Endomorphism/Minimal polynomial/Characteristic polynomial/Section

Corollary

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

$f\colon V\longrightarrow V$

be a linear mapping. Then the characteristic polynomial ${}\chi _{f}$ is a multiple of the minimal polynomial

${}\mu _{f}$ of ${}f$.

Proof

This follows directly from fact and fact.

$\Box$

In particular, the degree of the minimal polynomial of ${}\varphi \colon V\rightarrow V$ is bounded by the dimension of the vector space ${}V$. The minimal polynomial and the characteristic polynomial are related in several respects, for example, they have the same zeroes.

Lemma

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

$f\colon V\longrightarrow V$

be a linear mapping. Let ${}v\in V$ be an eigenvector of ${}f$ with eigenvalue ${}\lambda$, and let ${}P\in K[X]$ denote a polynomial. Then

${}(P(f))(v)=P(\lambda )v\,.$

In particular, ${}v$ is an eigenvector of ${}P(f)$ with eigenvalue ${}P(\lambda )$. The vector

${}v\neq 0$

belongs to the kernel of ${}P(f)$ if and only if ${}\lambda$ is a zero of ${}P$.

Proof

We have

${}(f^{k})(v)=\lambda ^{k}v\,.$

This implies the statement, since the assignment ${}P\mapsto P(f)$ is compatible with addition and scalar multiplication.

$\Box$

Lemma

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

$f\colon V\longrightarrow V$

be a linear mapping. Then the characteristic polynomial ${}\chi _{f}$ and the minimal polynomial

${}\mu _{f}$ have the same zeroes.

Proof

It follows directly from Cayley-Hamilton that the zeroes of the minimal polynomial are also zeroes of the characteristic polynomial.

To prove the other implication, let ${}\lambda \in K$ be a zero of the characteristic polynomial, and let ${}v\in V$ denote an eigenvector of ${}f$ with eigenvalue ${}\lambda$, its existence is guaranteed by fact. We write the minimal polynomial as

${}\mu _{f}=(X-\lambda _{1})^{m_{1}}\cdots (X-\lambda _{k})^{m_{k}}Q\,,$

where ${}Q$ has no zero. Then

${}{\begin{aligned}0&=\mu _{f}(f)\\&={\left((X-\lambda _{1})^{m_{1}}\cdots (X-\lambda _{k})^{m_{k}}Q\right)}(f)\\&=(f-\lambda _{1}\operatorname {Id} _{V})^{m_{1}}\cdots (f-\lambda _{k}\operatorname {Id} _{V})^{m_{k}}Q(f).\end{aligned}}$

We apply this mapping to ${}v$. Because of fact, the factors send the vector ${}v$ to ${}(\lambda -\lambda _{i})^{m_{i}}v$ or to ${}Q(\lambda )v$, respectively. Altogether, ${}v$ is sent to

$(\lambda -\lambda _{1})^{m_{1}}\cdots (\lambda -\lambda _{k})^{m_{k}}Q(\lambda )v.$

As the composed mapping is the zero mapping and ${}Q(\lambda )\neq 0$, we must have ${}\lambda _{i}=\lambda$ for some ${}i$.

$\Box$