6.4. DIAGONALIZABILITY 137

Example 6.4.5 The minimum polynomial for the matrix

A =

 10 12 −6−4 −4 33 4 −1

is λ

3−5λ2+8λ −4. This factors as (λ −2)2 (λ −1) and so the eigenvalues are 1,2. Find

the eigen-pairs. Then determine the matrix with respect to a basis of these eigenvectors ifpossible. If it is not possible to find a basis of eigenvectors, find a block diagonal matrixsimilar to the matrix. Note that from the above theorem, it is not possible to diagonalizethis matrix.

First find the eigenvectors for 2. You need to row reduce 10−2 12 −6 0−4 −4−2 3 03 4 −1−2 0

This yields  1 0 −3 0

0 1 32 0

0 0 0 0

Thus the eigenvectors which go with 2 are(

6z −3z 2z)T

, z ∈ R, z ̸= 0

The eigenvectors which go with 1 are

z(

2 −1 1)T

, z ∈ R, z ̸= 0

By Theorem 6.3.3, there are no other eigenvectors than those which correspond to eigen-values 1,2. Thus there is no basis of eigenvectors because the span of the eigenvectors hasdimension two.

However, we can consider

R3 = ker((A−2I)2

)⊕ker(A− I)

The second of these is just span((

2 −1 1)T). What is the first? We find it by row

reducing the following matrix which is the square of A− 2I augmented with a column ofzeros.  −2 0 6 0

1 0 −3 0−1 0 3 0

Row reducing this yields  1 0 −3 0

0 0 0 00 0 0 0

