# Minimal polynomial for diagonalizable operators

Let $V$ be the vector space of $n\times n$ matrices over the field $F$. Let $A$ be a fixed $n\times n$ matrix. Let $T$ be the linear operator on $V$ defined by $T(B)= AB$. How to show that the minimal polynomial for $T$ is the minimal polynomial for $A$? I have no idea!

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Note that $T^i(B) = A^iB$. If $f(t)=a_0+a_1t+\cdots+a_mt^m$ is the minimal polynomial of $A$, then $f(A) = a_0I + a_1A+\cdots + a_mA^m = \mathbf{0}$.
Now, $f(t)$ is the minimal polynomial of $T$ if and only if $f(T)$ is the zero transformation on $V$, and for every $g(t)$, if $g(T)$ is the zero transformation, then $f(t)|g(t)$.
For the first, what is $f(T)$? Well, if we evaluate it at $B$ we get \begin{align*} f(T)(B) &= \Bigl( a_0\mathrm{Id} + a_1T+\cdots a_mT^m\Bigr)(B)\\ &= a_0B + a_1T(B) +\cdots + a_mT^m(B)\\ &= a_0B + a_1AB + \cdots + a_mA^mB\\ &= (a_0I + a_1A+\cdots + a_mA^m)(B). \end{align*} Do we know how much that is?
Now if $g(T)$ is the zero linear transformation, then in particular $g(T)(I_n)=\mathbf{0}$. What does that tell you about $g(A)$? What does that tell you about the relationship between $f(t)$ and $g(t)$?