Eigenvectors and Eigenvalues

Apply a matrix to most vectors and they turn to point in a new direction. A few special vectors do not turn at all: the matrix only stretches them, shrinks them or flips them. Those directions reveal what the matrix really does, and finding them uses null spaces.

An eigenvector of an $n \times n$ matrix $A$ is a nonzero vector $\mathbf{v}$ with $$A\mathbf{v} = \lambda\mathbf{v}$$ for some number $\lambda$, called the eigenvalue for $\mathbf{v}$. The eigenvalue can be any number, including $0$ or a negative number (a flip).

The dashed purple lines are the eigenvector directions of $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$: along them, $A$ only stretches (by 5 and by 2).

Worked example: Checking candidates

Let $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$. Are $\mathbf{u} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$, $\mathbf{w} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$ and $\mathbf{z} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ eigenvectors?

$A\mathbf{u} = \begin{bmatrix} 5 \\ 5 \end{bmatrix} = 5\mathbf{u}$: an eigenvector with $\lambda = 5$.
$A\mathbf{w} = \begin{bmatrix} 2 \\ -4 \end{bmatrix} = 2\mathbf{w}$: an eigenvector with $\lambda = 2$.
$A\mathbf{z} = \begin{bmatrix} 4 \\ 2 \end{bmatrix}$, which is not a multiple of $(1, 0)$: not an eigenvector.

Answer: $\mathbf{u}$ ($\lambda = 5$) and $\mathbf{w}$ ($\lambda = 2$) are; $\mathbf{z}$ is not.

Why it works

To find eigenvectors for a given $\lambda$, rewrite $A\mathbf{v} = \lambda\mathbf{v}$ as $A\mathbf{v} - \lambda I\mathbf{v} = \mathbf{0}$, that is, $$(A - \lambda I)\mathbf{v} = \mathbf{0}.$$ (We write $\lambda I\mathbf{v}$ rather than $\lambda\mathbf{v}$ so that both terms are matrices times $\mathbf{v}$.) So the eigenvectors for $\lambda$ are the nonzero vectors in $\operatorname{Nul}(A - \lambda I)$, and $\lambda$ is an eigenvalue exactly when that null space contains a nonzero vector: when $A - \lambda I$ is not invertible.

Worked example: Finding eigenvectors for a given eigenvalue

Find the eigenvectors of $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ for $\lambda = 2$.

$A - 2I = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$ (subtract $2$ on the diagonal only).
Both rows say $2x_1 + x_2 = 0$, so $x_2 = -2x_1$ with $x_1$ free.
Eigenvectors: all nonzero multiples of $\begin{bmatrix} 1 \\ -2 \end{bmatrix}$. Check: $A\begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 2 \\ -4 \end{bmatrix} = 2\begin{bmatrix} 1 \\ -2 \end{bmatrix}$. ✓

Answer: $t\begin{bmatrix} 1 \\ -2 \end{bmatrix}$, $t \neq 0$

Worked example: Is 3 an eigenvalue?

Is $\lambda = 3$ an eigenvalue of the same $A$?

$A - 3I = \begin{bmatrix} 1 & 1 \\ 2 & 0 \end{bmatrix}$ has determinant $0 - 2 = -2 \neq 0$, so it is invertible.
Then $(A - 3I)\mathbf{v} = \mathbf{0}$ only for $\mathbf{v} = \mathbf{0}$, which does not count.

Answer: No.

Common mistake

Accepting the zero vector as an eigenvector. $A\mathbf{0} = \lambda\mathbf{0}$ is true for every number $\lambda$, so allowing it would make every number an eigenvalue and the idea would mean nothing. Eigenvectors must be nonzero; eigenvalues may be $0$. (And $\lambda = 0$ is an eigenvalue exactly when $A\mathbf{v} = \mathbf{0}$ for some $\mathbf{v} \neq \mathbf{0}$: when $A$ is not invertible.)

Any nonzero multiple of an eigenvector is again an eigenvector for the same eigenvalue, since $A(c\mathbf{v}) = cA\mathbf{v} = \lambda(c\mathbf{v})$. So we really find eigenvector directions, and when an answer needs one specific vector we fix a convenient entry, such as making the first entry $1$.

Key idea

$A\mathbf{v} = \lambda\mathbf{v}$ with $\mathbf{v} \neq \mathbf{0}$: along $\mathbf{v}$, $A$ just multiplies by $\lambda$. The eigenvectors for $\lambda$ are the nonzero solutions of $(A - \lambda I)\mathbf{v} = \mathbf{0}$, and $\lambda$ is an eigenvalue exactly when $A - \lambda I$ is not invertible.

Eigenvectors and Eigenvalues

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