Eigenvalue–Eigenvector Pair Definition

Scientific claim

Eigenvalue–Eigenvector Pair Definition

Claim

Let $V$ be a vector space over a scalar field $\mathbb F$ , and let

T:V\rightarrow V

be a linear operator.

A nonzero vector $v\in V$ is an eigenvector of $T$ with eigenvalue $\lambda\in\mathbb F$ if

T(v)=\lambda v.

Equivalently, $\lambda$ is an eigenvalue of $T$ if there exists some nonzero vector $v\in V$ satisfying

T(v)=\lambda v.

The pair $(\lambda,v)$ is called an eigenvalue–eigenvector pair of $T$ .

Theoretical: SupportedExperimental: Not applicable

Assumptions

$V$ is a vector space over $\mathbb F$ .
$T:V\rightarrow V$ is a linear operator.
$\lambda$ is a scalar belonging to $\mathbb F$ .
The eigenvector $v$ is nonzero.
The operator maps $V$ back into the same vector space.

Proof

Ordinarily, applying a linear operator $T$ to a vector $v$ can change both the vector’s magnitude and its direction.

An eigenvector is a special nonzero vector whose direction is preserved by the operator. If

T(v)=\lambda v,

then the output is only a scalar multiple of the input.

The scalar $\lambda$ describes how the operator acts along the direction defined by $v$ :

If $\lambda>1$ , the vector is stretched.
If $0<\lambda<1$ , the vector is shortened.
If $\lambda<0$ , its direction is reversed and its magnitude is scaled.
If $\lambda=0$ , the vector is mapped to the zero vector.
If $\lambda$ is complex, the scaling can include a complex phase.

The zero vector is excluded because

T(0)=\lambda0=0

for every scalar $\lambda$ . Allowing the zero vector would incorrectly make every scalar an eigenvalue.

Rearranging the eigenvalue equation gives

(T-\lambda I)v=0,

where $I$ is the identity operator on $V`.

Therefore, $\lambda$ is an eigenvalue exactly when the operator $T-\lambda I$ has a nontrivial kernel:

\ker(T-\lambda I)\neq\{0\}.

If $V$ is finite-dimensional and a basis is selected, $T$ is represented by a square matrix $A$ , while $v$ is represented by a column vector. The eigenvalue equation becomes

Av=\lambda v.

Equivalently,

(A-\lambda I)v=0.

A nonzero solution exists only when $A-\lambda I$ is not invertible. For a finite square matrix, this occurs when

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

This equation is called the characteristic equation of $A`.

Examples

Consider the linear operator represented by
$A= \begin{pmatrix} 2&0\\ 0&3 \end{pmatrix}.$
Let
$v_1= \begin{pmatrix} 1\\ 0 \end{pmatrix}.$
Then
$Av_1 = \begin{pmatrix} 2&0\\ 0&3 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix} = \begin{pmatrix} 2\\ 0 \end{pmatrix} = 2v_1.$
Therefore, $v_1$ is an eigenvector with eigenvalue $\lambda_1=2$ .
Similarly, for
$v_2= \begin{pmatrix} 0\\ 1 \end{pmatrix},$
we have
$Av_2 = \begin{pmatrix} 0\\ 3 \end{pmatrix} = 3v_2.$
Therefore, $v_2$ is an eigenvector with eigenvalue $\lambda_2=3$ .
Consider
$A= \begin{pmatrix} 2&1\\ 1&2 \end{pmatrix}$
and
$v= \begin{pmatrix} 1\\ 1 \end{pmatrix}.$
Applying $A$ gives
$Av = \begin{pmatrix} 2&1\\ 1&2 \end{pmatrix} \begin{pmatrix} 1\\ 1 \end{pmatrix} = \begin{pmatrix} 3\\ 3 \end{pmatrix}.$
Since
$\begin{pmatrix} 3\\ 3 \end{pmatrix} = 3 \begin{pmatrix} 1\\ 1 \end{pmatrix},$
the vector $v$ is an eigenvector of $A$ with eigenvalue $\lambda=3$ .
Eigenvectors are not restricted to ordinary column vectors. Functions can be vectors in a function space, and differential operators can act as linear operators.
Consider the differentiation operator
$D=\frac{d}{dx}$
acting on the function
$f(x)=e^{kx},$
where $k$ is a scalar.
Then
$D f(x) = \frac{d}{dx}e^{kx} = ke^{kx}.$
Therefore,
$D f=kf.$
The function $e^{kx}$ is an eigenfunction of the differentiation operator with eigenvalue $k`.
The term eigenfunction is used when the eigenvector belongs to a vector space of functions.
The one-dimensional quantum momentum operator is
$\hat P=-i\hbar\frac{d}{dx}.$
Consider the function
$\psi_p(x)=e^{ipx/\hbar},$
where $p$ is a real number.
Applying the momentum operator gives
$\hat P\psi_p(x) = -i\hbar\frac{d}{dx}e^{ipx/\hbar}.$
Since
$\frac{d}{dx}e^{ipx/\hbar} = \frac{ip}{\hbar}e^{ipx/\hbar},$
we obtain
$\hat P\psi_p(x) = p e^{ipx/\hbar} = p\psi_p(x).$
Therefore, $\psi_p$ is a generalized eigenfunction of the momentum operator with eigenvalue $p`.
It is called generalized because a plane wave is not square-integrable over the entire real line and therefore is not an ordinary vector in $L^2(\mathbb R)`.

Evidence

01

Eigenvectors and eigenvalues | Chapter 14, Essence of linear algebra3Blue1Brown · 2015-09-15 · supporting

02

Eigenvalues and EigenvectorsErik Cheever · 2026-08-24 · supporting

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