---
title: Eigenvectors and Eigenvalues
module: Eigenvalues and Eigenvectors
moduleNumber: 5
lessonNumber: 1
order: 501
summary: >
  An eigenvector of a square matrix is a nonzero vector the matrix only
  stretches; its eigenvalue is the stretch factor. The eigenspace of an
  eigenvalue is the null space of A minus lambda times the identity, the
  eigenvalues of a triangular matrix are its diagonal entries, and
  eigenvectors for distinct eigenvalues are linearly independent.
topics: [Eigenvalues and Eigenvectors]
sources:
  - book: Lay
    ref: "Ch. 5 — Eigenvalues and Eigenvectors; §5.1 Eigenvectors and Eigenvalues"
---

Multiplication by a square matrix $A$ moves most vectors in some new direction. For
a few special nonzero vectors the direction does not change at all: $A$ only lengthens
or shortens them, possibly reversing their sign. These are the eigenvectors of $A$.

> **Definition (Eigenvector and eigenvalue).** An **eigenvector** of an $n \times n$
> matrix $A$ is a nonzero vector $x$ such that $Ax = \lambda x$ for some scalar
> $\lambda$. A scalar $\lambda$ is an **eigenvalue** of $A$ if the equation
> $Ax = \lambda x$ has a nontrivial solution $x$; such an $x$ is an eigenvector
> **corresponding to** $\lambda$.

Two asymmetries in the definition matter. An eigenvector must be nonzero, since
$A0 = \lambda 0$ holds for every $\lambda$ and would make the notion vacuous. An
eigenvalue, on the other hand, is allowed to be $0$; the number $0$ is an eigenvalue
of $A$ exactly when $A$ has a nontrivial null space.

$$
% caption: The eigenvector $v$ keeps its direction under $A$ (here $Av = 2v$); the
% ordinary vector $u$ is rotated onto a new line.
\begin{tikzpicture}[>=stealth, scale=1.0, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-0.4,-0.4) grid (4.4,3.4);
  \draw[->, black] (-0.4,0) -- (4.6,0) node[right, black!70] {$x_1$};
  \draw[->, black] (0,-0.4) -- (0,3.6) node[above, black!70] {$x_2$};
  % eigenvector v and its image 2v
  \draw[->, acc, very thick] (0,0) -- (1,1) node[above left, acc] {$v$};
  \draw[->, acc, very thick] (0,0) -- (2,2) node[above right, acc] {$Av = 2v$};
  \draw[acc!45, dashed] (0,0) -- (3.2,3.2);
  % ordinary vector u and its rotated image
  \draw[->, black, very thick] (0,0) -- (2,1) node[below right] {$u$};
  \draw[->, black, very thick] (0,0) -- (1.4,2.6) node[above] {$Au$};
\end{tikzpicture}
$$

## Testing a vector and testing a scalar

Deciding whether a **given** vector is an eigenvector, or whether a **given** number
is an eigenvalue, needs no special theory. Both reduce to a single matrix product or
a single homogeneous system.

- **Is $x$ an eigenvector?** Compute $Ax$ and check whether it is a scalar multiple
  of $x$.
- **Is $\lambda$ an eigenvalue?** Check whether $(A - \lambda I)x = 0$ has a nontrivial
  solution, since $Ax = \lambda x$ is equivalent to $(A - \lambda I)x = 0$. This holds
  precisely when $A - \lambda I$ is singular.

> **Worked example.** For $A = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}$, decide
> whether $u = \begin{bmatrix} 6 \\ -5 \end{bmatrix}$ and
> $v = \begin{bmatrix} 3 \\ -2 \end{bmatrix}$ are eigenvectors.
>
> $$
> Au = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} 6 \\ -5 \end{bmatrix}
> = \begin{bmatrix} -24 \\ 20 \end{bmatrix} = -4u,
> \qquad
> Av = \begin{bmatrix} -9 \\ 11 \end{bmatrix}.
> $$
>
> $Au$ is a scalar multiple of $u$, so $u$ is an eigenvector with eigenvalue $-4$. But
> $Av$ is not a multiple of $v$, so $v$ is not an eigenvector.

