Eigenvalues, eigenvectors, and eigenspaces

Determinants tell you whether a square matrix is invertible. Eigenvalues ask a different question: which nonzero vectors keep their line when the matrix acts on them? If $Av=\lambda v$, then the line $\operatorname{span}\{v\}$ is mapped into itself. When $\lambda>0$, the orientation along that line is preserved; when $\lambda<0$, it is reversed; and when $\lambda=0$, the vector is sent to the zero vector, which has no direction. The line-invariance statement remains meaningful in all three cases.

A general matrix need not preserve every line. A scalar multiple of the identity maps every line into itself, including the zero-scalar case: $0I$ maps every vector to zero. For a fixed scalar $\lambda$, the nonzero eigenvectors become an eigenspace when $0$ is adjoined; across different eigenvalues, the collection of all eigenvectors need not be a subspace. An eigenvector is useful because the matrix action on it is completely described by one scalar. That scalar records stretching, shrinking, reversal, or collapse.

Find directions with a simple matrix action

When you solve $Ax=b$, you focus on one right-hand side and the whole system. When you study eigenvalues, you focus on the special vectors $v$ for which

$$Av=\lambda v.$$

This equation later lets us simplify powers of matrices, recognize diagonal matrices in a suitable basis, and describe invariant directions. The practical route is to turn it into a homogeneous system and then use the null-space, rank, and determinant tools already developed.

For a first picture, take $A=\operatorname{diag}(2,3)$. The horizontal vector $e_1$ becomes $2e_1$ and the vertical vector $e_2$ becomes $3e_2$: each stays on its own line. But $(1,1)^T$ becomes $(2,3)^T$, which is on a different line through the origin. Thus finding eigenvectors means identifying the exceptional directions where a matrix acts by a single scale factor.

Definition

Eigenvalue and eigenvector

Let $A$ be a real $n\times n$ matrix with $n\ge1$. Let $\lambda\in\mathbb R$ and let $v\in\mathbb R^n$ be nonzero. We say that $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ if

$$Av=\lambda v.$$

The condition $v\ne0$ is essential. The zero vector satisfies $A0=\lambda0$ for every scalar, so allowing it would make every scalar look like an eigenvalue. The eigenspace introduced below will contain $0$, but the word eigenvector always refers to a nonzero vector.

From an invariant line to a homogeneous system

The equation $Av=\lambda v$ is equivalent to $(A-\lambda I_n)v=0$. For a fixed eigenvalue $\lambda$, the nonzero solutions are eigenvectors; including zero gives the eigenspace.

Immediate consequences of the definition

Theorem

A fixed eigenvector has only one eigenvalue

Let $A$ be a square matrix and let $v\ne0$. If

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

then $\lambda=\mu$.

Subtracting the two equations gives $(\lambda-\mu)v=0$. Choose a coordinate $v_r\ne0$. Its $r$th coordinate says $(\lambda-\mu)v_r=0$. Since $v_r$ is a nonzero real number, cancellation gives $\lambda-\mu=0$, hence $\lambda=\mu$. The nonzero hypothesis is exactly what makes this cancellation valid.

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 $\lambda$.

Indeed, for $c\ne0$,

$$A(cv)=c(Av)=c(\lambda v)=\lambda(cv).$$

The vector $cv$ is still nonzero, so it satisfies both parts of the definition. Thus all nonzero points on the line through one eigenvector are eigenvectors, rather than just one isolated point.

Theorem

Linear combinations with a common eigenvalue

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

$$\alpha_1u_1+\cdots+\alpha_ku_k$$

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

Let $w=\alpha_1u_1+\cdots+\alpha_ku_k$. Linearity gives

$$Aw=\sum_{j=1}^k\alpha_jAu_j =\sum_{j=1}^k\alpha_j\lambda u_j =\lambda w.$$

If $w\ne0$, this is an eigenvector. The coefficients need not all be zero, but their combination can still cancel to $w=0$; that zero output belongs to the eigenspace but is not an eigenvector. This is why “nonzero combination” is part of the theorem. Components can cancel even when several coefficients are nonzero.

Definition

Eigenspace

For an $n\times n$ matrix $A$ and scalar $\lambda$, define

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

When $\lambda$ is an eigenvalue, this space contains all its eigenvectors and also the zero vector. If $\lambda$ is not an eigenvalue, the same null-space notation gives $E_A(\lambda)=\{0\}$.

