How to prove that characteristic polynomial divides the $n$-th power of the minimal polynomial?

Question

Let $M \in\textrm{Mat}_{n\times n}(\mathbb{C})$ be a matrix with complex coefficients, $\textrm{char}_M(X)$ its characteristic polynomial and $m_M(X)$ its minimal polynomial.

How do I prove that char$_M(X)$ divides $m_M(X)^n$ using that in $\mathbb{C}$, every monic polynomial of degree $d$ factors as $\prod_{i=1}^d(X-a_i)$?

Answer 1

Hint: You should use the fact that if $\lambda$ is an eigenvalue of $M$, then $(X - \lambda)$ divides $m_M(X)$. In other words, every linear factor of $\operatorname{char}_M$ divides $m_M$.

Answer 2

Since every polynomial splits, this is evident given that all the zeros of the characteristic polynomial are zeros of the minimal polynomial. Note that the algebraic multiplicity of a zero of the characteristic polynomial can't exceed $n$. So the $n$-th power of the minimal polynomial has all the factors of the characteristic polynomial in powers of greater than or equal order.