Minimal polynomial Let $V$ be the vector space of square matrices of order $n$ over the field $F$. Let $A$ be  a fixed square matrix of $n$ and let $T$ be a linear operator on $V$ such that $T(B) = AB$. Show that the minimal polynomial for $T$ is the minimal polynomial for $A$.
Thank you for your time.

This content  is actually supposed to be a "Proof verification Question".I thought of posting this as a question but then i realized there is a page having same question. So, I thought to make this as a "partial answer" and ask for help instead of getting a "duplicate question tag"
I have $V$- vector space of all $n\times n$ matrices over a field $F$ and let $A$ be a fixed matrix in $V$.
Define $T : V\rightarrow V$ as $T(B)=AB$
Question is to prove that minimal polynomial of $T$ and $A$ are same.
Suppose $f(x)=\sum_{i=0}^n a_i x^i $ be characteristic polynomial for $T$
i.e., $\sum_{i=0}^n a_i T^i=0$ i.e., $\sum_{i=0}^n a_i T^i(B)=0$ for all $B\in V$
in particular, $\sum_{i=0}^n a_i T^i(I)=0$
Now, $T(I)=A$ and $T^2(I)=T(T(I))=T(A)=AA=A^2$ for similar reasons, $T^n(I)=A^n$
So, we have $\sum_{i=0}^n a_i T^i(I)=0$ implies $\sum_{i=0}^n a_i A^i=0$.
So, $f(x)$ is characteristic polynomial for $A$
Suppose $f(x)=\sum_{i=0}^n a_i x^i $ be characteristic polynomial for $A$
i.e., $\sum_{i=0}^n a_i A^i=0$ but, $T^i(I)=A^i$ So, $\sum_{i=0}^n a_i T^i(I)=0$ i..e, $(\sum_{i=0}^n a_i T^i)(I)=0$
multiplying by arbitrary $B\in V$ we get $(\sum_{i=0}^n a_i T^i)(I)(B)=0.B=0$ i.e., $(\sum_{i=0}^n a_i T^i)(B)=0$ for all $B\in V$ Thus, $f(x)$ is characteristic polynomial for $T$
So, we concluded that characteristic polynomials of $T$ and $A$ are same.
I did not understand how to proceed to show its minimal polynomials are same.
