteristic equation being

(λ −3)(λ −6)2 = 0.

It remains to find the eigenvectors for these eigenvalues. First consider the eigenvectors forλ = 3. You must solve

 2 −2 −1−2 −1 −214 25 14

−3

 1 0 00 1 00 0 1

 x

yz

=

 000

 .

The augmented matrix is  −1 −2 −1 | 0−2 −4 −2 | 014 25 11 | 0

and the row reduced echelon form is 1 0 −1 0

0 1 1 00 0 0 0

so the eigenvectors are nonzero vectors of the form(

t −t t)T

= t(

1 −1 1)T

Next consider the eigenvectors for λ = 6. This requires you to solve 2 −2 −1−2 −1 −214 25 14

−6

 1 0 00 1 00 0 1

 x

yz

=

 000

and the augmented matrix for this system of equations is −4 −2 −1 | 0

−2 −7 −2 | 014 25 8 | 0

