[ALGEBRA] Eigenvalues and eigenvectors

This page defines eigenvalues, eigenvectors and eigenspaces and gives the determinant test for finding them. A matrix representation depends on the chosen basis and the most useful basis vectors are those that the map sends to scalar multiples of themselves.

Enough such vectors form an eigenbasis. That basis separates matrix powers, linear ODE systems and Markov systems into scalar modes.

Invariant directions

[DEFINITION]

Linear map

Let $V$ and $W$ be vector spaces over the same field $F$ . A function $T:V\to W$ is a linear map if

T(u+v)=T(u)+T(v)\qquad\text{for all }u,v\in V,

T(\omega u)=\omega T(u)\qquad\text{for all }\omega\in F,\ u\in V.

[1]:Eigenvalue
[2]:Matrix maps are linear
[3]:The matrix of a linear map on standard coordinates
[4]:A dot-product map
18 more

[THEOREM]

Matrix representation of a linear map

Let $T:V\to W$ be a linear map, where $\dim(V)=n$ and $\dim(W)=m$ . For ordered bases

B_V=(v_1,\ldots,v_n),\qquad B_W=(w_1,\ldots,w_m),

form the coordinate columns

a_j=[T(v_j)]_{B_W}\qquad (j=1,\ldots,n).

Then

A=\begin{pmatrix}a_1&\cdots&a_n\end{pmatrix}

is the unique $m\times n$ matrix satisfying

[T(v)]_{B_W}=A[v]_{B_V}\qquad\text{for every }v\in V.

We call $A$ the matrix representation of $T$ relative to $B_V$ and $B_W$ and may write it as $[T]_{B_V}^{B_W}$ .

With this notation, the subscript names the domain basis and the superscript names the codomain basis, matching the direction $[v]_{B_V}\mapsto[T(v)]_{B_W}$ in the displayed coordinate identity.

[1]:Linear map
[2]:Basis
[3]:Coordinate vector
[4]:Values on a basis determine a linear map
[5]:Diagonalisation
[6]:Diagonalisation
[7]:Characteristic polynomial
[8]:First-order linear system
9 more

[DEFINITION]

Kernel

If $T:V\to W$ is a linear map, its kernel is

\ker(T)=\{u\in V:T(u)=0\}.

The same set is also called the null space. If $A\in\mathbb F^{m\times n}$ is regarded as the map $\mathbb F^n\to\mathbb F^m$ , then

\ker(A)=\{x\in\mathbb F^n:Ax=0\}.

[DEFINITION]

Eigenvalue

Let $T:V\to V$ be a linear map over a field $F$ . A scalar $\lambda\in F$ is an eigenvalue of $T$ if there is a non-zero vector $v\in V$ such that

T(v)=\lambda v.

The equation says that the line through $v$ is mapped into itself, with every vector on that line scaled by $\lambda$ . A negative real $\lambda$ reverses direction and $\lambda=0$ collapses the line to the origin. For a matrix $A$ , the same condition is

Av=\lambda v.

[1]:Linear map
[2]:Eigenvectors for distinct eigenvalues
[3]:A real matrix with complex eigenvalues
[4]:Invertibility and eigenvalues
[5]:Diagonalisation
16 more

[DEFINITION]

Eigenvector

If $T(v)=\lambda v$ for a non-zero vector $v$ , then $v$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda$ .

The zero vector is excluded because $T(0)=\lambda 0$ holds for every scalar, so it says nothing about a direction. Any non-zero scalar multiple of an eigenvector is another eigenvector for the same eigenvalue.

[1]:Eigenvalue
[2]:Eigenvectors for distinct eigenvalues
[3]:Eigenbasis
[4]:A real matrix with complex eigenvalues
[5]:Diagonalisation
8 more

[EXAMPLE]

An eigenvector of differentiation

Let $V=C^\infty(\mathbb R,\mathbb R)$ , the vector space of smooth real-valued functions and let $T:V\to V$ be differentiation, $T(f)=f'$ . For any real $\lambda$ , the function

f(x)=e^{\lambda x}

satisfies

T(f)=f'=\lambda e^{\lambda x}=\lambda f.

Thus $e^{\lambda x}$ is an eigenvector of differentiation for the eigenvalue $\lambda$ . For a matrix, the same test reads $Av=\lambda v$ , which is the form used throughout diagonalisation.

