Find Minimal Polynomial of $T_p = P^{-1}AP$ I got this problem: The transformation $T_p:M_{2\times 2}^{\mathbb{C}} \to M_{2\times 2}^{\mathbb{C}}$ is defined by: $T_p(X)=P^{-1}XP$, for any $X \in M_{2\times 2}^{\mathbb{C}}$.
$P=\begin{pmatrix}
i & 2\\ 
-1 & -i
\end{pmatrix}$
Find the minimal polynomial of $T_p$
So the minimal polynomial of $P$ is $\lambda^2+3$, but I don't really understand how to apply it to $T_p$. $P$ is the change of basis matrix for $X$, and the minimal polynomial of $T_p$ appears to depend on $X$.
For example:


*

*If $X = I$ then $T_p(X) = I$, which makes $m_{T_p} = \lambda-1$.

*If $X = P$ then $T_p(X) = P$, which makes $m_{T_p} = \lambda^2+3$.


How can I even begin to find the minimal polynomial of $T_p$ without knowing anything about $X$?
 A: There is an interpretation when viewing the map $T_p$ as a map from the 4 dimensional vector space of 2 by 2 matrices to itself. In that way $T_p$ corresponds to a 4 by 4 matrix which may be calculated. The characteristic polynomial of that is then of degree 4. Representing a 2 by 2 matrix 
$$ X=\left( \begin{matrix} x_1 & x_2 \\ x_3 & x_4 \end{matrix} \right)$$
as a vector 
$$ \vec{X}= \left( \begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix} \right)$$
One finds that $\vec{T_p({X})}=M\vec{X}$ with a matrix (after a lengthy calculation):
$$ M=\frac13 \left( \begin{matrix} 1 & i & -2i & 2\\ -2i & -1 & -4 & 2i\\ 
  i & -1 & -1 & -i \\ 2 & -i & 2i & 1 \end{matrix} \right) $$
The characteristic polynomial is $\lambda^4 -2\lambda^2+1$ and the minimal is $\lambda^2 -1$. This (miraculously) is also explained by $P^2=-3I$ which  yields $T_p(T_p(X)) = X$. To see this note that
$$ T_p(T_p(X)) = P^{-1}( P^{-1} X P) P = (-\frac13 I) X  (-3I) = X$$
In 4D space this corresponds to precisely a (minimal) polynomial of $\lambda^2-1$. In retrospect one could have tried to see that form the start, which might have been the intention of the problem setter...
