Endomorphism/Generalized eigenspace/Introduction/Section

Definition

For a linear mapping ${}\varphi$ on a ${}K$-vector space ${}V$ and an eigenvalue ${}\lambda \in K$,

${}\operatorname {GeEig} _{\lambda }(\varphi )=\bigcup _{n\in \mathbb {N} }\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{n}\,$

is called generalized eigenspace

of ${}\varphi$ for this eigenvalue.

If ${}V$ is finite-dimensional, then the chain

${}\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}\subseteq \operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{2}\subseteq \operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{3}\subseteq ...\,$

becomes stationary, that is, there exists some ${}r\in \mathbb {N}$ such that

${}\operatorname {GeEig} _{\lambda }(\varphi )=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{r}\,.$

Generalized eigenspaces are, due to exercise, invariant under the linear mapping. By definition, we have

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

and for ${}\varphi$ diagonalizable we have equality, see exercise. We want to understand trigonalizable mappings via their generalized eigenspaces.

Lemma

Let ${}\varphi \colon V\rightarrow V$ be a linear mapping on a finite-dimensional ${}K$-vector space ${}V$, and let

${}\chi _{\varphi }=P\cdot Q\,$

be a factorization of the characteristic polynomial in coprime polynomials ${}P,Q\in K[X]$. Then we have the direct sum decomposition

${}V=\operatorname {kern} P(\varphi )\oplus \operatorname {kern} Q(\varphi )\,,$

where these linear subspaces are ${}\varphi$-invariant.

The restriction of ${}P(\varphi )$ onto ${}\operatorname {kern} Q(\varphi )$ is bijective.

Proof

Due to the Lemma of Bezout, there exist polynomials ${}S,T\in K[T]$ such that

${}SP+TQ=1\,.$

Set ${}U=\operatorname {kern} P(\varphi )$ and ${}W=\operatorname {kern} Q(\varphi )$. Let ${}v\in V$. Due to the Theorem of Cayley-Hamilton , we have

${}0=\chi _{\varphi }(\varphi )=(P(\varphi )\circ Q(\varphi ))(v)=P(\varphi )(Q(\varphi )(v))\,.$

Therefore, the image of ${}Q(\varphi )$ belongs to the kernel of ${}P(\varphi )$ and vice versa. From

${}{\begin{aligned}v&=\operatorname {Id} _{V}(v)\\&=(SP+TQ)(\varphi )(v)\\&=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))\\&=P(\varphi )(S(\varphi )(v))+Q(\varphi )(T(\varphi )(v))\end{aligned}}$

we can read off that the left-hand summand belongs to ${}\operatorname {Im} P(\varphi )\subseteq \operatorname {kern} Q(\varphi )$ and the right-hand summand belongs to ${}\operatorname {Im} Q(\varphi )\subseteq \operatorname {kern} P(\varphi )$. Therefore, we have a sum decomposition, which is direct, since ${}P(\varphi )(v)=Q(\varphi )(v)=0$ implies ${}v=0$. For the ${}\varphi$-invariance of these spaces, see exercise. For ${}v\in \operatorname {kern} Q(\varphi )$, we have

${}v=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))=S(\varphi )(P(\varphi )(v))=P(\varphi )(S(\varphi )(v))\,,$

that is, we have ${}\operatorname {Im} P(\varphi )=\operatorname {kern} Q(\varphi )$. Therefore, the restriction of ${}P(\varphi )$ to the kernel of ${}Q(\varphi )$ is surjective, thus bijective.

