Let be a square matrix of order and let be the corresponding linear transformation. In this section, we ask the question ``does there exist a basis of such that the matrix of the linear transformation is in the simplest possible form."

We know that, the simplest form for a matrix is the identity matrix and the diagonal matrix. In this section, we show that for a certain class of matrices we can find a basis such that is a diagonal matrix, consisting of the eigenvalues of This is equivalent to saying that is similar to a diagonal matrix. To show the above, we need the following definition.

- Let Then has no real eigenvalue (see Example 6.1.8 and hence doesn't have eigenvectors that are vectors in Hence, there does not exist any non-singular real matrix such that is a diagonal matrix.
- In case,
the two complex eigenvalues of
are
and the corresponding eigenvectors are
and
respectively. Also,
and
can be taken as a basis of
Define a
complex matrix by
Then

Or equivalently, Let Then implies that

Since 's are the columns of a non-singular matrix they are non-zero and so for we get the eigenpairs of Since, 's are columns of the non-singular matrix using Corollary 4.3.9, we get are linearly independent.

Thus we have shown that if is diagonalisable then has linearly independent eigenvectors.

Conversely, suppose
has
linearly independent eigenvectors
with eigenvalues
Then
Let
Since
are linearly independent, by
Corollary 4.3.9,
is non-singular. Also,

Therefore the matrix is diagonalisable. height6pt width 6pt depth 0pt

Now suppose that for each . Then for each , we can choose linearly independent eigenvectors. Also by Corollary 6.1.17, the eigenvectors corresponding to distinct eigenvalues are linearly independent. Hence has linearly independent eigenvectors. Hence by Theorem 6.2.4, is diagonalisable. height6pt width 6pt depth 0pt

- Let Then Hence, has eigenvalues It is easily seen that and are the only eigenpairs. That is, the matrix has exactly one eigenvector corresponding to the repeated eigenvalue Hence, by Theorem 6.2.4, the matrix is not diagonalisable.
- Let
Then
Hence,
has eigenvalues
It can be easily
verified that
and
correspond to the eigenvalue
and
corresponds to the eigenvalue
Note that the set
consisting of
eigenvectors corresponding to the eigenvalue
are not orthogonal.
This set can be replaced by the orthogonal set
which still consists of eigenvectors corresponding to the eigenvalue
as
.
Also, the set
forms a basis of
So, by Theorem 6.2.4, the matrix
is diagonalisable.
Also, if
is the corresponding unitary
matrix then
Observe that the matrix is a symmetric matrix. In this case, the eigenvectors are mutually orthogonal. In general, for any real symmetric matrix there always exist eigenvectors and they are mutually orthogonal. This result will be proved later.

- By finding the eigenvalues of the following matrices,
justify whether or not
for some real non-singular matrix
and a real diagonal matrix

for any with - Are the two matrices and diagonalisable?
- Find the eigenvalues and eigenvectors of , where if and otherwise.
- Let be an matrix and an matrix. Suppose Then show that is diagonalisable if and only if both and are diagonalisable.
- Let
be a linear
transformation with
and
- determine the eigenvalues of
- find the number of linearly independent eigenvectors corresponding to each eigenvalue?
- is diagonalisable? Justify your answer.

- Let
be a non-zero square matrix such that
Show that
cannot be diagonalised. [
*Hint: Use Remark 6.2.2.*] - Are the following matrices diagonalisable?

A K Lal 2007-09-12