3.6 Eigenvalue problem

For an n x n square matrix A, we seek a scalar $\lambda$ and vector x such that:

We want a nontrivial solution ($\vv{x} \ne \vv{0}$). This pair represents a vector x that, when multiplied into A, does not change its direction. However, it may adopt a new magnitude $\lambda$. Borrowing from a German word for “own”, we call x an eigenvector and $\lambda$ an eigenvalue of A.

To find the eigenvalues of A, rearrange:

If $\vv{A} - \lambda \vv{I}$ is invertible, we have only $\vv{x} = \vv{0}$ as a solution. Hence, we require this matrix to be singular! This occurs when its determinant is zero:

This equation creates a characeristic polynomial of degree n for $\lambda$ that can be solved. Then, the eigenvector x that corresponds to each root $\lambda$ can be determined. For example, to find the eigenvalues of:

First compute the determinant:

Hence, the eigenvalues of A are $\lambda_1 = -1$ and $\lambda_2 = -6$. Note that the particular ordering of the eigenvalues is not important, and we are only labeling them to make it convenient to refer to a particular eigenvalue later.

Next, we seek the eigenvector $\vv{x}_1$ that corresponds to $\lambda_1$. This vector must solve the system:

We can solve for $\vv{x}_1$ using row reduction. Since the last column of the augmented matrix would be only zeros, it is not necessary to include it, and we row reduce only the matrix itself:

Note that this reduced matrix is equivalent to $x_1 = x_2/2$ (using x to represent elements of $\vv{x}_1$ for convenience) with $x_2$ being free. This means that the eigenvector is not unique, which makes sense both from the original equation we are trying to solve and us forcing the matrix to be singular! Equivalently, eigenvectors can be scaled up or down by an arbitrary nonzero constant. We may then choose $x_2$ so that the eigenvector has either nice values (like like integers) or a nice norm (1 is conventional). For example, choosing $x_2 = 2$ gives:

Let’s use the same strategy to get the eigenvector for $\lambda_2 = -6$. We’ll jump straight to the matrix we need:

using the swapping trick on the first row.

Multiple eigenvalues¶

An n x n matrix has n eigenvalues, but they may not be distinct! For example, consider the matrix:

We need to evaluate the determinant:

The roots of this cubic polynomial are $\lambda_1 = 5$ and $\lambda_2 = \lambda_3 = -3$. The eigenvector for $\lambda_1 = 5$ is obtained from

This matrix is equivalent to the system of equations:

with $x_3$ free. Choosing $x_3 = -1$ gives

The eigenvectors for $\lambda_2 = \lambda_3 = -3$ are obtained from

This matrix is equivalent to

with $x_2$ and $x_3$ free. Choosing $x_2 = 1$ and $x_3 = 0$ gives one eigenvector

while choosing an independent combination $x_2 = 0$ and $x_3 = 1$ gives

Complex eigenvalues¶

Matrices may have complex eigenvaules. They always come in conjugate pairs! For example, for the matrix:

We evaluate the determinant:

For $\lambda_1 = i$:

This matrix is equivalent to

with $x_2$ free. Choosing $x_2 = i$ gives

For $\lambda_2 = -i$:

This matrix is equivalent to

with $x_2$ free. Choosing $x_2 = -i$ gives

Useful tricks¶

• The eigenvaules of upper triangular matrices are the values on the diagonal. For example, the matrix

  $$\begin{bmatrix} 1 & 4 & 5 \\ 0 & 2 & 6 \\ 0 & 0 & 3 \end{bmatrix}$$

  (31)

  has $\lambda_1 = 1$, $\lambda_2 = 2$, and $\lambda_3 = 3$ as eigenvalues.

• The transpose $\vv{A}^{\rm T}$ has the same eigenvalues as $\vv{A}$. For example, the matrix:

  $$\begin{bmatrix} 1 & 0 & 0 \\ 4 & 2 & 0 \\ 5 & 6 & 3 \end{bmatrix}$$

  (32)

  also has $\lambda_1 = 1$, $\lambda_2 = 2$, and $\lambda_3 = 3$ as eigenvalues because it is the transpose of the matrix above.

  Note that this means that the eigenvalues of lower triangular matrices are also the values on the diagonal!

• If A is symmetric ($\vv{A}^{\rm T} = \vv{A}$), its eigenvalues are real.

• If A is skew-symmetric ($\vv{A}^{\rm T} = -\vv{A}$), its eigenvalues are purely imaginary.

Skill builder problems¶

Find the eigenvalues and eigenvectors for: