Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 22

Relations between eigenspaces

We have seen in Lemma 21.7 that the eigenspace to ${}0$ is the kernel of the endomorphism. More general, the following characterization holds.

Lemma

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

$\varphi \colon V\longrightarrow V$

a linear mapping. Let ${}\lambda \in K$. Then

${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\operatorname {ker} {\left(\lambda \cdot \operatorname {Id} _{V}-\varphi \right)}\,.$

Proof

Let ${}v\in V$. Then ${}v\in \operatorname {Eig} _{\lambda }{\left(\varphi \right)}$ if and only if ${}\varphi (v)=\lambda v$, and this is the case if and only if ${}\lambda v-\varphi (v)=0$ holds, which means ${}{\left(\lambda \cdot \operatorname {Id} _{V}-\varphi \right)}(v)=0$.

$\Box$

In particular, ${}\lambda \in K$ is an eigenvalue of ${}\varphi$ if and only if ${}\lambda \operatorname {Id} _{V}-\varphi$ is not injective. For a given ${}\lambda$, this property can be checked with the help of a linear system (or the determinant), and the eigenspace can be determined. However, it is not a linear problem to decide whether ${}\varphi$ has eigenvalues at all and how those can be determined. We will continue to study a linear mapping ${}\varphi \colon V\rightarrow V$ by considering the differences to homotheties ${}\lambda \operatorname {Id} _{V}-\varphi$ for various ${}\lambda$.

For an ${}n\times n$-matrix ${}M$, we have to determine the kernel of the matrix ${}\lambda E_{n}-M$. If, for example, we want t know whether the matrix ${}{\begin{pmatrix}4&7\\-2&5\end{pmatrix}}$ has the eigenvalue ${}3$, then

${}3E_{2}-{\begin{pmatrix}4&7\\-2&5\end{pmatrix}}=3{\begin{pmatrix}1&0\\0&1\end{pmatrix}}-{\begin{pmatrix}4&7\\-2&5\end{pmatrix}}={\begin{pmatrix}-1&-7\\2&-2\end{pmatrix}}\,$

shows that this is not the case.

Lemma

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

$\varphi \colon V\longrightarrow V$

a linear mapping. Let ${}\lambda _{1}\neq \lambda _{2}$ be elements in ${}K$. Then

${}\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\cap \operatorname {Eig} _{\lambda _{2}}{\left(\varphi \right)}=0\,.$

Proof

Let ${}v\in \operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\cap \operatorname {Eig} _{\lambda _{2}}{\left(\varphi \right)}$. Then

${}\lambda _{1}v=\varphi (v)=\lambda _{2}v\,.$

Therefore,

${}(\lambda _{1}-\lambda _{2})v=0\,,$

and this implies, because of ${}\lambda _{1}\neq \lambda _{2}$, ${}v=0$.

$\Box$

Lemma

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

$\varphi \colon V\longrightarrow V$

a linear mapping. Let ${}v_{1},\ldots ,v_{n}$ be eigenvectors for (pairwise) different eigenvalues ${}\lambda _{1},\ldots ,\lambda _{n}\in K$. Then ${}v_{1},\ldots ,v_{n}$ are

linearly independent.

Proof

We prove the statement by induction on ${}n$. For ${}n=0$, the statement is true. Suppose now that the statement is true for less than ${}n$ vectors. We consider a representation of ${}0$, say

${}a_{1}v_{1}+\cdots +a_{n}v_{n}=0\,.$

We apply ${}\varphi$ to this and get, on one hand,

${}a_{1}\varphi (v_{1})+\cdots +a_{n}\varphi (v_{n})=\lambda _{1}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.$

On the other hand, we multiply the equation with ${}\lambda _{n}$ and get

${}\lambda _{n}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.$

We look at the difference of the two equations, and get

${}(\lambda _{n}-\lambda _{1})a_{1}v_{1}+\cdots +(\lambda _{n}-\lambda _{n-1})a_{n-1}v_{n-1}=0\,.$

By the induction hypothesis, we get for the coefficients ${}(\lambda _{n}-\lambda _{i})a_{i}=0$, ${}i=1,\ldots ,n-1$. Because of ${}\lambda _{n}-\lambda _{i}\neq 0$, we get ${}a_{i}=0$ for ${}i=1,\ldots ,n-1$, and because of ${}v_{n}\neq 0$, we also get ${}a_{n}=0$.

