Find the minimal polynomial of $T(B)=B^T$ 
Let $T:M_3(F)\rightarrow M_3(F)$ be a linear transformation defined by $T(B)=B^T$.
Find the minimal polynomial of $T$.

My try:
$T(B)=B^T\implies T^2(B)=B \implies T^3(B)=B^T\implies T^3=T \implies T^3-T=0$
Thus, the minimal polynomial is: $x(x^2-1)=x(x-1)(x+1)$
Since $\ker T=\{0\},$ one can conclude that $T$ is invertible, which means $0$ is not an eigenvalue of $T$.
But my result for the minimal polynomial shows that $0$ is an eigenvalue.
Could someone please point at my mistake?

You have found one polynomial, $p(x)=x^3-x$, such that $p(T)=0$. Thus, the minimal polynomial $\mu_T(x)$ divides it.
Notice, however, that you had already found another polynomial $q$ of degree $2$ such that $q(T)=0$ when you wrote $T^2(B)=B$; it was $q(x)=x^2-1$.
Now, you must prove that no divisor $\widetilde q$ of $x^2-1$ satisfies $\widetilde q(T)=0$. This is, however, rather easy.