First examples

Worked example

A 2×2 matrix with two distinct eigenvalues

Let

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

For

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

Thus $u_1$ has eigenvalue $1$. For $u_2=\begin{bmatrix}2\\-1\end{bmatrix}$,

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

so $u_2$ has eigenvalue $-2$. The same matrix can therefore preserve two different lines, with different scaling factors.

The common-eigenvalue combination theorem does not apply to $u_1+u_2$, because their eigenvalues differ. In fact,

$$u_1+u_2=\begin{bmatrix}7\\-3\end{bmatrix},\qquad A(u_1+u_2)=\begin{bmatrix}1\\0\end{bmatrix}.$$

If this sum were an eigenvector, the second coordinate would force $\lambda=0$, while the first would require $1=0\cdot7$, a contradiction.

Worked example

One eigenvalue can have more than one direction

Let

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

For $x=(x_1,x_2,x_3)^T$, write $s=x_1+x_2+x_3$. Then

$$Bx=(s+x_1,\ s+x_2,\ s+x_3)^T.$$

For $\lambda=1$, the equation $Bx=x$ reduces in every coordinate to $s=0$. Therefore

$$E_B(1)=\{x\in\mathbb R^3:x_1+x_2+x_3=0\} =\operatorname{span}\left\{\begin{bmatrix}1\\-1\\0\end{bmatrix}, \begin{bmatrix}1\\0\\-1\end{bmatrix}\right\}.$$

The two displayed vectors are linearly independent: if $a(1,-1,0)^T+b(1,0,-1)^T=0$, then the second coordinate gives $a=0$ and the third gives $b=0$. They form a basis for the two-dimensional eigenspace.

For $\lambda=4$, the equations $Bx=4x$ say $s=3x_1=3x_2=3x_3$, so $x_1=x_2=x_3$. Hence

$$E_B(4)=\operatorname{span}\left\{\begin{bmatrix}1\\1\\1\end{bmatrix}\right\}.$$

This example separates two ideas. One fixed nonzero vector cannot have two eigenvalues, but one eigenvalue can have a whole plane of eigenvectors. The plane includes $0$ as a vector in the eigenspace; its nonzero points are the eigenvectors.

Eigenvalues are a null-space question

Move all terms in $Av=\lambda v$ to one side:

$$Av-\lambda v=0, \qquad\text{so}\qquad (A-\lambda I_n)v=0.$$

The identity matrix is needed because $\lambda v=\lambda I_nv$. This rewrite makes the eigenvalue problem a homogeneous-system problem.

Theorem

Eigenvalue criterion via a homogeneous system

Let $A$ be an $n\times n$ matrix, let $\lambda$ be a scalar, and let $v\ne0$. Then the following are equivalent:

1. $v$ is an eigenvector of $A$ with eigenvalue $\lambda$.
2. $v$ is a nontrivial solution of $(A-\lambda I_n)x=0$.

The forward direction is just rearrangement of $Av=\lambda v$. Conversely, a nontrivial solution is nonzero and rearranges to $Av=\lambda v$, so it meets the definition.

The null-space description gives closure directly. If $x,y\in E_A(\lambda)$ and $a,b$ are scalars, then

$$(A-\lambda I_n)(ax+by)=a(A-\lambda I_n)x+b(A-\lambda I_n)y=0.$$

Thus $E_A(\lambda)$ is a subspace. It has positive dimension exactly when $\lambda$ is an eigenvalue; otherwise its dimension is zero.

The number $\dim E_A(\lambda)$ is called the geometric multiplicity of $\lambda$. It counts independent eigenvector directions for that eigenvalue: in the matrix $B$ above, the eigenvalue $1$ has geometric multiplicity $2$, whereas $4$ has geometric multiplicity $1$. This dimension is obtained by solving the null-space equations, after the candidate scalar has been found.

Concept lensStructural

A fixed-eigenvalue space is not all eigenvectors at once

For a fixed eigenvalue $\lambda$, the solutions of $(A-\lambda I_n)x=0$ form a null space. This solution set already contains zero and is closed under addition and scalar multiplication; the eigenvectors are precisely its nonzero members. The union of eigenspaces for different eigenvalues need not be closed under addition. In the first example, $u_1+u_2$ is an explicit non-eigenvector sum. Keep each fixed-$\lambda$ eigenspace separate from the collection of all eigenvectors.