$\Box$

Corollary

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

$\varphi \colon V\longrightarrow V$

a linear mapping. Then there exist at most ${}\dim _{K}{\left(V\right)}$ many eigenvalues

for ${}\varphi$.

Proof

See Exercise 22.3 .

$\Box$

In particular, an endomorphism on a finite-dimensional vector space has only finitely many eigenvalues.

Geometric multiplicity

The restriction of a linear mapping to an eigenspace is the homothety with the corresponding eigenvalue, thus a very simple linear mapping. For a 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 standard basis has the property that every basis vector is an eigenvector for the linear mapping given by the matrix. In this case, it is easy to describe the eigenspaces, see Example 21.4 ; the eigenspace for ${}d$ consists of all linear combinations of the standard vectors ${}e_{i}$, for which ${}d_{i}$ equals ${}d$. In particular, the dimension of the eigenspace equals the number how often ${}d$ occurs as an diagonal element. In general, the dimensions of the eigenspaces are important invariants of an endomorphism.

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 call

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

the geometric multiplicity

of ${}\lambda$.

In particular, a number ${}\lambda \in K$ is an eigenvalue of ${}\varphi$ if and only if its geometric multiplicity is at least ${}1$. It is easy to give examples where the geometric multiplicity of an eigenvalue is any number between ${}1$ and the dimension of the space.

Lemma

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

be a linear mapping. Then the sum of the eigenspaces ${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}$ is direct, and we have

${}\sum _{\lambda \in K}\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\leq \dim _{K}{\left(V\right)}\,.$

Proof

This follows directly from Lemma 22.3 .

$\Box$

Diagonalizability

Definition

Let ${}K$ denote a field, let ${}V$ denote a vector space, and let

$\varphi \colon V\longrightarrow V$

denote a linear mapping. Then ${}\varphi$ is called diagonalizable, if ${}V$ has a basis consisting of eigenvectors

for ${}\varphi$.

Theorem

Let ${}K$ denote a field, and let ${}V$ denote a finite-dimensional vector space. Let

$\varphi \colon V\longrightarrow V$

denote a

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is diagonalizable.
2. There exists a basis ${}{\mathfrak {v}}$ of ${}V$ such that the describing matrix ${}M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )$ is a diagonal matrix.
3. For every describing matrix ${}M=M_{\mathfrak {w}}^{\mathfrak {w}}(\varphi )$ with respect to a basis ${}{\mathfrak {w}}$, there exists an invertible matrix ${}B$ such that

   $BMB^{-1}$

   is a diagonal matrix.

Proof

The equivalence between (1) and (2) follows from the definition, from Example 21.4 , and the correspondence between linear mappings and matrices. The equivalence between (2) and (3) follows from Corollary 11.12 .

$\Box$

If ${}\varphi$ is diagonalizable and if the eigenvalues together with their geometric multiplicities are known, then we can simply write down a corresponding diagonal matrix: just take the diagonal matrix such that, in the diagonal, the eigenvalues occur as often as the geometric multiplicities. In particular, the corresponding diagonal matrix of a diagonalizable mapping is, up to the ordering of the diagonal entries, uniquely determined.

Example

We continue with Example 21.5 . There exists the two eigenvectors ${}{\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}$ and ${}{\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}$ for the different eigenvalues ${}{\sqrt {5}}$ and ${}-{\sqrt {5}}$, so that the mapping is diagonalizable, due to Corollary 22.10 . With respect to the basis ${}{\mathfrak {u}}$, consisting of these eigenvectors, the linear mapping is described by the diagonal matrix

${\begin{pmatrix}{\sqrt {5}}&0\\0&-{\sqrt {5}}\end{pmatrix}}.$

The transformation matrix, from the basis ${}{\mathfrak {u}}$ to the standard basis ${}{\mathfrak {v}}$, consisting of ${}e_{1}$ and ${}e_{2}$, is simply

