1 Matrices with repeated eigenvalues

So far we have considered the diagonalization of matrices with distinct (i.e. non-repeated) eigenvalues. We have accomplished this by the use of a non-singular modal matrix P (i.e. one where det P ≠ 0 and hence the inverse P − 1 exists). We now want to discuss briefly the case of a matrix A with at least one pair of repeated eigenvalues . We shall see that for some such matrices diagonalization is possible but for others it is not.

The crucial question is whether we can form a non-singular modal matrix P with the eigenvectors of A as its columns.

Example

Consider the matrix

A = 1 0 − 4 1

which has characteristic equation

det ( A − λ I ) = ( 1 − λ ) ( 1 − λ ) = 0.

So the only eigenvalue is 1 which is repeated or, more formally, has multiplicity 2.

To obtain eigenvectors of A corresponding to λ = 1 we proceed as usual and solve

A X = 1 X

or

1 0 − 4 1 x y = x y

implying

x = x  and − 4 x + y = y

from which x = 0 and y is arbitrary.

Thus possible eigenvectors are

0 − 1 , 0 1 , 0 2 , 0 3 …

However, if we attempt to form a modal matrix P from any two of these eigenvectors,

e.g. 0 − 1  and 0 1 then the resulting matrix P = 0 0 − 1 1 has zero determinant.

Thus P − 1 does not exist and the similarity transformation P − 1 A P that we have used previously to diagonalize a matrix is not possible here.

The essential point, at a slightly deeper level, is that the columns of P in this case are not linearly independent since

0 − 1 = ( − 1 ) 0 1

i.e. one is a multiple of the other.

This situation is to be contrasted with that of a matrix with non-repeated eigenvalues. Earlier, for example, we showed that the matrix

A = 2 3 3 2

has the non-repeated eigenvalues λ 1 = − 1 , λ 2 = 5 with associated eigenvectors

X 1 = 1 − 1 X 2 = 1 1

These two eigenvectors are linearly independent .

since 1 − 1 ≠ k 1 1 for any value of k ≠ 0 .

Here the modal matrix

P = 1 1 − 1 1

has linearly independent columns: so that det P ≠ 0 and P − 1 exists.

The general result, illustrated by this example, is given in the following Key Point.

Key Point 4

Eigenvectors corresponding to distinct eigenvalues are always linearly independent.

It follows from this that we can always diagonalize an n × n matrix with n distinct eigenvalues since it will possess n linearly independent eigenvectors. We can then use these as the columns of P , secure in the knowledge that these columns will be linearly independent and hence P − 1 will exist.

It follows, in considering the case of repeated eigenvalues, that the key problem is whether or not there are still n linearly independent eigenvectors for an n × n matrix.

We shall now consider two 3 × 3 cases as illustrations.

Task!

Let A = − 2 0 1 1 1 0 0 0 − 2

  1. Obtain the eigenvalues and eigenvectors of A .
  2. Can three linearly independent eigenvectors for A be obtained?
  3. Can A be diagonalized?
  1. The characteristic equation of A is det ( A − λ I ) = − 2 − λ 0 1 1 1 − λ 0 0 0 − 2 − λ = 0

    i.e. ( − 2 − λ ) ( 1 − λ ) ( − 2 − λ ) = 0 which gives λ = 1 , λ = − 2 , λ = − 2 .

    For λ = 1 the associated eigenvectors satisfy − 2 0 1 1 1 0 0 0 − 2 x y z = x y z from which x = 0 , z = 0 and y is arbitrary. Thus an eigenvector is X = 0 α 0 where α is arbitrary, α ≠ 0 .

    For the repeated eigenvalue λ = − 2 we must solve A Y = ( − 2 ) Y for the eigenvector Y :

    − 2 0 1 1 1 0 0 0 − 2 x y z = − 2 x − 2 y − 2 z from which z = 0 , x + 3 y = 0 so the eigenvectors are of the form Y = − 3 β β 0 = β − 3 1 0 where β ≠ 0 is arbitrary.

  2. X and Y are certainly linearly independent (as we would expect since they correspond to distinct eigenvalues.) However, there is only one independent eigenvector of the form Y corresponding to the repeated eigenvalue − 2 .
  3. The conclusion is that since A is 3 × 3 and we can only obtain two linearly independent eigenvectors then A cannot be diagonalized .
Task!

The matrix A = 5 − 4 4 12 − 11 12 4 − 4 5 has eigenvalues − 3 , 1 , 1 . The eigenvector corresponding to the eigenvalue − 3 is X = 1 3 1 or any multiple .

Investigate carefully the eigenvectors associated with the repeated eigenvalue λ = 1 and deduce whether A can be diagonalized.

We must solve A Y = ( 1 ) Y for the required eigenvector

 i.e. 5 − 4 4 12 − 11 12 4 − 4 5 x y z = x y z

Each equation here gives on simplification   x − y + z = 0.   So we have just one equation in three unknowns so we can choose any two values arbitrarily. The choices x = 1 , y = 0 (and hence z = − 1 ) and x = 0 , y = 1 (and hence z = 1 ) for example, give rise to linearly independent eigenvectors Y 1 = 1 0 − 1 Y 2 = 0 1 1

We can thus form a non-singular modal matrix P from Y 1 and Y 2 together with X (given)

P = 1 1 0 3 0 1 1 − 1 1

We can then indeed diagonalize A through the transformation

P − 1 A P = D = − 3 0 0 0 1 0 0 0 1

Key Point 5

An n × n matrix with repeated eigenvalues can be diagonalized provided we can obtain n linearly independent eigenvectors for it. This will be the case if, for each repeated eigenvalue λ i of multiplicity m i > 1 , we can obtain m i linearly independent eigenvectors.