Math Core

A matrix $A$ usually knocks a vector off its line: it rotates it, shears it, and stretches it all at once. But for many matrices a few special directions come through untouched, and along those directions $A$ does nothing more complicated than multiply by a number. Those directions, and the numbers that go with them, explain the long-run behavior of dynamical systems, the shape of quadratic forms, the vibration modes of structures, and much more.

Vectors that stay on their line

Take

A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}, \qquad \mathbf{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad \mathbf{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.

Then $A\mathbf{u} = (2, 1)$ , which points in a new direction. But $A\mathbf{v} = (3, 3) = 3\mathbf{v}$ : the image of $\mathbf{v}$ lies on the same line through the origin as $\mathbf{v}$ , just three times as long. The same is true of every multiple of $\mathbf{v}$ , so the whole line $y = x$ is simply stretched by a factor of $3$ . You can check that the line $y = -x$ is also special: $A(1, -1) = (1, -1)$ , so vectors on that line are left exactly where they are.

Along the eigen-direction y = x, A stretches v = (1, 1) to Av = (3, 3) on the same line. The vector u = (1, 0) is not an eigenvector: Au = (2, 1) leaves its line.Open in grapher →

Definition

Eigenvector and eigenvalue

Let $A$ be an $n \times n$ matrix. A nonzero vector $\mathbf{x}$ is an eigenvector of $A$ if

A\mathbf{x} = \lambda \mathbf{x}

for some scalar $\lambda$ . The scalar $\lambda$ is called an eigenvalue of $A$ , and $\mathbf{x}$ is an eigenvector corresponding to $\lambda$ .

The requirement $\mathbf{x} \ne \mathbf{0}$ matters. Since $A\mathbf{0} = \lambda\mathbf{0}$ for every $\lambda$ , allowing the zero vector would make every number an eigenvalue. The eigenvalue itself, however, may be $0$ : that happens exactly when $A\mathbf{x} = \mathbf{0}$ has a nontrivial solution.

Only square matrices have eigenvalues, because $A\mathbf{x}$ and $\mathbf{x}$ must live in the same space to be compared.

Testing a vector

To decide whether a given $\mathbf{x}$ is an eigenvector, multiply. If $A\mathbf{x}$ is a scalar multiple of $\mathbf{x}$ , it is, and the scalar is the eigenvalue. If not, it isn't.

Worked example: Checking candidates

Let $A = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} 6 \\ -5 \end{bmatrix}$ and $\mathbf{w} = \begin{bmatrix} 3 \\ -2 \end{bmatrix}$ . Are $\mathbf{u}$ and $\mathbf{w}$ eigenvectors?

A\mathbf{u} = \begin{bmatrix} 6 - 30 \\ 30 - 10 \end{bmatrix} = \begin{bmatrix} -24 \\ 20 \end{bmatrix} = -4\begin{bmatrix} 6 \\ -5 \end{bmatrix}, \qquad A\mathbf{w} = \begin{bmatrix} 3 - 12 \\ 15 - 4 \end{bmatrix} = \begin{bmatrix} -9 \\ 11 \end{bmatrix}.

So $\mathbf{u}$ is an eigenvector with eigenvalue $-4$ . For $\mathbf{w}$ , the ratios $-9/3 = -3$ and $11/(-2)$ disagree, so $A\mathbf{w}$ is not a multiple of $\mathbf{w}$ and $\mathbf{w}$ is not an eigenvector.

Testing a number

Deciding whether a number $\lambda$ is an eigenvalue turns into a question you already know how to answer. Rewrite $A\mathbf{x} = \lambda\mathbf{x}$ as $A\mathbf{x} - \lambda I\mathbf{x} = \mathbf{0}$ , that is,

(A - \lambda I)\mathbf{x} = \mathbf{0}.

(You need the identity matrix $I$ here; $A - \lambda$ makes no sense.) So $\lambda$ is an eigenvalue exactly when this homogeneous system has a nontrivial solution, which happens exactly when $A - \lambda I$ is not invertible. The eigenvectors are the nonzero solutions.

Eigenspaces

$\lambda$ is an eigenvalue of $A$ if and only if $(A - \lambda I)\mathbf{x} = \mathbf{0}$ has a nontrivial solution. The set of all solutions,

\operatorname{Nul}(A - \lambda I),

is a subspace of $\mathbb{R}^n$ called the eigenspace of $A$ corresponding to $\lambda$ . It consists of the zero vector together with all eigenvectors for $\lambda$ .

Because an eigenspace is a null space, you find a basis for it by row reducing $A - \lambda I$ and writing the solution in parametric vector form, exactly as in earlier units.

Worked example: Is 7 an eigenvalue?

With $A = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}$ again, is $7$ an eigenvalue?

A - 7I = \begin{bmatrix} -6 & 6 \\ 5 & -5 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}.