$\Box$

Example

We consider the permutation matrix

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

over ${}\mathbb {R}$, the characteristic polynomial is

${}\chi _{M}=X^{3}-1=(X-1){\left(X^{2}+X+1\right)}=P\cdot Q\,,$

where the two factors are coprime. We want to check fact in this example. We have

${}P(M)=M-E_{3}={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}\,$

with

${}\operatorname {kern} P(M)=\operatorname {Eig} _{1}{\left(M\right)}=\mathbb {R} {\begin{pmatrix}1\\1\\1\end{pmatrix}}\,,$

and

${}Q(M)={\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}}+{\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}+{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}={\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}\,$

with

${}\operatorname {kern} Q(M)=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,.$

We have

${}\mathbb {R} ^{3}=\mathbb {R} {\begin{pmatrix}1\\1\\1\end{pmatrix}}\oplus \langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,.$

Moreover, we have

${}{\begin{aligned}P(M){\begin{pmatrix}1\\-1\\0\end{pmatrix}}&={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}{\begin{pmatrix}1\\-1\\0\end{pmatrix}}\\&={\begin{pmatrix}-1\\2\\-1\end{pmatrix}}\\&=-{\begin{pmatrix}1\\-1\\0\end{pmatrix}}+{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\end{aligned}}$

and

${}{\begin{aligned}P(M){\begin{pmatrix}0\\1\\-1\end{pmatrix}}&={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\\&={\begin{pmatrix}-1\\-1\\2\end{pmatrix}}\\&=-{\begin{pmatrix}1\\-1\\0\end{pmatrix}}-2{\begin{pmatrix}0\\1\\-1\end{pmatrix}}.\end{aligned}}$

From this, we can read off that the restriction of ${}P(M)$ to ${}\operatorname {kern} Q(M)$ is bijective. The representation of the ${}1$ from example yields the matrix equation

${}{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}={\frac {1}{3}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}-{\frac {1}{3}}{\begin{pmatrix}2&0&1\\1&2&0\\0&1&2\end{pmatrix}}{\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}\,.$

Theorem

Let

$\varphi \colon V\longrightarrow V$

be an endomorphism on the finite-dimensional ${}K$-vector space ${}V$, and let ${}\lambda \in K$. Then the dimension of the generalized eigenspace ${}\operatorname {GeEig} _{\lambda }(\varphi )$ equals the algebraic multiplicity

of ${}\lambda$.

Proof

We write the characteristic polynomial of ${}\varphi$ as

${}\chi _{\varphi }=(X-\lambda )^{k}Q\,,$

where ${}(X-\lambda )$ does not occur in ${}Q$ as a linear factor, that is, ${}k$ is the algebraic multiplicity of ${}\lambda$. Then, ${}P=(X-\lambda )^{k}$ and ${}Q$ are coprime, and, due to fact, we have the decompositon

${}V=\operatorname {kern} P(\varphi )\oplus \operatorname {kern} Q(\varphi )\,,$

and

$P(\varphi )={\left(\varphi -\lambda \operatorname {Id} \right)}^{k}\colon \operatorname {kern} Q(\varphi )\longrightarrow \operatorname {kern} Q(\varphi )$

is a bijection. Moreover,

${}H:=\operatorname {GeEig} _{\lambda }(\varphi )=\operatorname {kern} P(\varphi )\,,$

where the inclusion ${}\supseteq$ is clear, and the other inclusion follows from the fact that higher powerse of ${}X-\lambda$ do not annihilate further elements, by the bijectivity on ${}\operatorname {kern} Q(\varphi )$ just mentioned. For the characteristic polynomial, we have, due to the direct sum decomposition according to fact, the relation

${}\chi _{\varphi }=\chi _{1}\cdot \chi _{2}\,,$

where ${}\chi _{1}$ is the characteristic polynomial of ${}\varphi {|}_{H}$ and ${}\chi _{2}$ is the characteristic polynomial of ${}\varphi {|}_{\operatorname {kern} Q(\varphi )}$. Since ${}(\varphi -\lambda )^{k}$ restricted to ${}H$ is the zero mapping, the minimal polynomial of ${}\varphi {|}_{H}$ and, hence, also the characteristic polynomial ${}\chi _{1}$ are some power of ${}(X-\lambda )$, say

${}\chi _{1}=(X-\lambda )^{d}\,,$

where

${}d=\dim _{K}{\left(H\right)}\,.$

In particular, ${}d\leq k$, as ${}\chi _{1}$ is a divisor of ${}\chi _{\varphi }$. Assume that ${}d<k$. Then ${}\lambda$ is a zero of ${}\chi _{2}$ and ${}\lambda$ is an eigenvalue of ${}\varphi {|}_{\operatorname {kern} Q(\varphi )}$. But this is a contradiction to the fact that ${}P(\varphi )$ is a bijection on this space.

