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.
 A: You need to show that every polynomial that kills $A$ kills $T$.  But then to show minimality, you need to show that every polynomial that fails to kill $A$ fails to kill $T$.
$$f(A)=0$$
$$f(T)B = \left(\sum_{k=0}^n c_k T^k \right)B = \sum_{k=0}^n c_k (T^k B).\tag1$$
$$
T^k(B) = T^{k-1}(T(B)) = T^{k-1}(AB) = T^{k-2} (A(AB)) = T^{k-2} (AA(B)) = \cdots.
$$
In other words, show by induction on $k$ that $T^k (B) = A^k B$ and then apply $(1)$.
That should suggest how to do the other part.
A: 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.
A: On checks immediately that $T^k(B)=A^k\cdot B$ for all $k$ and $B$, and so by linearity $P[T](B)=P[A]\cdot B$ for all polynomials$~P$.
Now if $P[A]=0$ then for all $B$ one has $P[T](B)=P[A]\cdot B=0\cdot B=0$ so every polynomial annihilating $A$ annihilates $T$. Conversely if $P[T](B)=0$ for all $B$ then taking $B=I$ gives $0=P[T](I)=P[A]\cdot I=P[A]$, so so every polynomial annihilating $T$ annihilates $A$. Thus their minimal polynomials are the same.
A: Let $T$ be a linear operator(or a matrix $A$ over the field $\mathbb{F}$) on a vector space  $V(\mathbb{F})$, then the minimal polynomial of $T$ (or of matrix $A$) is the monic generator of the P.I.D. $$\{p(x)\in\mathbb{F}[x]:p(T)=0\}$$ Now just show that $$\{p(x)\in\mathbb{F}[x]:p(T)=0\}=\{p(x)\in\mathbb{F}[x]:p(A)=0\}$$ So that they have same monic generators i.e. same minimal polynomials. 
