Let $K$ be a Field. Let $A\in Mat(n\times n,K)$ and $I:=\{f(x)\in K[x]\mid f(A)=0\}$
1)Show that it exists exactly one monic polynomial $m(x)\in$ $I$ of least possible degree.
2)Let $f(x)\in I$. Show that it exists a polynomial $g(x)\in K[x]$ with $f(x)=m(x)g(x)$
(1) $I$ is non-empty since $\operatorname{Mat}(n\times n, K)$ has dimension $n^2$ implies that $I,A,A^2,...,A^{n^2}$ must be linearly dependent. Dividing elements of $\operatorname{Mat}(n\times n, K)$ by their leading coefficients you get that there are monic polynomials in $I$.
Assume that $m_1(x)$ and $m_2(x)$ are monic of the same minimal degree such that $m_1(A)=m_2(A)=0$. Then $m_3(x)=m_1(x)-m_2(x)$ satisfies $m_3(A)=0$ and has smaller degree than $m_1$ and $m_2$. Dividing by its leading coefficient we can make $m_3$ monic too. This contradicts that the degree of $m_1,m_2$ was minimal.
(2) If $m\in I$ is the monic of minimal degree, then $f(x)=m(x)g(x)+r(x)$ with the degree of $r$ strictly less than the degree of $f$. Then $0=f(A)=m(A)g(A)+r(A)=r(A)$ implies that $r\in I$. Since $m$ was of minimal degree, then $r=0$.