8.1 Eigenvalues, eigenvectors, and eigenspaces

Determinants tell you whether a square matrix is invertible. Eigenvalues ask a different question: which vectors keep their direction when the matrix acts on them?

Most vectors are bent into new directions by matrix multiplication. An eigenvector is exceptional. It is only stretched, shrunk, or reversed by a scalar factor. That is why eigenvalues reveal the internal geometry of a matrix rather than only its solvability properties.

Why this section matters

When you solve $Ax=b$ , you focus on the whole system. When you study eigenvalues, you focus on the special vectors v for which the matrix action collapses to the simpler rule

Av=\lambda v.

If you can recognize those vectors, then later you can simplify powers of matrices, understand diagonalization, and describe invariant directions.

Definition

Eigenvalue and eigenvector

Let $A$ be an $n\times n$ square matrix. Let $\lambda$ be a scalar, and let v be a nonzero column vector in $\mathbb{R}^n$ .

We say that v is an eigenvector of $A$ with eigenvalue $\lambda$ if

Av=\lambda v.

The vector v must be nonzero. The zero vector satisfies $A0=\lambda0$ for every scalar, so it would destroy the meaning of the definition if we allowed it.

The point is that the matrix action and scalar multiplication agree on that vector. The direction of v is preserved, although the magnitude may change and the sign may flip.

Immediate consequences of the definition

Theorem

A fixed eigenvector has only one eigenvalue

Let $A$ be a square matrix, and let v be a nonzero vector. If

Av=\lambda v \qquad\text{and}\qquad Av=\mu v,

then $\lambda=\mu$ .

The reason is simple. Subtract the two equations:

(\lambda-\mu)v=0.

Since $v\neq0$ , the scalar factor must be 0, so $\lambda=\mu$ .

Theorem

Nonzero scalar multiples stay in the same eigen-direction

If v is an eigenvector of $A$ with eigenvalue $\lambda$ , then every nonzero scalar multiple cv is also an eigenvector of $A$ with eigenvalue $\lambda$ .

That is why an eigenvector never comes alone. It naturally represents a whole line through the origin.

Theorem

Linear combinations with a common eigenvalue

Suppose $u_1,\dots,u_k$ are eigenvectors of $A$ , all with the same eigenvalue $\lambda$ . Then every nonzero linear combination

\alpha_1u_1+\cdots+\alpha_ku_k

is also an eigenvector of $A$ with eigenvalue $\lambda$ .

This is the first hint that eigenvectors belonging to the same eigenvalue form a subspace once the zero vector is added back in.

First examples

Worked example

A 2×2 matrix with two distinct eigenvalues

Let

A= \begin{bmatrix} 13&30\\ -6&-14 \end{bmatrix}.

Check the vector

u_1= \begin{bmatrix} 5\\ -2 \end{bmatrix}.

Then

Au_1= \begin{bmatrix} 13&30\\ -6&-14 \end{bmatrix} \begin{bmatrix} 5\\ -2 \end{bmatrix} = \begin{bmatrix} 5\\ -2 \end{bmatrix} =1\cdot u_1.

So $u_1$ is an eigenvector with eigenvalue 1.

Now try

u_2= \begin{bmatrix} 2\\ -1 \end{bmatrix}.

Then

Au_2= \begin{bmatrix} -4\\ 2 \end{bmatrix} =-2 \begin{bmatrix} 2\\ -1 \end{bmatrix} =-2u_2.

So $u_2$ is an eigenvector with eigenvalue $-2$ .

Worked example

One eigenvalue can have more than one direction

Let

B= \begin{bmatrix} 2&1&1\\ 1&2&1\\ 1&1&2 \end{bmatrix}.

The vector

u_1= \begin{bmatrix} 1\\1\\1 \end{bmatrix}

satisfies $Bu_1=4u_1$ , so 4 is one eigenvalue.

Now consider

u_2= \begin{bmatrix} 1\\-1\\0 \end{bmatrix}, \qquad u_3= \begin{bmatrix} 1\\0\\-1 \end{bmatrix}.

Both satisfy

Bu_2=u_2, \qquad Bu_3=u_3.

So the same eigenvalue 1 has at least two linearly independent eigenvectors. That does not contradict the uniqueness theorem above. The theorem says one fixed nonzero vector cannot correspond to two different eigenvalues. It does not say one eigenvalue can have only one eigenvector direction.

