Tests for Diagonalizability
How do we know if a matrix is diagonalizable?
In all the discussions in this unit, we will be only concerned with real vector spaces. The matrices, their eigenvalues and eigenvectors, if they exist, will be real.
Negative test
Are all matrices diagonalizable? Not really. As a counterexample, consider the following matrix:
\[ \mathbf{A} =\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]
The characteristic polynomial for \(\mathbf{A}\) is:
\[ \begin{aligned} |\mathbf{A} -\lambda \mathbf{I} | & =\begin{vmatrix} -\lambda & 1\\ -1 & -\lambda \end{vmatrix}\\ & \\ & =\lambda ^{2} +1 \end{aligned} \]
We see that the polynomial has no real roots. Thus the matrix \(\mathbf{A}\) has no real eigenvalues. Using this example, we can formulate the following negative test for diagonalizability:
If the characteristic polynomial of a matrix \(\mathbf{A}\) does not have \(n\) eigenvalues, then the matrix is not diagonalizable.
Note that when we talk about \(n\) eigenvalues, we include the multiplicity of eigenvalues. For example, the \(n\times n\) identity matrix \(\mathbf{I}\) has \(n\) eigenvalues even though it has only one unique eigenvalue, namely \(\lambda =1\).
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.
The eigenvectors corresponding to distinct eigenvalues of \(\mathbf{A}\) are linearly independent.
We shall prove this for the case of two eigenvalues. Let \((\lambda _{1} ,\mathbf{v}_{1} )\) and \((\lambda _{2} ,\mathbf{v}_{2} )\) be two eigenpairs of \(A\) such that \(\lambda _{1} \neq \lambda _{2}\). Consider any linear combination of \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\):
\[ \alpha _{1}\mathbf{v}_{1} +\alpha _{2}\mathbf{v}_{2} =0 \]
Pre-multiplying by \(\mathbf{A} -\lambda _{1}\mathbf{I}\) on both sides:
\[ \begin{aligned} (\mathbf{A} -\lambda _{1}\mathbf{I} )(\alpha _{1}\mathbf{v}_{1} +\alpha _{2}\mathbf{v}_{2} ) & =0\\ & \\ \alpha _{2} (\lambda _{2} -\lambda _{1} )\mathbf{v}_{2} & =0 \end{aligned} \]
Since \(\lambda _{1} \neq \lambda _{2}\), and since the eigenvector \(\mathbf{v}_{2} \neq 0\), we have \(\alpha _{2} =0\). From this it follows that \(\alpha _{1} =0\). So, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly independent. This argument can be extended to more than two eigenvalues. So, one positive test for diagonalizability is this:
Positive test: If an \(n\times n\) matrix \(\mathbf{A}\) 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 a complete set of \(n\) eigenvalues, there is no guarantee that \(\mathbb{R}^{n}\) will have a basis of eigenvectors. For example, consider the matrix:
\[ \mathbf{A} =\begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \]
The characteristic polynomial is \((2-\lambda )^{2}\). So, \(\lambda =2\) is the only eigenvalue, but repeated twice. Still, \(\mathbf{A}\) does have two eigenvalues. To get the eigenvectors, we solve \((\mathbf{A} -2\mathbf{I} )\mathbf{v} =0\):
\[ \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\begin{bmatrix} v_{1}\\ v_{2} \end{bmatrix} =\begin{bmatrix} 0\\ 0 \end{bmatrix} \]
This gives us \(v_{2} =0\) and \(v_{1} =k\), where \(k\) is some non-zero real number. Thus, every eigenvector of \(\mathbf{A}\) is some non-zero vector in \(\text{span}\{( 1,0)\}\). Since we don’t have two linearly independent eigenvectors, \(\mathbf{A}\) is not diagonalizable. Recall that we are employing the second definition here. For \(\mathbf{A}\) to be diagonalizable, \(\mathbb{R}^{2}\) needs to have a basis of eigenvectors of \(\mathbf{A}\). But with all eigenvectors of \(\mathbf{A}\) lying in a one-dimensional subspace, this can never materialize.
Though this is all we can say at present, this discussion is by no means complete. We will soon learn about a rich class of matrices that are diagonalizable: symmetric matrices. This requires a detour into the world of complex matrices, which is what we will take up in future documents.
Summary
Not all matrices are diagonalizable. There are two tests for diagonalizability that we have seen so far. For an \(n \times n\) matrix \(\mathbf{A}\):
Negative test: if \(\mathbf{A}\) doesn’t have \(n\) eigenvalues, it is not diagonalizable.
Positive test: if \(\mathbf{A}\) has \(n\) distinct eigenvalues, then it is diagonalizable.
If \(\mathbf{A}\) has \(n\) eigenvalues, but with repetitions, then we can’t comment on the diagonalizability without more information.