11.1. EIGENVALUES AND EIGENVECTORS 189

Thus, as explained earlier, the last column is−5 times the first added to 5 times the second.Thus (

2 11 3

)2

=−5

(1 00 1

)+5

(2 11 3

)You can see from the row reduced echelon form that no smaller linear combination relatingthe matrices I,A,A2 is possible. Hence the minimal polynomial is

λ2−5λ +5

The eigenvalues are therefore, the roots of this polynomial. They are

52+

12

√5,

52− 1

2

√5

Now one can find eigenvectors associated with these. Consider the first of them. Wewant a nonzero vector x such that

(A−

(52 +

12

√5)

I)x= 0. Thus we need consider the

augmented matrix  2−(

52 +

12

√5)

1 0

1 3−(

52 +

12

√5)

0

We row reduce this to obtain (

1 12 −

12

√5 0

0 0 0

)Thus the eigenvectors are of the form

y

(12

√5− 1

21

),y ∈ C

Example 11.1.6 Find the minimum polynomial for

A =

 0 −2 −22 5 4−1 −2 −1

We look for linear combinations for A0,A,A2,A3 . These are the matrices, listed in

order of decreasing powers are −6 −14 −1414 29 28−7 −14 −13

 ,

 −2 −6 −66 13 12−3 −6 −5

 ,

 0 −2 −22 5 4−1 −2 −1

 ,

 1 0 00 1 00 0 1

We can arrange them as column vectors in C9 as done earlier, but it might be easier tosimply look at the entries in a single row or column. Lets pick the first column of each.Thus the augmented matrix to solve would be 1 0 −2 −6 0

0 2 6 14 00 −1 −3 −7 0