Eigenvalues are a null-space question

The equation $Av=\lambda v$ becomes much easier to analyze when you move all terms to one side:

Av-\lambda v=0.

Since $\lambda v=\lambda I_nv$ , this is equivalent to

(A-\lambda I_n)v=0.

Theorem

Eigenvalue criterion via a homogeneous system

Let $A$ be an $n\times n$ matrix, $\lambda$ a scalar, and $v\neq0$ .

Then the following are equivalent:

v is an eigenvector of $A$ with eigenvalue $\lambda$ .
v is a nontrivial solution of the homogeneous system

(A-\lambda I_n)x=0.

This recasts the eigenvalue problem as an ordinary linear-system problem. The vector v must live in the null space of $A-\lambda I$ .

Definition

Eigenspace

Let $A$ be an $n\times n$ matrix, and let $\lambda$ be an eigenvalue of $A$ . The eigenspace of $A$ corresponding to $\lambda$ is

E_A(\lambda)=N(A-\lambda I_n).

So the eigenspace contains all eigenvectors for $\lambda$ together with the zero vector.

Because it is a null space, $E_A(\lambda)$ is automatically a subspace.

Equivalent formulations for an eigenvalue

Once the null-space formulation is in place, the invertibility dictionary immediately gives several equivalent tests.

Theorem

Equivalent tests for λ to be an eigenvalue

Let $A$ be an $n\times n$ matrix and $\lambda$ a scalar. The following statements are equivalent:

$\lambda$ is an eigenvalue of $A$ .
$(A-\lambda I_n)x=0$ has a nontrivial solution.
$N(A-\lambda I_n)\neq\{0\}$ .
$A-\lambda I_n$ is not invertible.
$\det(A-\lambda I_n)=0$ .

This theorem is the bridge from eigenvalues back to determinants. It is also the theorem that later produces the characteristic polynomial.

Worked example

Find eigenvalues and eigenspaces by row reduction

Let

C= \begin{bmatrix} 3&2\\ 3&-2 \end{bmatrix}.

To find the eigenvalues, solve

\det(C-\lambda I)=0.

We get

\det \begin{bmatrix} 3-\lambda&2\\ 3&-2-\lambda \end{bmatrix} =(3-\lambda)(-2-\lambda)-6 =\lambda^2-\lambda-12.

So the eigenvalues are the roots:

\lambda=4,\qquad \lambda=-3.

For $\lambda=4$ ,

C-4I= \begin{bmatrix} -1&2\\ 3&-6 \end{bmatrix} \sim \begin{bmatrix} 1&-2\\ 0&0 \end{bmatrix}.

Thus $x_1=2x_2$ , so one basis vector is

\begin{bmatrix} 2\\1 \end{bmatrix}.

For $\lambda=-3$ ,

C+3I= \begin{bmatrix} 6&2\\ 3&1 \end{bmatrix} \sim \begin{bmatrix} 3&1\\ 0&0 \end{bmatrix}.

Thus $3x_1+x_2=0$ , so one basis vector is

\begin{bmatrix} 1\\-3 \end{bmatrix}.

Therefore

E_C(4)=\operatorname{span}\left\{ \begin{bmatrix} 2\\1 \end{bmatrix} \right\}, \qquad E_C(-3)=\operatorname{span}\left\{ \begin{bmatrix} 1\\-3 \end{bmatrix} \right\}.

Important properties

Theorem

Zero as an eigenvalue detects noninvertibility

For a square matrix $A$ , the following are equivalent:

0 is an eigenvalue of $A$ .
$A$ is not invertible.

Equivalently, $A$ is invertible if and only if 0 is not an eigenvalue.

This is just the previous equivalence theorem with $\lambda=0$ .

Theorem

Eigenvalues under simple matrix operations

If $\lambda$ is an eigenvalue of $A$ , then:

$k\lambda$ is an eigenvalue of kA;
$\lambda^m$ is an eigenvalue of $A^m$ for every nonnegative integer m;
$\lambda$ is an eigenvalue of $A^T$ ;
if $A$ is invertible, then $\lambda^{-1}$ is an eigenvalue of $A^{-1}$ .

These statements are not mysterious. They all come from applying the relevant matrix operation to the defining equation $Av=\lambda v$ .

