Characteristic Polynomial
How do we find the eigenvalues of a matrix?
Necessary condition
If \(\mathbf{x}\) is an eigenvector of a matrix \(\mathbf{A}\) with eigenvalue \(\lambda\), then we have:
\[ \begin{equation*} \mathbf{Ax} =\lambda \mathbf{x} \end{equation*} \]
Let us rearrange this a bit. Observe that if \(\mathbf{I}\) is the identity matrix, then:
\[ \begin{equation*} \mathbf{Ix} =\mathbf{x} \end{equation*} \]
Using this fact, we can express the eigenvalue equation as:
\[ \begin{equation*} \begin{aligned} \mathbf{Ax} & =\lambda \mathbf{Ix}\\ & \\ \mathbf{Ax} -\lambda \mathbf{Ix} & =\mathbf{0}\\ & \\ (\mathbf{A} -\lambda \mathbf{I} )\mathbf{x} & =0 \end{aligned} \end{equation*} \]
If \((\lambda ,\mathbf{x} )\) is an eigenpair, then \((\mathbf{A} -\lambda \mathbf{I} )\mathbf{x} =0\). Another way of expressing this is:
\((\mathbf{A} -\lambda \mathbf{I} )\mathbf{x} =0\) is a necessary condition for \(\mathbf{x}\) to be an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\).
Sufficient condition
Let us try the other direction. What if \((\mathbf{A} -\lambda \mathbf{I} )\mathbf{x} =\mathbf{0}\) for \(\mathbf{x} \neq \mathbf{0}\)? We have:
\[ \begin{equation*} \begin{aligned} (\mathbf{A} -\lambda \mathbf{I} )\mathbf{x} & =\mathbf{0}\\ & \\ \mathbf{Ax} -\lambda \mathbf{Ix} & =\mathbf{0}\\ & \\ \mathbf{Ax} & =\lambda \mathbf{x} \end{aligned} \end{equation*} \]
From the above result and with \(\mathbf{x} \neq \mathbf{0}\), we see that \(\mathbf{x}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\).
\((\mathbf{A} -\lambda \mathbf{I} )\mathbf{x} =\mathbf{0}\) and \(\mathbf{x} \neq \mathbf{0}\) is a sufficient condition for \(\mathbf{x}\) to be an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\).
Nullspace argument
Thus we see that for a non-zero vector \(\mathbf{x}\), \((\lambda ,\mathbf{x} )\) is an eigenpair of \(\mathbf{A}\) if and only if \((\mathbf{A} -\lambda \mathbf{I} )\mathbf{x} =\mathbf{0}\). From this, we deduce that \(\mathbf{x}\) has to be in the nullspace of \(\mathbf{A} -\lambda \mathbf{I}\). Thus the nullspace of \(\mathbf{A} -\lambda \mathbf{I}\) is non-trivial. If the nullspace is non-trivial, we can conclude that the matrix is not invertible. To see why this is true, think about what happens if the matrix is invertible. If the matrix is invertible, then \(\mathbf{0}\) would be the only element in the nullspace.
If the matrix is not invertible, then its determinant is zero. So, we have the following result:
\(\lambda\) is an eigenvalue of \(\mathbf{A}\) if and only if \(|\mathbf{A} -\lambda \mathbf{I} |=0\)
Notice how we end up with a result that doesn’t involve the eigenvector and only the matrix and its eigenvalues. We can now characterize the collection of all eigenvectors along with the zero vector, the eigenspace, as follows:
The eigenspace corresponding to an eigenvalue \(\lambda\) is the nullspace of \(\mathbf{A} -\lambda \mathbf{I}\). If we denote the eigenspace corresponding to the eigenvalue \(\lambda\) as \(E(\lambda)\), we have: \[ E(\lambda) = \mathcal{N}(\mathbf{A} - \lambda \mathbf{I}) \]
The Polynomial
At this stage, let us take up an example:
\[ \begin{equation*} \mathbf{A} =\begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \end{equation*} \]
For \(\lambda\) to be an eigenvalue of the matrix \(\mathbf{A}\), \(|\mathbf{A} -\lambda \mathbf{I} |=0\). This translates to:
\[ \begin{equation*} \begin{vmatrix} 3-\lambda & 1\\ 0 & 2-\lambda \end{vmatrix} =0 \end{equation*} \]
Expanding the determinant, we get:
\[ \begin{equation*} \begin{aligned} (3-\lambda )(2-\lambda )=0 \end{aligned} \end{equation*} \]
The expression on the LHS is a polynomial of degree \(2\) and is called the characteristic polynomial of \(\mathbf{A}\). We see that \(\lambda =2,3\) are the eigenvalues of the matrix \(\mathbf{A}\).
For a \(n \times n\) matrix \(\mathbf{A}\), the determinant \(|\mathbf{A} - \lambda \mathbf{I}|\) is a polynomial of degree \(n\) and is termed the characteristic polynomial. The roots or zeros of this polynomial are the eigenvalues of \(\mathbf{A}\).
Properties
There are two properties of the polynomial that are worth knowing:
- The sum of the eigenvalues of the matrix \(\mathbf{A}\) is equal to its trace.
- The product of the eigenvalues of the matrix \(\mathbf{A}\) is equal to its determinant.
Summary
To find the eigenvalues of a matrix \(\mathbf{A}\), we first compute the characteristic polynomial \(|\mathbf{A} -\lambda \mathbf{I} |\). The roots of this polynomial are the eigenvalues of the matrix \(\mathbf{A}\). There are two important properties:
- sum of the eigenvalues is equal to the trace of the matrix
- product of the eigenvalues is equal to the determinant of the matrix