> **Worked example.** Confirm that $7$ is an eigenvalue of the same $A$ and find its
> eigenvectors.
>
> Form $A - 7I$ and row reduce the homogeneous system $(A - 7I)x = 0$:
>
> $$
> A - 7I = \begin{bmatrix} -6 & 6 \\ 5 & -5 \end{bmatrix}
> \sim
> \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}.
> $$
>
> The columns of $A - 7I$ are linearly dependent, so the system has a free variable and
> $7$ is an eigenvalue. The general solution is
> $x = x_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix}$: every nonzero multiple of $(1, 1)$ is
> an eigenvector for $\lambda = 7$.

> **Warning.** Row reduction finds **eigenvectors** once an eigenvalue is known, but
> it does not find eigenvalues. An echelon form of $A$ generally does not display the
> eigenvalues of $A$, because row operations change them.

## The eigenspace

The set of solutions of $(A - \lambda I)x = 0$ is the null space of $A - \lambda I$,
a subspace of $\mathbb{R}^n$. When $\lambda$ is an eigenvalue this subspace has more
than just the zero vector in it.

> **Definition (Eigenspace).** For an eigenvalue $\lambda$ of $A$, the **eigenspace**
> of $A$ corresponding to $\lambda$ is
> $$
> \operatorname{Nul}(A - \lambda I) = \{\, x : Ax = \lambda x \,\}.
> $$
> It consists of the zero vector together with all eigenvectors for $\lambda$.

For the matrix $A = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}$, the eigenspace for
$\lambda = 7$ is the line through $(1,1)$, and the eigenspace for $\lambda = -4$ is the
line through $(6,-5)$. On the first line $A$ acts as multiplication by $7$; on the
second, as multiplication by $-4$. Each eigenspace is an **invariant** line: $A$ maps
it into itself.

$$
% caption: The two eigenspaces of $A$; on each line $A$ acts as pure scaling, by $7$
% on the line through $(1,1)$ and by $-4$ on the line through $(6,-5)$.
\begin{tikzpicture}[>=stealth, scale=0.7, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \definecolor{red}{HTML}{B0413E}
  \draw[->, black] (-4.4,0) -- (4.4,0) node[right, black!70] {$x_1$};
  \draw[->, black] (0,-4.4) -- (0,3.2) node[above, black!70] {$x_2$};
  % eigenspace for 7: line through (1,1)
  \draw[acc!55, thick] (-3,-3) -- (3,3);
  \draw[->, acc, very thick] (0,0) -- (1,1);
  \draw[->, acc, very thick] (0,0) -- (2,2);
  \node[acc, anchor=west] at (2.5,1.3) {scale by 7};
  \node[acc] at (2.6,3.0) {eigenspace, 7};
  % eigenspace for -4: line through (6,-5) scaled down
  \draw[red!60, thick] (-3.6,3) -- (3.6,-3);
  \draw[->, red, very thick] (0,0) -- (1.2,-1);
  \draw[->, red, very thick] (0,0) -- (-1.2,1);
  \node[red, anchor=west] at (-4.3,1.05) {scale by 4};
  \node[red] at (3.4,-3.4) {eigenspace, mirrored};
\end{tikzpicture}
$$

Eigenspaces can be more than one-dimensional.

> **Worked example.** The matrix
> $$
> A = \begin{bmatrix} 4 & -1 & 6 \\ 2 & 1 & 6 \\ 2 & -1 & 8 \end{bmatrix}
> $$
> has $2$ as an eigenvalue. Find a basis for the corresponding eigenspace.
>
> Row reduce $(A - 2I)x = 0$:
> $$
> \begin{bmatrix} 2 & -1 & 6 \\ 2 & -1 & 6 \\ 2 & -1 & 6 \end{bmatrix}
> \sim
> \begin{bmatrix} 2 & -1 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.
> $$
> Two free variables remain, so the general solution is
> $$
> x = x_2 \begin{bmatrix} 1/2 \\ 1 \\ 0 \end{bmatrix}
>   + x_3 \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}.
> $$
> The eigenspace for $\lambda = 2$ is the plane through the origin spanned by these two
> vectors, and $A$ dilates every vector in it by the factor $2$.

