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 ${}U\subset V$

of dimension ${}n-1$.

Proof

Because of the condition and fact, the mapping ${}\varphi -\lambda \operatorname {Id} _{V}$ has a nontrival 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

(with respect to range and target).

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

of ${}\varphi$ is a multiple of the minimal polynomial of ${}\varphi {|}_{U}$.

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.