Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 21

Eigenvalues and eigenvectors

For a reflection at an axis in the plane, certain vectors behave particularly simply. The vectors on the axis are sent to themselves, and the vectors which are orthogonal to the axis are sent to their negatives. For all these vectors, the image under this linear mapping lies on the line spanned by these vectors. In the theory of eigenvalues and eigenvectors, we want to know whether, for a given linear mapping, there exist lines (one-dimensional linear subspaces), which are mapped to themselves. The goal is to find, for the linear mapping, a basis such that the describing matrix is quite simple. Here, an important application is to find solutions for a system of linear differential equations.

A reflection at an axis has two eigenlines, the axis with eigenvalue ${}1$ and the orthogonal line with eigenvalue ${}-1$.

Definition

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and

$\varphi \colon V\longrightarrow V$

a linear mapping. Then an element ${}v\in V$, ${}v\neq 0$, is called an eigenvector of ${}\varphi$ (for the eigenvalue ${}\lambda$), if

${}\varphi (v)=\lambda v\,$

for some ${}\lambda \in K$

holds.

To paraphrase, an eigenvector is a vector ${}v\neq 0$ which is linearly dependent to ${}\varphi (v)$.

A shear mapping has one eigenline for the eigenvalue ${}1$, and no further eigenvalues.

Definition

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and

$\varphi \colon V\longrightarrow V$

a linear mapping. Then an element ${}\lambda \in K$ is called an eigenvalue for ${}\varphi$, if there exists a vector ${}v\in V$, ${}v\neq 0$ such that

${}\varphi (v)=\lambda v\,.$

The set of all eigenvalues is called the spectrum of ${}\varphi$.

Definition

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and

$\varphi \colon V\longrightarrow V$

a linear mapping. For ${}\lambda \in K$, we denote by

${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}:={\left\{v\in V\mid \varphi (v)=\lambda v\right\}}\,$

the eigenspace of ${}\varphi$ for the value ${}\lambda$.

Thus we allow arbitrary values (not only eigenvalues) in the definition of an eigenspace. We will see in Lemma 21.6 that they are linear subspaces. In particular, ${}0$ belongs to every eigenspace, though it is never an eigenvector. The linear subspace generated by an eigenvector is called an eigenline. For most (in fact all up to finitely many, in case the vector space has finite dimension) ${}\lambda$, the eigenspace is just the zero space.

In case of a homothety with factor ${}a$, every vector ${}v\neq 0$ is an eigenvector for the eigenvalue ${}a$. The eigenspace of the eigenvalue ${}a$ is the total space. Also, we can restrict an endomorphism to any eigenspace (as source and target), yielding the mapping

$\varphi {|}_{\operatorname {Eig} _{\lambda }{\left(\varphi \right)}}\colon \operatorname {Eig} _{\lambda }{\left(\varphi \right)}\longrightarrow \operatorname {Eig} _{\lambda }{\left(\varphi \right)}.$

This mapping is just the homothety with factor ${}\lambda$.

For matrices, we use the same concepts. If ${}\varphi \colon V\rightarrow V$ is a linear mapping, and ${}M$ is a describing matrix with respect to a basis, then for an eigenvalue ${}\lambda$ and an eigenvector ${}v\in V$ with corresponding coordinate tuple ${}{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}$ with respect to the basis, we have the relation

${}M{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}=\lambda {\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\,.$

Example

We consider the linear mapping

$\varphi \colon K^{n}\longrightarrow K^{n},e_{i}\longmapsto d_{i}e_{i},$

given by the diagonal matrix

${\begin{pmatrix}d_{1}&0&\cdots &\cdots &0\\0&d_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1}&0\\0&\cdots &\cdots &0&d_{n}\end{pmatrix}}.$

The diagonal entries ${}d_{i}$ are the eigenvalues of ${}\varphi$, and the ${}i$-th standard vector ${}e_{i}$ is a corresponding eigenvector. The eigenspaces are

${}{\begin{aligned}\operatorname {Eig} _{d}{\left(\varphi \right)}&={\left\{v\in K^{n}\mid v{\text{ is a linear combination of those }}e_{i},{\text{ for which }}d=d_{i}{\text{ holds}}\right\}}\\\end{aligned}}$

These spaces are not ${}0$ if and only if ${}d$ equals one of the diagonal entries. The dimension of the eigenspace ${}\operatorname {Eig} _{d}{\left(\varphi \right)}$ is given by the number how often the value ${}d$ occurs in the diagonal. The sum of all these dimension gives ${}n$.