## Eigenvalues of a triangular matrix

For most matrices, finding eigenvalues means solving the
[characteristic equation](/linear-algebra/eigenvalues/the-characteristic-equation), a
polynomial equation in $\lambda$. Triangular matrices are the one large family where
the answer is immediate.

> **Theorem (Eigenvalues of a triangular matrix).** The eigenvalues of a triangular matrix are the entries on its main
> diagonal.

The reason is transparent from the structure of $A - \lambda I$. If $A$ is upper
triangular, then

$$
A - \lambda I =
\begin{bmatrix}
a_{11} - \lambda & a_{12} & a_{13} \\
0 & a_{22} - \lambda & a_{23} \\
0 & 0 & a_{33} - \lambda
\end{bmatrix},
$$

and this matrix has a free variable (so $(A - \lambda I)x = 0$ has a nontrivial
solution) exactly when one of the diagonal entries $a_{ii} - \lambda$ is zero, that is,
when $\lambda$ equals one of $a_{11}, a_{22}, a_{33}$.

$$
% caption: In $A - \lambda I$ for a triangular matrix, a zero appears on the diagonal
% precisely when $\lambda$ equals a diagonal entry, producing a free variable.
\begin{tikzpicture}[scale=0.7, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \i in {0,1,2} {
    \foreach \j in {0,1,2} {
      \ifnum\j<\i
        \node[minimum size=8mm] at (\j,-\i) {0};
      \else
        \node[minimum size=8mm] at (\j,-\i) {$*$};
      \fi
    }
  }
  % highlight diagonal
  \foreach \d in {0,1,2} \draw[acc, thick] (\d-0.42,-\d-0.42) rectangle (\d+0.42,-\d+0.42);
  \node[acc] at (1,-3.1) {diagonal entries = eigenvalues};
  \draw[black] (-0.55,0.55) rectangle (2.55,-2.55);
\end{tikzpicture}
$$

For example, the matrices

$$
A = \begin{bmatrix} 3 & 6 & -8 \\ 0 & 0 & 6 \\ 0 & 0 & 2 \end{bmatrix},
\qquad
B = \begin{bmatrix} 4 & 0 & 0 \\ -2 & 1 & 0 \\ 5 & 3 & 4 \end{bmatrix}
$$

have eigenvalues $\{3, 0, 2\}$ and $\{4, 1\}$ respectively, read straight off their
diagonals. Note that $A$ has $0$ as an eigenvalue, consistent with $A$ being singular.

> **Eigenvalue $0$ and invertibility.** The number $0$ is an eigenvalue of $A$ if and
> only if $Ax = 0$ has a nontrivial solution, i.e. if and only if $A$ is **not**
> invertible. This fact joins the
> [Invertible Matrix Theorem](/linear-algebra/matrix-algebra/matrix-inverse-and-invertibility).

## Independence of eigenvectors

Eigenvectors belonging to different eigenvalues cannot combine trivially, so distinct
eigenvalues supply independent directions.

> **Theorem (Independence of eigenvectors for distinct eigenvalues).** If $v_1, \dots, v_r$ are eigenvectors corresponding to **distinct**
> eigenvalues $\lambda_1, \dots, \lambda_r$ of an $n \times n$ matrix $A$, then
> $\{v_1, \dots, v_r\}$ is linearly independent.

