# 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
$\varphi$
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.

## 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
$\lambda$

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

### 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 $\varphi$.
$\Box$