12.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX 263

That is, the columns are the eigenvectors. Then

S−1 =

 −14

12

14

12 1 1

214

12 − 1

4

 .

Then S−1AS = −14

12

14

12 1 1

214

12 − 1

4

 2 0 0

1 4 −1−2 −4 4

 −2 1 0

1 0 10 1 −2

=

 2 0 00 2 00 0 6

 .

We know the result from the above theorem, but it is nice to see it work in a specific examplejust the same. You may wonder if there is a need to find S−1. The following is an exampleof a situation where this is needed. It is one of the major applications of diagonalizability.

Example 12.1.26 Here is a matrix. A =

 2 1 00 1 0−1 −1 1

 Find A50.

Sometimes this sort of problem can be made easy by using diagonalization. In this casethere are eigenvectors,  0

01

 ,

 −110

 ,

 −101

 ,

the first two corresponding to λ = 1 and the last corresponding to λ = 2. Then let theeigenvectors be the columns of the matrix, S. Thus

S =

 0 −1 −10 1 01 0 1

Then also

S−1 =

 1 1 10 1 0−1 −1 0

and S−1AS = 1 1 1

0 1 0−1 −1 0

 2 1 0

0 1 0−1 −1 1

 0 −1 −1

0 1 01 0 1

=

 1 0 00 1 00 0 2

= D

Now it follows

A = SDS−1 =

 0 −1 −10 1 01 0 1

 1 0 0

0 1 00 0 2

 1 1 1

0 1 0−1 −1 0

 .