The Eigenvalues of a Matrix

This page is intended to be a part of the Numerical Analysis section of Math Online. Similar topics can also be found in the Linear Algebra section of the site.

The Eigenvalues of a Matrix

Often times when we're dealing with linear operators we can use an $n \times n$ matrix $A$ to describe such an operator. Sometimes such a matrix will have eigenvalues which are critically important in applications to linear operators. We formally define an eigenvalue of a matrix below.

 Definition: If $A$ is an $n \times n$ matrix, then $\lambda$ is an eigenvalue of $A$ if $Av = \lambda v$ for some nonzero column vector $v$. The vectors $v$ which satisfy this equation are called the corresponding Eigenvectors to the eigenvalue $\lambda$.

For example, consider one of the simplest of matrices, the $2 \times 2$ identity matrix $I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$, and consider the equation $Iv = \lambda v$. We note that $Iv = v$, so this equation reduces to $v = \lambda v$, so $\lambda = 1$ is an eigenvalue of the identity matrix, and every vector is an eigenvector corresponding to the eigenvalue $\lambda$.

For another example, consider the $3 \times 3$ matrix $A = \begin{bmatrix} 3 & 1 & 1\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix}$. The eigenvalues of this matrix are $\lambda_1 = 1$, $\lambda_2 = 2$, and $\lambda_3 = 3$. Let's prove this.

First consider the eigenvalue $\lambda_1 = 1$. We need to show that there exists a nonzero vector $v$ such that $Av = 1v = v$, that is, consider the equation:

(1)
\begin{align} \quad Av = \quad \begin{bmatrix} 3 & 1 & 1\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} \end{align}

From this matrix equation above we have that:

(2)
\begin{align} \quad 3v_1 + v_2 + v_3 = v_1 \\ \quad 2v_2 = v_2 \\ \quad v_3 = v_3 \end{align}

We have that the second equation implies that $v_2 = 0$. Meanwhile, the third equation implies that $v_3$ can equal any number, say $v_3 = a \in \mathbb{R}$. Thus equation 1 implies that $3v_1 + a = v_1$, so $2v_1 = -a$ and $v_1 = \frac{-a}{2}$. Therefore, the corresponding eigenvectors to the eigenvalue $\lambda_1 = 1$ are $\begin{bmatrix} \frac{-a}{2}\\ 0\\ a \end{bmatrix}$.

Now let's consider the eigenvalue $\lambda_2 = 2$ and consider the following equation:

(3)
\begin{align} \quad Av = \quad \begin{bmatrix} 3 & 1 & 1\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 2v_1\\ 2v_2\\ 2v_3 \end{bmatrix} \end{align}

From this matrix equation above we have the following system of equations:

(4)
\begin{align} \quad 3v_1 + v_2 + v_3 = 2v_1 \\ \quad 2v_2 = 2v_2 \\ \quad v_3 = 2v_3 \end{align}

The third equation above implies that $v_3 = 0$. Furthermore, the second equation implies that $v_2$ can be any number, say $v_2 = b \in \mathbb{R}$. The first equation implies that $3v_1 + b = 2v_1$ so $v_1 = -b$. Therefore, the corresponding eigenvectors to the eigenvalue $\lambda_2 = 2$ are $\begin{bmatrix} -b \\ b\\ 0 \end{bmatrix}$.

Lastly, consider the the eigenvalue $\lambda_3 = 3$ and consider the following equation:

(5)
\begin{align} Av = \quad \begin{bmatrix} 3 & 1 & 1\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 3v_1\\ 3v_2\\ 3v_3 \end{bmatrix} \end{align}

From this matrix equation above we have the following system of equation:

(6)
\begin{align} \quad 3v_1 + v_2 + v_3 = 3v_1 \\ \quad 2v_2 = 3v_2 \\ \quad v_3 = 3v_3 \end{align}

The second and third equations imply that $v_2 = 0$ and $v_3 = 0$. Plugging this into the first equation and we get that $3v_1 = 3v_1$, so $v_1$ can be any number, say $v_1 = c \in \mathbb{R}$, and so the corresponding eigenvectors for the eigenvalue $\lambda_3 = 3$ are $\begin{bmatrix} c \\ 0\\ 0 \end{bmatrix}$.

Note that in the previous example, the eigenvalues of $A$ were the entries on the main diagonal. For triangular matrices, this is always true as summarized with the following theorem.

 Theorem 1: If $A$ is an $n \times n$ triangular matrix, then the eigenvalues of $A$ are precisely the entries on the main diagonal of $A$.

We will soon look at some more methods for quickly computing the eigenvalues associated with a matrix.