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 $A$.
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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.