${}M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\,.$

The inverse matrix is

${}{\frac {1}{2{\sqrt {5}}}}{\begin{pmatrix}1&{\sqrt {5}}\\-1&{\sqrt {5}}\end{pmatrix}}={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}\,.$

Because of Corollary 11.12 , we have the relation

${}{\begin{aligned}{\begin{pmatrix}{\sqrt {5}}&0\\0&-{\sqrt {5}}\end{pmatrix}}&={\begin{pmatrix}{\frac {1}{2}}&{\frac {\sqrt {5}}{2}}\\{\frac {1}{2}}&{\frac {-{\sqrt {5}}}{2}}\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}.\end{aligned}}$

Corollary

Let ${}K$ denote a field, and let ${}V$ denote a finite-dimensional vector space. Let

$\varphi \colon V\longrightarrow V$

denote a linear mapping. Suppose that there exists ${}n$ different eigenvalues. Then ${}\varphi$ is

diagonalizable.

Proof

Because of Lemma 22.3 , there exist ${}n$ linearly independent eigenvectors. These form, due to Corollary 8.10 , a basis.

$\Box$

Lemma

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

be a linear mapping. Then ${}\varphi$ is diagonalizable if and only if ${}V$ is the direct sum of the

eigenspaces.

Proof

If ${}\varphi$ is diagonalizable, then there exists a basis ${}v_{1},\ldots ,v_{n}$ of ${}V$ consisting of eigenvectors. Hence,

${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\langle v_{i},\,{\text{the eigenvalue of }}v_{i}{\text{ is }}\lambda \rangle \,.$

Therefore,

${}V=\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{k}}{\left(\varphi \right)}\,.$

That the sum is direct follows from Lemma 22.2 . If, the other way round,

${}V=\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{k}}{\left(\varphi \right)}\,$

holds, then we can choose in every eigenspace a basis. These bases consist of eigenvectors and yield together a basis of ${}V$.

$\Box$

Example

We consider ${}2\times 2$-shearing matrices

${\begin{pmatrix}1&a\\0&1\end{pmatrix}}$

with ${}a\in K$. The condition for some ${}\lambda \in K$ to be an eigenvalue means

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

This yields the equations

$x+ay=\lambda x{\text{ and }}y=\lambda y.$

For ${}\lambda \neq 1$, we get ${}y=0$ and hence also ${}x=0$, that is, only ${}1$ can be an eigenvalue. In this case, the second equation is fulfilled, and the first equation becomes

$x+ay=x{\text{ and }}ay=0.$

For ${}a\neq 0$, we get ${}y=0$ and thus ${}{\begin{pmatrix}x\\0\end{pmatrix}}$ is the eigenspace for the eigenvalue ${}1$, and ${}{\begin{pmatrix}1\\0\end{pmatrix}}$ is an eigenvector which spans this eigenspace. For ${}a=0$, we have the identity matrix, and the eigenspace for the eigenvalue ${}1$ is the total plane. For ${}a\neq 0$, there is a one-dimensional eigenspace, and the mapping is not diagonalizable.

The product of two diagonal matrices is again a diagonal matrix. The following example shows that the product of two diagonalizable matrices is not necessarily diagonalizable.

Example

Let ${}G_{1}$ and ${}G_{2}$ denote two lines in ${}\mathbb {R} ^{2}$ through the origin, and let ${}\varphi _{1}$ and ${}\varphi _{2}$ denote the reflections at these axes. A reflection at an axis is always diagonalizable, the axis and the line orthogonal to the axis are eigenlines (with eigenvalues ${}1$ and ${}-1$). The composition

${}\psi =\varphi _{2}\circ \varphi _{1}\,$

of the reflections is a plane rotation, the angle of rotation being twice the angle between the two lines. However, a rotation is only diagonalizable if the angle of rotation is ${}0$ or ${}180$ degree. If the angle between the axes is different from ${}0,90$ degree, then ${}\psi$ does not have any eigenvector.

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