[routine]

[1]:Eigenvalue
[2]:Eigenvector

[DEFINITION]

Eigenspace

For $A\in M_n(F)$ and an eigenvalue $\lambda\in F$ , the eigenspace corresponding to $\lambda$ is

E_\lambda=\ker_F(A-\lambda I).

Its non-zero vectors are precisely the eigenvectors of $A$ corresponding to $\lambda$ .

To find the eigenspace, solve the homogeneous system

(A-\lambda I)v=0.

The kernel description makes the eigenspace a subspace, so it contains the zero vector even though the zero vector is not an eigenvector.

[1]:Eigenvalue
[2]:Eigenvector
[3]:Kernel
[4]:A repeated eigenvalue without enough modes
[5]:The eigenvalue criterion
[6]:The matrix with all entries equal to one
[7]:Geometric multiplicity

The characteristic polynomial

[DEFINITION]

Characteristic polynomial

For an $n\times n$ matrix $A$ over $F$ , the characteristic polynomial is

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

The equation $p_A(\lambda)=0$ is the associated characteristic equation.

The characteristic polynomial turns the search for a non-zero solution of $(A-\lambda I)v=0$ into root-finding over the chosen field. A polynomial with no roots in $F$ may gain roots when the scalars are extended, for example from $\mathbb R$ to $\mathbb C$ .

Some texts define the polynomial as $\det(\lambda I-A)$ instead. For an $n\times n$ matrix this is $(-1)^n p_A(\lambda)$ , so both conventions give the same eigenvalues.

Two names clash with Calculus. The characteristic equation of a scalar ODE is a different object with the same name. In constrained extrema, the letter $\lambda$ denotes a Lagrange multiplier rather than an eigenvalue.

[THEOREM]

The eigenvalue criterion

A scalar $\lambda\in F$ is an eigenvalue of $A$ if and only if

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

For such a $\lambda$ , the eigenvectors are the non-zero vectors in $\ker(A-\lambda I)$ .

This determinant test is how eigenpairs are computed. Diagonalisation then uses the full list of roots to build an eigenbasis.

[1]:Eigenvalue
[2]:Eigenvector
[3]:Eigenspace
[4]:Characteristic polynomial
[5]:Proof
[6]:The matrix with all entries equal to one

[PROOF]

The equation $Av=\lambda v$ is equivalent to $(A-\lambda I)v=0$ . A non-zero solution exists exactly when $A-\lambda I$ is singular and a square matrix is singular exactly when its determinant is zero. The description of the eigenvectors then follows from the definition of the kernel.

[1]:The eigenvalue criterion

[THEOREM]

Counting eigenvalues

The characteristic polynomial of an $n\times n$ matrix is degree $n$ . Over $\mathbb C$ , it has $n$ roots counted with multiplicity and these roots are the eigenvalues of the matrix.

The roots exist by the fundamental theorem of algebra. Their moduli and real parts later govern growth in stability of systems.

Computing eigenpairs

[PROPOSITION]

Eigenvalues of a triangular matrix

For a triangular matrix, the eigenvalues are its diagonal entries, counted with their multiplicities.

The matrix $A-\lambda I$ is still triangular, so its determinant is the product of its diagonal entries and no full expansion is needed.

[EXAMPLE]

The matrix with all entries equal to one

Let

A= \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}.

The characteristic polynomial is

p_A(\lambda) =\det\begin{pmatrix}1-\lambda&1\\1&1-\lambda\end{pmatrix} =(1-\lambda)^2-1 =\lambda(\lambda-2).

Thus the eigenvalues are $0$ and $2$ . For $\lambda=0$ ,

A v=0 \quad\Longrightarrow\quad v=\alpha\begin{pmatrix}1\\-1\end{pmatrix},

E_0=\operatorname{span}\left\{\begin{pmatrix}1\\-1\end{pmatrix}\right\}.

For $\lambda=2$ ,

(A-2I)v=0 \quad\Longrightarrow\quad v=\alpha\begin{pmatrix}1\\1\end{pmatrix},

E_2=\operatorname{span}\left\{\begin{pmatrix}1\\1\end{pmatrix}\right\}.

The two eigenspaces are perpendicular lines: $A$ doubles the component parallel to $(1,1)^{\mathsf T}$ and removes the component parallel to $(1,-1)^{\mathsf T}$ . The roots $0$ and $2$ alone do not show this. The eigenspaces are what describe the action of the matrix and supply an eigenbasis.

[routine]

[1]:The eigenvalue criterion
[2]:Eigenspace

Eigenvalues and eigenvectors

Linear map

Matrix representation of a linear map

Kernel

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