Equivalent formulations for an eigenvalue

Theorem

Equivalent tests for λ to be an eigenvalue

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

1. $\lambda$ is an eigenvalue of $A$.
2. $(A-\lambda I_n)x=0$ has a nontrivial solution.
3. $N(A-\lambda I_n)\ne\{0\}$.
4. $A-\lambda I_n$ is not invertible.
5. $\det(A-\lambda I_n)=0$.

The implications are the familiar homogeneous-system dictionary. A nontrivial solution is exactly a nonzero null-space vector, so (1), (2), and (3) are the same condition. A square matrix has a nontrivial null space exactly when it is not invertible, giving (3) $\Longleftrightarrow$ (4). The determinant criterion then gives (4) $\Longleftrightarrow$ (5). In particular, an eigenvalue forces $\operatorname{rank}(A-\lambda I_n)\le n-1$. The determinant equation is a way to discover candidate eigenvalues; the null space is still required to find the corresponding eigenvectors.

Reading an eigenspace from the equations

Once a candidate $\lambda$ is known, form $A-\lambda I_n$ and row reduce it. The free variables in the resulting homogeneous system parameterize every vector in $E_A(\lambda)$. The number of free variables is the dimension of that eigenspace, so it tells you how many independent eigenvector directions are available. Every nonzero vector in the resulting span is an eigenvector, while the zero vector remains part of the space only because a subspace must contain zero. This is why an eigenvalue list is incomplete without its eigenspaces. For a repeated eigenvalue, the determinant records algebraic multiplicity, whereas the null-space calculation records geometric multiplicity; comparing them is a key question for the diagonalization developed in the next section.

Worked example

Find eigenvalues and eigenspaces by row reduction

Let

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

First find the eigenvalues from

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

Factoring gives $\lambda=4$ and $\lambda=-3$. For $\lambda=4$,

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

so $x_1=2x_2$ and

$$E_C(4)=\operatorname{span}\left\{\begin{bmatrix}2\\1\end{bmatrix}\right\}.$$

For $\lambda=-3$,

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

so $3x_1+x_2=0$ and

$$E_C(-3)=\operatorname{span}\left\{\begin{bmatrix}1\\-3\end{bmatrix}\right\}.$$

The determinant calculation discovers the two scalars. Each row reduction then solves a different homogeneous system; substituting the basis vectors verifies $C(2,1)^T=4(2,1)^T$ and $C(1,-3)^T=-3(1,-3)^T$.

Diagonal and triangular matrices

For a diagonal matrix $D=\operatorname{diag}(d_1,\ldots,d_n)$, the eigenvector equation is especially transparent:

$$(D-\lambda I_n)x=0 \quad\Longleftrightarrow\quad (d_j-\lambda)x_j=0\quad\text{for every }j.$$

Thus

$$E_D(\lambda)=\operatorname{span}\{e_j:d_j=\lambda\},$$

where the span is $\{0\}$ if no diagonal entry equals $\lambda$. Repeated diagonal entries therefore increase the dimension of an eigenspace. For the zero matrix every nonzero vector is an eigenvector with eigenvalue $0$; for the identity matrix every nonzero vector is an eigenvector with eigenvalue $1$.

If $T$ is upper or lower triangular, then $T-\lambda I_n$ is triangular, so

$$\det(T-\lambda I_n)=\prod_{j=1}^n(t_{jj}-\lambda).$$

The eigenvalues are exactly the diagonal entries, counted with their algebraic multiplicity. The diagonal entries do not by themselves determine the eigenvectors. For the upper triangular matrix

$$T=\begin{bmatrix}1&4\\0&2\end{bmatrix},$$

$Te_1=e_1$, so $e_1$ spans $E_T(1)$. For $\lambda=2$, $(T-2I)x=0$ gives $-x_1+4x_2=0$, hence

$$E_T(2)=\operatorname{span}\left\{\begin{bmatrix}4\\1\end{bmatrix}\right\}.$$

Changing the upper-right entry would keep the diagonal and eigenvalues while changing this eigenspace. The diagonal data alone are therefore insufficient.

Important properties

Theorem

Zero as an eigenvalue detects noninvertibility

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

1. $0$ is an eigenvalue of $A$.
2. $A$ is not invertible.

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

This is the equivalent-tests theorem with $\lambda=0$.

For completeness, if $0$ is an eigenvalue then some nonzero $v$ satisfies $Av=0$, so the null space is nontrivial and $A$ cannot be invertible. Conversely, if $A$ is not invertible, the square-matrix dictionary supplies a nonzero vector in $N(A)$; it satisfies $Av=0\cdot v$, so $0$ is an eigenvalue. The two directions use the same homogeneous-system criterion.

