Tests for Diagonalizability

Question

How do we know if a matrix is diagonalizable?

Negative test

Are all matrices diagonalizable¹? Not really. As a counterexample, consider the following matrix:

\[ A = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]

The characteristic polynomial for \(A\) is:

\[ \begin{aligned} |A - \lambda I| &= \begin{vmatrix} -\lambda & 1\\ -1 & -\lambda \end{vmatrix}\\\\ &= \lambda^2 + 1 \end{aligned} \]

We see that the polynomial has no real roots. Thus the matrix \(A\) has no real eigenvalues. Using this example, we can formulate the following negative test for diagonalizability:

Negative test

If the characteristic polynomial of a matrix \(A\) does not have \(n\) eigenvalues, then the matrix is not diagonalizable.

Positive test

Now that we know that not all matrices are diagonalizable, is there a way to find out which matrices are? We will look at one such test which is based on the following observation.

Note

The eigenvectors corresponding to distinct eigenvalues of \(A\) are linearly independent.

We shall prove this for the case of two eigenvlaues. Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\) such that \(\lambda_1 \neq \lambda_2\). Consider any linear combination of \(v_1\) and \(v_2\):

\[ \alpha_1v_1 + \alpha_2 v_2 = 0 \]

Pre-multiplying by \(A - \lambda_1 I\) on both sides:

\[ \begin{aligned} (A - \lambda_1I)(\alpha_1 v_1 + \alpha_2 v_2) &= 0\\\\ \alpha_2(\lambda_2 - \lambda_1) &= 0 \end{aligned} \]

Since \(\lambda_1 \neq \lambda_2\), we have \(\alpha_2 = 0\). From this it follows that \(\alpha_1 = 0\). So, \(v_1\) and \(v_2\) are linearly independent. So, one positive test for diagonalizability is this:

Positive test

If an \(n \times n\) matrix \(A\) has \(n\) distinct eigenvalues, then it is diagonalizable.

No man's land

A natural question now arises. What about the case where a matrix has \(n\)​ eigenvalues, but in which some eigenvalues repeat. Is it always diagonalizable? Even if a matrix has eigenvalues, there is no guarantee that it will have a basis of eigenvectors. For example, consider the matrix:

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \]

The characteristic polynomial is \((\lambda - 2)^2\). So, \(\lambda = 2\) is the only eigenvalue, but repeated twice. To get the eigenvectors, we solve \((A - 2I)v = 0\):

\[ \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} \]

This gives us \(v_2 = 0\) and \(v_1 = k\), where \(k\) is some non-zero real number. Thus, every eigenvector of \(A\) is some non-zero vector in \(\text{span}\left( \left\{ \begin{bmatrix}1\\0 \end{bmatrix} \right\} \right)\). Since we don't have two linearly independent eigenvectors, \(A\) is not diagonalizable.

Summary

Not all matrices are diagonalizable. There are two tests for diagonalizability:

• Negative test: if a matrix doesn't have \(n\) eigenvalues, it is not diagonalizable.
• Positive test: if a matrix has \(n\) distinct eigenvalues, then it is diagonalizable.

If a matrix has \(n\) eigenvalues, but with repetitions, then the test is inconclusive.

---

1. In all the discussions in this unit, we will be only concerned with real vector spaces. The matrices and their eigenvalues, if they exist, will be real. ↩