Example

We consider the linear mapping

$\varphi \colon \mathbb {Q} ^{2}\longrightarrow \mathbb {Q} ^{2},{\begin{pmatrix}x\\y\end{pmatrix}}\longmapsto {\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}5y\\x\end{pmatrix}},$

given by the matrix

${}M={\begin{pmatrix}0&5\\1&0\end{pmatrix}}\,.$

The question whether this mapping has eigenvalues, leads to the question whether there exists some ${}\lambda \in \mathbb {Q}$, such that the equation

${}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}\,$

has a nontrivial solution ${}(x,y)\neq (0,0)$. For a given ${}\lambda$, this is a linear problem and can be solved with the elimination algorithm. However, the question whether there exist eigenvalues at all, leads, due to the variable "eigenvalue parameter“ ${}\lambda$, to a nonlinear problem. The system of equations above is

$5y=\lambda x{\text{ and }}x=\lambda y.$

For ${}y=0$, we get ${}x=0$, but the null vector is not an eigenvector. Hence, suppose that ${}y\neq 0$. Both equations combined yield the condition

${}5y=\lambda x=\lambda ^{2}y\,,$

hence ${}5=\lambda ^{2}$. But in ${}\mathbb {Q}$, the number ${}5$ does not have a square root, therefore there is no solution, and that means that ${}\varphi$ has no eigenvalues and no eigenvectors.

Now we consider the matrix ${}M$ as a real matrix, and look at the corresponding mapping

$\psi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2},{\begin{pmatrix}x\\y\end{pmatrix}}\longmapsto {\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}5y\\x\end{pmatrix}}.$

The same computations as above lead to the condition ${}5=\lambda ^{2}$, and within the real numbers, we have the two solutions

$\lambda _{1}={\sqrt {5}}\,\,{\text{ and }}\,\,\lambda _{2}=-{\sqrt {5}}.$

For both values, we have now to find the eigenvectors. First, we consider the case ${}\lambda ={\sqrt {5}}$, which yields the linear system

${}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\sqrt {5}}{\begin{pmatrix}x\\y\end{pmatrix}}\,.$

We write this as

${}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}{\sqrt {5}}&0\\0&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}\,$

and as

${}{\begin{pmatrix}{\sqrt {5}}&-5\\-1&{\sqrt {5}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}\,.$

This system can be solved easily, the solution space has dimension one, and

${}v={\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}\,$

is a basic solution.

For ${}\lambda =-{\sqrt {5}}$, we do the same steps, and the vector

${}w={\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}\,$

is a basic solution. Thus over ${}\mathbb {R}$, the numbers ${}{\sqrt {5}}$ and ${}-{\sqrt {5}}$ are eigenvalues, and the corresponding eigenspaces are

$\operatorname {Eig} _{\sqrt {5}}{\left(\psi \right)}={\left\{s{\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}\,\,{\text{ and }}\,\,\operatorname {Eig} _{-{\sqrt {5}}}{\left(\psi \right)}={\left\{s{\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}.$

Lemma

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and

$\varphi \colon V\longrightarrow V$

a

linear mapping. Then the following statements hold.

1. Every eigenspace

   $\operatorname {Eig} _{\lambda }{\left(\varphi \right)}$

   is a linear subspace of ${}V$.

2. ${}\lambda$ is an eigenvalue for ${}\varphi$, if and only if the eigenspace ${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}$ is not the null space.
3. A vector ${}v\in V,\,v\neq 0$, is an eigenvector for ${}\lambda$, if and only if ${}v\in \operatorname {Eig} _{\lambda }{\left(\varphi \right)}$.

Proof

(1). Let ${}u,v\in \operatorname {Eig} _{\lambda }{\left(\varphi \right)}$ and let ${}w=au+bv$. Then

${}\varphi (w)=a\varphi (u)+b\varphi (v)=a\lambda u+b\lambda v=\lambda (au+bv)=\lambda w\,.$

(2) and (3) follow directly from the definitions.

$\Box$

Kernel and fixed space

Lemma

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and

$\varphi \colon V\longrightarrow V$

a linear mapping. Then

${}\operatorname {ker} {\left(\varphi \right)}=\operatorname {Eig} _{0}{\left(\varphi \right)}\,.$

In particular, ${}0$ is an

eigenvalue of ${}\varphi$ if and only if ${}\varphi$ is not

injective.

Proof

See Exercise 21.9 .