Theorem

Eigenvalues under simple matrix operations

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

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

For (1), if $Av=\lambda v$ then $(kA)v=k\lambda v$; this also covers $k=0$, because the zero matrix has every nonzero vector as a $0$-eigenvector. For powers with $m\ge1$, induction gives $A^mv=\lambda^mv$: the induction step multiplies $A^mv=\lambda^mv$ by $A$ and uses $Av=\lambda v$. The case $m=0$ is separate: $A^0=I$, and the same nonzero $v$ has eigenvalue $1$ for $I$. We do not interpret this as the expression $0^0$.

For (4), invertibility implies $\lambda\ne0$. Multiplying $Av=\lambda v$ by $A^{-1}$ gives $v=\lambda A^{-1}v$, so division by $\lambda$ gives $A^{-1}v=\lambda^{-1}v$. For (3), use the determinant identity

$$\det(A^T-\lambda I_n)=\det((A-\lambda I_n)^T) =\det(A-\lambda I_n)=0.$$

This proves the transpose statement through the eigenvalue criterion. It is not valid to transpose $Av=\lambda v$ and conclude that $A^Tv=\lambda v$. For the matrix $T=\begin{bmatrix}1&4\\0&2\end{bmatrix}$, $Te_1=e_1$ but

$$T^Te_1=\begin{bmatrix}1&0\\4&2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}1\\4\end{bmatrix}\ne e_1.$$

Optional complex extension: the field matters

The core definition above is over $\mathbb R$. Consider the real matrix

$$R=\begin{bmatrix}1&-1\\1&1\end{bmatrix}.$$

For a real scalar $\lambda$,

$$\det(R-\lambda I_2)=(1-\lambda)^2+1=(\lambda-1)^2+1>0.$$

Therefore $R$ has no real eigenvalues or real eigenvectors. Over $\mathbb C$, the same matrix has eigenvalues $1+i$ and $1-i$. With

$$u_+=\begin{bmatrix}i\\1\end{bmatrix},\qquad u_-=\begin{bmatrix}-i\\1\end{bmatrix},$$

we verify

$$Ru_+=\begin{bmatrix}i-1\\i+1\end{bmatrix}=(1+i)u_+, \qquad Ru_- =\begin{bmatrix}-1-i\\1-i\end{bmatrix}=(1-i)u_-.$$

Solving the first row of $(R-(1+i)I)x=0$ gives $x=iy$, while the first row for $1-i$ gives $x=-iy$. Hence over $\mathbb C$ the eigenspaces are $\operatorname{span}_{\mathbb C}\{u_+\}$ and $\operatorname{span}_{\mathbb C}\{u_-\}$. This example marks the field boundary: nonreal roots are not real eigenvalues, but they become valid after the scalar field is explicitly changed.

Common mistake

Common mistake

The zero vector is never an eigenvector

The equation $A0=\lambda0$ holds for every scalar, but the definition excludes $0$. An eigenspace is a subspace and therefore contains $0$; its nonzero members are the eigenvectors. Also remember that an eigenvalue belongs to a specified matrix and scalar field.

Quick check

Checkpoint

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

Assume $v\ne0$.

Solution · Answer

Yes. $A(3v)=3Av=3\lambda v=\lambda(3v)$, and $3v\ne0$.

Checkpoint

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

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

Solution · Answer

If $0$ is an eigenvalue, then $Ax=0$ has a nontrivial solution. The null space of $A$ is nontrivial, so $A$ is not invertible.

Checkpoint

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

Use the definition above.

Solution · Answer

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

It is the null space of $A-\lambda I_n$, including the zero vector.

Exercises

Checkpoint

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

$$\begin{bmatrix}2&1\\1&2\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix} =\begin{bmatrix}3\\3\end{bmatrix}=3\begin{bmatrix}1\\1\end{bmatrix}.$$

The vector is an eigenvector with eigenvalue $3$.

Checkpoint

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

$$A-4I=\begin{bmatrix}-3&0\\0&0\end{bmatrix}, \qquad x_1=0.$$

Thus $x_2$ is free and

$$E_A(4)=\operatorname{span}\left\{\begin{bmatrix}0\\1\end{bmatrix}\right\}.$$

Checkpoint

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

No. Invertibility of $A$ implies invertibility of $A^T$. By the zero-eigenvalue criterion, $A^T$ cannot have $0$ as an eigenvalue.

Related notes

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 shape of a matrix.

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