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 $\prod_{i=1}^d(X-a_i)$?
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 $m_M$.
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.
Denote by $p(x)$ the characteristic polynomial of $M\in C^{n\times n}$. If $f(x)$ is a polynomial different from the zero polynomial such that $f(M)=0$ then $p(x)\mid f^n(x)$.
Proof:
${\rm deg} \ p(x)=n$ and ${\rm deg} \ f^n(x)\geq n$. Note that ${\rm deg}\ f(x)\geq 1$, because if it was order $0$, then $f(x)$ would be a constant and hence $f(M)\neq 0$, and $f(x)\neq 0$, as given. Hence ${\rm deg}\ f^n(x)\geq {\rm deg}\ p(x)$, and accordingly $f^n(x)$ can be divided by $p(x)$ using the division theorem:
$$f^n(x)=q(x)p(x)+r(x)$$
Let $m(x)$ be the minimal polynomial of $M$. Then $m(x)\mid p(x)$, or
$$p(x)=q_1(x)m(x)$$.
Therefore
$$f^n(x)=q(x)q_1(x)m(x)+r(x)$$
Say that $r(x)\neq 0$, i.e. say that $r(x)$ is not the zero polynomial. Then we obtain that $m(x)\nmid f^n(x)$. This is a contradiction because any polynomial annihilated by $M$ is divided by the minimal polynomial. Hence $r(x)=0$, $\therefore$
$$f^n(x)=q(x)p(x)$$
Hence $p(x)\mid f^n(x)$.