$\Box$

Corollary

For a linear mapping ${}\varphi \colon V\rightarrow V$ on a finite-dimensional ${}K$-vector space ${}V$ and two eigenvalues ${}\lambda \neq \delta$, the corresponding generalized eigenspace have trivial intersection, that is

${}\operatorname {GeEig} _{\lambda }(\varphi )\cap \operatorname {GeEig} _{\delta }(\varphi )=0\,.$

Proof

The characteristic polynomial of ${}\varphi$ has the form

${}\chi _{\varphi }=(X-\lambda )^{k}(X-\delta )^{\ell }F\,,$

where neither ${}\lambda$ nor ${}\delta$ is a zero of ${}F$. Because of fact, applied to ${}Q=(X-\delta )^{\ell }F$, we have

${}\operatorname {GeEig} _{\lambda }(\varphi )\cap \operatorname {kern} Q(\varphi )=0\,.$

Because of ${}\operatorname {GeEig} _{\delta }(\varphi )\subseteq \operatorname {kern} Q(\varphi )$, this implies immediately

${}\operatorname {GeEig} _{\lambda }(\varphi )\cap \operatorname {GeEig} _{\delta }(\varphi )=0\,.$

$\Box$

Theorem

Let

$\varphi \colon V\longrightarrow V$

be a trigonalizable ${}K$-endomorphism on the finite-dimensional ${}K$-vector space ${}V$. Then ${}V$ is the direct sum of the generalized eigenspaces, that is,

${}V=\operatorname {GeEig} _{\lambda _{1}}(\varphi )\oplus \cdots \oplus \operatorname {GeEig} _{\lambda _{m}}(\varphi )\,,$

where ${}\lambda _{1},\ldots ,\lambda _{m}$ are the different eigenvalues of ${}\varphi$, and ${}\varphi$ is the direct sum of the restrictions

$\varphi _{i}=\varphi {|}_{H_{i}}\colon H_{i}\longrightarrow H_{i}$

on the generalized eigenspaces.

Proof

Let

${}\chi _{\varphi }=(X-\lambda _{1})^{k_{1}}\cdots (X-\lambda _{m})^{k_{m}}\,$

be the characteristic polynomial, which splits into linear factors according to fact, where the ${}\lambda _{i}$ are different. We do induction over ${}m$. For ${}m=1$, there is only one eigenvalue ${}\lambda$ and only one generalized eigenspace. Due to fact, the minimal polynomial is of the form ${}(X-\lambda )^{s}$ and thus ${}V=\operatorname {GeEig} _{\lambda }(\varphi )$. Suppose that the statement is already proven for smaller ${}m$. We set ${}P=(X-\lambda _{1})^{k_{1}}$ and ${}Q=(X-\lambda _{2})^{k_{2}}\cdots (X-\lambda _{m})^{k_{m}}$. We are then in the situation of fact and fact. Therefore, we have a direct sum decomposition into ${}\varphi$-invariant linear subspaces

${}V=\operatorname {GeEig} _{\lambda _{1}}(\varphi )\oplus \operatorname {kern} Q(\varphi )\,.$

The characteristic polynomial is, according to fact, the product of the characteristic polynomials of the restrictions to the spaces. Because of fact, the polynomial ${}(X-\lambda _{1})^{k_{1}}$ is the characteristic polynomial of the restriction to the first generalized eigenspace, hence, ${}Q$ is the characteristic polynomial of the restriction to ${}\operatorname {kern} Q(P)$. In particular, this restriction is also trigonalizable. By the induction hypothesis, ${}\operatorname {kern} Q(P)$ is the direct sum of the generalized eigenspaces for ${}\lambda _{2},\ldots ,\lambda _{m}$. Altogether, this implies the direct sum decomposition of ${}V$ and of ${}\varphi$.

$\Box$