# Diagonalisation

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.

DEFINITION 6.2.1 (Matrix Diagonalisation)   A matrix is said to be diagonalisable if there exists a non-singular matrix such that is a diagonal matrix.

Remark 6.2.2   Let be an diagonalisable matrix with eigenvalues By definition, is similar to a diagonal matrix Observe that as similar matrices have the same set of eigenvalues and the eigenvalues of a diagonal matrix are its diagonal entries.

EXAMPLE 6.2.3   Let Then we have the following:
1. 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.
2. 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

THEOREM 6.2.4   let be an matrix. Then is diagonalisable if and only if has linearly independent eigenvectors.

Proof. Let be diagonalisable. Then there exist matrices and such that

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

COROLLARY 6.2.5   let be an matrix. Suppose that the eigenvalues of are distinct. Then is diagonalisable.

Proof. As is an matrix, it has eigenvalues. Since all the eigenvalues of are distinct, by Corollary 6.1.17, the eigenvectors are linearly independent. Hence, by Theorem 6.2.4, is diagonalisable. height6pt width 6pt depth 0pt

COROLLARY 6.2.6   Let be an matrix with as its distinct eigenvalues and as its characteristic polynomial. Suppose that for each divides but does not divides for some positive integers . Then

Or equivalently

Proof. As is diagonalisable, by Theorem 6.2.4, has linearly independent eigenvalues. Also, as . Hence, for each eigenvalue , has exactly linearly independent eigenvectors. Thus, for each , the homogeneous linear system has exactly linearly independent vectors in its solution set. Therefore, . Indeed for follows from a simple counting argument.

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

EXAMPLE 6.2.7
1. 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.
2. 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.

EXERCISE 6.2.8
1. 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
2. Are the two matrices and diagonalisable?
3. Find the eigenvalues and eigenvectors of , where if and otherwise.
4. Let be an matrix and an matrix. Suppose Then show that is diagonalisable if and only if both and are diagonalisable.
5. Let be a linear transformation with and

Then
1. determine the eigenvalues of
2. find the number of linearly independent eigenvectors corresponding to each eigenvalue?