Minimal polynomial and characteristic polynomial/Same zeroes/Fact/Proof

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 ${}i$.

To fact