# Endomorphism/Trigonalizable/Invariant linear subspaces/Section

A trigonalizable endomorphism is described, with respect to a suitable basis, by a matrix of the form
    $M={\begin{pmatrix}_{1}&\ast &\cdots &\cdots &\ast \\0&_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&_{n-1}&\ast \\0&\cdots &\cdots &0&_{n}\end{pmatrix}}\,.$
A property which hold for such an upper triangular matrix and which can be described as a property of the linear mapping (being independent of a chosen basis), must hold for any trigonalizable mapping. We want to understand such properties. By an upper triangular matrix, the $j$-th standard vector $e_{j}$ is sent to
    $Me_{j}=a_{1j}e_{1}+\cdots +a_{jj}e_{j}\,.$
In particular, $e_{1}$ is a eigenvector with the eigenvalue $a_{11}$. It is typical for a trigonalizable mapping that the linear subspace
    $V_{j}=\langle e_{1},\ldots ,e_{j}\rangle \,$
is mapped by $M$ into itself. That is, the $V_{j}$ are $M$-invariant linear subspaces, which are contained in each other and those dimensions are $j$. We will show, after some preparations, that these properties characterize trigonalizable mappings.

## Lemma

Let $K$ denote a field, and let $V$ be a $n$-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping and let $\lambda \in K$ be an eigenvalue of $\varphi$. Then there exists a $\varphi$-invariant linear subspace $U\subset V$
$n-1$

### Proof

Because of the condition and
fact, the mapping $\varphi -\lambda \operatorname {Id} _{V}$ has a non-trival kernel. Hence, this mapping is not injective and, due to fact, also not surjective. Therefore,
    $B:=\operatorname {Im} {\left(\varphi -\lambda \operatorname {Id} _{V}\right)}\subset V\,$
is a strict linear subspace of $V$. It follows that there exists also a linear subspace $U\subset V$ of dimension $n-1$, which contains $B$. For $u\in U$, we have
    $\varphi (u)=\lambda u+(\varphi -\lambda \operatorname {Id} _{V})u\in U+B\subseteq U\,.$
Hence the image of $U$ belongs to $U$, that is, $U$ is $\varphi$-invariant.
$\Box$
When $U\subseteq V$ is an $\varphi$-invariant linear subspace, and $P\in K[X]$ is a polynomial, then $U$ is also $P(\varphi )$-invariant, see exercise. In this situation, the following identity holds.

## Lemma

Let $K$ be a field, $V$ a $K$-vector space and
    $\varphi \colon V\longrightarrow V$
a linear mapping. Let $U\subseteq V$ be an $\varphi$-invariant linear subspace. Then, for every polynomial $P\in K[X]$, the relation
    $P{\left(\varphi {|}_{U}\right)}={\left(P(\varphi )\right)}{|}_{U}\,$
holds, where here $\varphi {|}_{U}$ denotes the restricted mapping

### Proof

This can be checked directly for the powers $X^{n}$ and for linear combinations of powers.
$\Box$

## Corollary

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. Let $U\subseteq V$ be a $\varphi$-invariant linear subspace and
    $\varphi {|}_{U}\colon U\longrightarrow U$
the restriction to $U$ (also in the target). Then the minimal polynomial
$\varphi$
$\varphi {|}_{U}$

### Proof

Let $\mu$ be the minimal polynomial of $\varphi$. For $u\in U$, we have
    $\mu {\left(\varphi {|}_{U}\right)}(u)=\mu (\varphi )(u)=0\,,$
due to fact. Therefore, $\mu$ annihilates the restricted endomorphism $\varphi {|}_{U}$, and so $\mu$ is a multiple of the minimal polynomial of $\varphi {|}_{U}$.
$\Box$

## Example

We consider the permutation matrix
    ${\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}.$
The line $K{\begin{pmatrix}1\\1\\1\end{pmatrix}}$ is the eigenspace for the eigenvalue $1$. Moreover,
    $U=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,$
is an invariant linear subspace (which, over $\mathbb {C}$, according to fact, can be decomposed further into smaller eigenspaces). With respect to the given basis, the restriction of the linear mapping to $U$ has the describing matrix
    ${\begin{pmatrix}0&-1\\1&-1\end{pmatrix}}.$
Therefore, the characteristic polynomial of this matrix is
    $X(X+1)+1=X^{2}+X+1\,.$
This is also the minimal polynomial of the restriction. The minimal polynomial of the permutation matrix is $X^{3}-1$, and indeed we have
    $X^{3}-1=(X-1)(X^{2}+X+1)\,$
in accordance with fact.