Common mistake

The zero vector is never an eigenvector

Students often notice that $A0=\lambda0$ for every scalar and conclude that the zero vector is an eigenvector for every eigenvalue. That is exactly why the definition excludes 0. An eigenvector must point in a genuine direction, and the zero vector has no direction to preserve.

Quick check

If `v` is an eigenvector of $A$ with eigenvalue $\lambda$ , is `3v` also an eigenvector of $A$ ?

Assume $v\neq0$ .

Solution

Answer

Quick check

Why does `0` being an eigenvalue force $A$ to be noninvertible?

Use the system $(A-0I)x=0$ .

Solution

Answer

Quick check

What is the eigenspace $E_A(\lambda)$ in null-space language?

Use the definition introduced above.

Solution

Answer

Exercises

Quick check

Show that $\begin{bmatrix}1\\1\end{bmatrix}$ is an eigenvector of $\begin{bmatrix}2&1\\1&2\end{bmatrix}$ and find its eigenvalue.

Multiply first, then compare the output with the original vector.

Solution

Guided solution

Quick check

Find the eigenspace of $A=\begin{bmatrix}1&0\\0&4\end{bmatrix}$ for the eigenvalue `4`.

Solve $(A-4I)x=0$ .

Solution

Guided solution

Quick check

If $A$ is invertible, can `0` appear as an eigenvalue of $A^T$ ?

Use the transpose property and the invertibility criterion together.

Solution

Guided solution

Keep 7.2 Row operations, products, and invertibility nearby, because the test $\det(A-\lambda I)=0$ depends directly on determinant and invertibility reasoning.

Continue with 8.2 Diagonalization and similarity to see how a full basis of eigenvectors changes the whole shape of a matrix.

The subspace viewpoint here also leans on 6.5 Basis and dimension.

8.1 Eigenvalues, eigenvectors, and eigenspaces

MATH1030: Linear algebra I

Why this section matters

Eigenvalue and eigenvector

Immediate consequences of the definition

A fixed eigenvector has only one eigenvalue

Nonzero scalar multiples stay in the same eigen-direction

Linear combinations with a common eigenvalue

First examples

A 2×2 matrix with two distinct eigenvalues

One eigenvalue can have more than one direction

Eigenvalues are a null-space question

Eigenvalue criterion via a homogeneous system

Eigenspace

Equivalent formulations for an eigenvalue

Equivalent tests for λ to be an eigenvalue

Find eigenvalues and eigenspaces by row reduction

Important properties

Zero as an eigenvalue detects noninvertibility

Eigenvalues under simple matrix operations

Common mistake

The zero vector is never an eigenvector

Quick check

If v is an eigenvector of AAA with eigenvalue λ\lambdaλ, is 3v also an eigenvector of AAA?

Answer

Why does 0 being an eigenvalue force AAA to be noninvertible?

Answer

What is the eigenspace EA(λ)E_A(\lambda)EA​(λ) in null-space language?

Answer

Exercises

Show that [11]\begin{bmatrix}1\\1\end{bmatrix}[11​] is an eigenvector of [2112]\begin{bmatrix}2&1\\1&2\end{bmatrix}[21​12​] and find its eigenvalue.

Guided solution

Find the eigenspace of A=[1004]A=\begin{bmatrix}1&0\\0&4\end{bmatrix}A=[10​04​] for the eigenvalue 4.

Guided solution

If AAA is invertible, can 0 appear as an eigenvalue of ATA^TAT?

Guided solution

Related notes

Section mastery checkpoint

Prerequisites

Key terms in this unit

More notes in this series

If `v` is an eigenvector of $A$ with eigenvalue $\lambda$ , is `3v` also an eigenvector of $A$ ?

Why does `0` being an eigenvalue force $A$ to be noninvertible?

What is the eigenspace $E_A(\lambda)$ in null-space language?

Show that $\begin{bmatrix}1\\1\end{bmatrix}$ is an eigenvector of $\begin{bmatrix}2&1\\1&2\end{bmatrix}$ and find its eigenvalue.

Find the eigenspace of $A=\begin{bmatrix}1&0\\0&4\end{bmatrix}$ for the eigenvalue `4`.

If $A$ is invertible, can `0` appear as an eigenvalue of $A^T$ ?