$\Box$

Remark

Beside the eigenspace for ${}0\in K$, which is the kernel of the linear mapping, the eigenvalues ${}1$ and ${}-1$ are in particular interesting. The eigenspace for ${}1$ consists of all vectors which are sent to themselves. Restricted to this linear subspace, the mapping is just the identity, it is called the fixed space. The eigenspace for ${}-1$ consists in all vector which are sent to their negative. On this linear subspace, the mapping acts like the reflection at the origin.

Definition

Let ${}K$ be a field, ${}V$ a ${}K$-vector space and

$\varphi \colon V\longrightarrow V$

a linear mapping. The fixed space of ${}\varphi$ is the eigenspace for the eigenvalue

${}1$, that is, the set ${}{\left\{v\in V\mid \varphi (v)=v\right\}}$.

Eigenvalues under base change

Lemma

Let

$\varphi \colon V\longrightarrow V$

denote an endomorphism on a ${}K$-vector space ${}V$, and let

$f\colon V\longrightarrow W$

denote an isomorphism of ${}K$-vector spaces. Set

${}\psi =f\circ \varphi \circ f^{-1}\,.$

Then the following hold.

1. A vector ${}v\in V$ is an eigenvector of ${}\varphi$ for the eigenvalue ${}a\in K$ if and only if ${}f(v)$ is an eigenvector of ${}\psi$ for the eigenvalue ${}a$.
2. ${}\varphi$ and ${}\psi$ have the same eigenvalues.
3. The mapping ${}f$ induces for every ${}a\in K$ an isomorphism

   $f\colon \operatorname {Eig} _{a}{\left(\varphi \right)}\longrightarrow \operatorname {Eig} _{a}{\left(\psi \right)}.$

Proof

(1). Let ${}v\in V$ be an eigenvector of ${}\varphi$ for the eigenvalue ${}a$. Set

${}w:=f(v)\,.$

Then

${}\psi (w)=(f\circ \varphi \circ f^{-1})(f(v))=(f\circ \varphi )(v)=f(\varphi (v))=f(av)=af(v)=aw\,.$

The reverse statement holds in the same way. (2) and (3) follow directly from (1).

$\Box$

If we have an endomorphism on a finite-dimensional vector space, which is described by the matrix ${}M$ with respect to a given basis, then the eigenvalues and the eigenvectors correspond to each other. Then eigenvector of the matrix is the Coordinate tuple of the corresponding eigenvector with respect to the basis. The eigenvalues do not depend on the chosen basis, but the eigenvectors do.

Corollary

Let ${}\varphi \colon V\rightarrow V$ be an endomorphism on the finite-dimensional ${}K$-vector space ${}V$, and let ${}{\mathfrak {u}}=u_{1},\ldots ,u_{n}$ denote a basis of ${}V$. Let ${}M=M_{\mathfrak {u}}^{\mathfrak {u}}$ be the describing matrix of ${}\varphi$ with respect to the basis. Then ${}v\in V$ is an eigenvector of ${}\varphi$ for the eigenvalue ${}a$ if and only if the coordinate tuple

of ${}v$ with respect to the basis is an eigenvector of ${}M$ for the eigenvalue ${}a$.

Proof

This follows directly from Lemma 21.10   (1), using the diagram

${\begin{matrix}V&{\stackrel {\varphi }{\longrightarrow }}&V&\\\!\!\!\!\!\!\!\!\!\!\!\!\!\psi _{\mathfrak {u}}^{-1}\downarrow &&\downarrow \!\psi _{\mathfrak {u}}^{-1}\!\!\!\!\!&\\K^{n}&{\stackrel {M_{\mathfrak {u}}^{\mathfrak {u}}(\varphi )}{\longrightarrow }}&K^{n}\,.&\!\!\!\!\!\\\end{matrix}}$

$\Box$

Corollary

Let ${}M$ be an ${}n\times n$-matrix over a field ${}K$, and let ${}B$ denote an invertible ${}n\times n$-matrix. Let ${}a\in K$. Then an ${}n$-tuple ${}{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}$ is an eigenvector of ${}M$ for the eigenvalue ${}a$ if and only if

${}{\begin{pmatrix}x'_{1}\\\vdots \\x'_{n}\end{pmatrix}}=B{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\,$

is an eigenvector of the matrix ${}BMB^{-1}$ for the eigenvalue ${}a$. In particular, ${}M$ and ${}BMB^{-1}$

have the same eigenvalues.

Proof

This follows from Lemma 21.10 .

$\Box$

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