There is a free variable, so nontrivial solutions exist and $7$ is an eigenvalue. The solutions satisfy $x_1 = x_2$ , so the eigenspace is spanned by $(1, 1)$ . Check: $A(1, 1) = (7, 7)$ .

Worked example: A two-dimensional eigenspace

Let $A = \begin{bmatrix} 4 & -1 & 2 \\ 1 & 2 & 2 \\ 1 & -1 & 5 \end{bmatrix}$ . Given that $3$ is an eigenvalue, find a basis for its eigenspace.

A - 3I = \begin{bmatrix} 1 & -1 & 2 \\ 1 & -1 & 2 \\ 1 & -1 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.

The single equation $x_1 - x_2 + 2x_3 = 0$ leaves $x_2$ and $x_3$ free. Writing $x_1 = x_2 - 2x_3$ ,

\mathbf{x} = x_2\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}.

So $\{(1, 1, 0),\ (-2, 0, 1)\}$ is a basis, and the eigenspace is a plane through the origin. Every vector in that plane is simply tripled by $A$ .

Common mistake

Row reducing $A$ itself tells you nothing about eigenvalues: row operations change them. Always row reduce $A - \lambda I$ , for one specific $\lambda$ at a time. And remember that an eigenvector must be nonzero, even though the eigenspace contains $\mathbf{0}$ .

Two quick facts

Triangular matrices. If $A$ is upper or lower triangular, then $A - \lambda I$ is triangular with diagonal entries $a_{ii} - \lambda$ . A triangular matrix fails to be invertible exactly when some diagonal entry is zero, so:

Eigenvalues you can read off

The eigenvalues of a triangular matrix are the entries on its main diagonal. In addition, $0$ is an eigenvalue of $A$ if and only if $A$ is not invertible.

Worked example: Reading eigenvalues

$B = \begin{bmatrix} 5 & 0 & 0 \\ 2 & -1 & 0 \\ 8 & 4 & 0 \end{bmatrix}$ is lower triangular, so its eigenvalues are $5$ , $-1$ and $0$ . Because $0$ is an eigenvalue, $B$ is not invertible, which you can confirm from its zero third column.

Independence. Eigenvectors that belong to different eigenvalues can never be linearly dependent. If $\mathbf{v}_1, \dots, \mathbf{v}_r$ are eigenvectors for distinct eigenvalues $\lambda_1, \dots, \lambda_r$ , then $\{\mathbf{v}_1, \dots, \mathbf{v}_r\}$ is linearly independent. The idea of the proof: take a shortest dependence relation $c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p = \mathbf{0}$ among them. Multiplying by $A$ and, separately, by $\lambda_p$ and subtracting kills the $\mathbf{v}_p$ term and leaves a shorter relation with coefficients $c_i(\lambda_i - \lambda_p)$ , which are not all zero because the eigenvalues are distinct. That contradicts minimality. This fact is the engine behind diagonalization later in the unit.

Tip

If $A\mathbf{x} = \lambda\mathbf{x}$ , then $A^2\mathbf{x} = A(\lambda\mathbf{x}) = \lambda^2\mathbf{x}$ , and in general $A^k\mathbf{x} = \lambda^k\mathbf{x}$ . For invertible $A$ , $A^{-1}\mathbf{x} = \lambda^{-1}\mathbf{x}$ . Eigenvectors turn matrix algebra into ordinary arithmetic on numbers.

Practice

Practice 1

Let $A = \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}$ and $\mathbf{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ . Given that $\mathbf{v}$ is an eigenvector of $A$ , find its eigenvalue.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Which vector is an eigenvector of $A = \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}$ ?

Practice 3

Find all eigenvalues of $\begin{bmatrix} 2 & 5 & -1 \\ 0 & -3 & 4 \\ 0 & 0 & 7 \end{bmatrix}$ .

Separate answers with commas, e.g. 2, -5

Practice 4

$A = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}$ has eigenvalue $\lambda = -2$ . An eigenvector for $\lambda = -2$ has the form $(3, k)$ . Find $k$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Find the value of $c$ for which $4$ is an eigenvalue of $A = \begin{bmatrix} 1 & c \\ 2 & 3 \end{bmatrix}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find the dimension of the eigenspace of $A = \begin{bmatrix} 3 & 0 & 1 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ corresponding to $\lambda = 3$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

$A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$ is not invertible, so $0$ is an eigenvalue. Find the eigenvector for $\lambda = 0$ whose first entry is $2$ .

Enter a point like (2, -3)

Practice 8

Suppose $A\mathbf{x} = 4\mathbf{x}$ for some nonzero $\mathbf{x}$ . Then $\mathbf{x}$ is also an eigenvector of $B = A^2 - 3A + I$ . Find the corresponding eigenvalue of $B$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.