> **Proof.** Suppose the set were dependent. Since $v_1 \neq 0$, one vector is a linear
> combination of the ones before it; let $p$ be the least index for which $v_{p+1}$ is a
> combination of the (independent) vectors $v_1, \dots, v_p$:
>
> $$
> c_1 v_1 + \cdots + c_p v_p = v_{p+1}. \tag{$*$}
> $$
>
> Multiply both sides by $A$ and use $Av_k = \lambda_k v_k$:
>
> $$
> c_1 \lambda_1 v_1 + \cdots + c_p \lambda_p v_p = \lambda_{p+1} v_{p+1}.
> $$
>
> Now multiply $(*)$ by $\lambda_{p+1}$ and subtract:
>
> $$
> c_1(\lambda_1 - \lambda_{p+1}) v_1 + \cdots + c_p(\lambda_p - \lambda_{p+1}) v_p = 0.
> $$
>
> Because $\{v_1, \dots, v_p\}$ is independent, every coefficient is zero. The factors
> $\lambda_i - \lambda_{p+1}$ are nonzero (the eigenvalues are distinct), so each
> $c_i = 0$. Then $(*)$ forces $v_{p+1} = 0$, contradicting that $v_{p+1}$ is an
> eigenvector. Hence the set is independent. $\blacksquare$

$$
% caption: Three eigenvectors for three distinct eigenvalues point along three
% different invariant lines; no one of them lies in the span of the others.
\begin{tikzpicture}[>=stealth, scale=0.85, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black!45, thick] (-2.2,0) -- (2.2,0);
  \draw[black!45, thick] (-1.4,-1.8) -- (1.4,1.8);
  \draw[black!45, thick] (2,-1.4) -- (-2,1.4);
  \draw[->, very thick] (0,0) -- (1.8,0) node[right] {$v_1$};
  \draw[->, very thick] (0,0) -- (0.95,1.22) node[above] {$v_2$};
  \draw[->, very thick] (0,0) -- (-1.7,1.19) node[left] {$v_3$};
  \node[black] at (0,-1.5) {three distinct eigenvalues, three independent lines};
\end{tikzpicture}
$$

A useful consequence: an $n \times n$ matrix has at most $n$ distinct eigenvalues,
because more than $n$ eigenvectors for distinct eigenvalues would be an independent set
of more than $n$ vectors in $\mathbb{R}^n$, which is impossible.

## Eigenvectors and difference equations

The reason eigenvalues govern long-term behavior appears already in a first-order
difference equation

$$
x_{k+1} = A x_k \qquad (k = 0, 1, 2, \dots).
$$

Here $\{x_k\}$ is a sequence of vectors in $\mathbb{R}^n$ produced by applying $A$
repeatedly. A **solution** is an explicit formula for $x_k$ that does not require
computing the earlier terms. If $x_0$ happens to be an eigenvector with eigenvalue
$\lambda$, one solution is immediate:

$$
x_k = \lambda^k x_0,
$$

since $A x_k = A(\lambda^k x_0) = \lambda^k (A x_0) = \lambda^{k+1} x_0 = x_{k+1}$.
The dynamics along an eigenspace are pure geometric growth or decay at rate $\lambda$.
Linear combinations of such solutions are again solutions, so once $x_0$ is written
in an eigenvector basis the whole trajectory decouples into independent scalar
sequences. The same construction underlies
[difference equations and Markov chains](/linear-algebra/vector-spaces/difference-equations-and-markov)
and the classification of
[dynamical systems](/linear-algebra/eigenvalues/dynamical-systems).

## Summary

| Question | Reduces to | Tool |
| --- | --- | --- |
| Is $x$ an eigenvector? | Is $Ax$ a multiple of $x$? | one matrix product |
| Is $\lambda$ an eigenvalue? | Is $A - \lambda I$ singular? | test $\operatorname{Nul}(A - \lambda I) \neq \{0\}$ |
| Find eigenvectors for known $\lambda$ | Solve $(A - \lambda I)x = 0$ | row reduction |
| Eigenvalues of a triangular matrix | Read the diagonal | none |
| Eigenvalue $0$ present? | Is $A$ singular? | invertibility test |

Finding the eigenvalues of a general (non-triangular) matrix requires turning
$\det(A - \lambda I) = 0$ into a polynomial equation, the **characteristic
equation**.[^lay51]

[^lay51]: **Lay**, _Linear Algebra and Its Applications_, 5th ed., §5.1 — Eigenvectors and Eigenvalues: the definitions, the eigenspace as the null space of $A - \lambda I$, Theorem 1 on triangular matrices, and Theorem 2 on independence of eigenvectors for distinct eigenvalues.