# Counterexample involving the minimal polynomial of a linear operator

Let $F$ be a field and let $V = F^{4\times4}$ be the vector space of $4x4$ matrices over $F$. For $A \in F^{4\times4}$, define $T_A : V \rightarrow V$ by $T_A(B) = AB$ for each $B \in V$.

Question: True or false "The minimal polynomial of $T_A$ is never equal to the characteristic polynomial of $T_A$"

I know how to prove that the minimal polynomial of $T_A$ is equal to the minimal polynomial of $A$. I was thinking I could pick a suitable diagonal matrix $A$ to make the question false but after testing a couple of examples I am failing to come up with the correct diagonal matrix.

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I know how to prove that the minimal polynomial of $T_A$ is equal to the minimal polynomial of $A$.
Good, then you are almost done. The minimal polynomial of $A$ has degree at most [blank], while the characteristic polynomial of $T_A$ has degree equal to [blank]. You can fill in the blanks by looking at the dimensions of the spaces which $A$ and $T_A$ act on.
Thanks for your help. I do not understand your last statement about considering the dimension of the spaces that $A$ and $T_A$ act on could you please clarify. –  user7980 Aug 8 '11 at 7:50
@user7980: If you have an $n$-by-$n$ matrix, then you can think of it as a linear transformation on $F^n$. You know the degree of the characteristic polynomial of an $n$-by-$n$ matrix? By the Cayley-Hamilton theorem, the minimal polynomial has equal or lesser degree. You can apply this to $A$. Now $T_A$ is not explicitly defined as a matrix, but as a linear tranformation on a given vector space $F^{4\times 4}$. What's the degree of the char. polynomial of a linear transformation acting on a space of this size? (It may help to think of the size of the corresponding matrix of $T_A$) –  Jonas Meyer Aug 8 '11 at 8:01