Finding the minimal polynomial of a linear operator Let $P=\begin{pmatrix}
i & 2\\ 
-1 & -i
\end{pmatrix}$ and $T_P\colon M_{2\times 2}^{\mathbb{C}} \to M_{2\times 2}^{\mathbb{C}}$ a linear map defined by $T_P(X)=P^{-1}XP$. I need to find the minimal polynomial of $T_P$. 
I was able to find the minimal polynomial using the matrix $[T_P]_E$ (which represents $T_P$ in the standard basis $E$) and calculating its chracteristic polynomial (the minimal polynomial is $(x-1)(x+1)$), but according to a hint which I was given - there's no need to find $[T_P]_E$. It appears that I must use somehow the fact that $P^2+3I=0$ but I don't know how. I noticed that $T_P(P)=P$ which means $\lambda=1$ is an eigenvalue (and thus the term $(x-1)$ must appear in the minimal polynomial) and that's it. But how can I deduce that $\lambda=-1$ is an eigenvalue as well (without actually plugging in different matrices and hoping to get the desired eigenvector)? Also, how can I ensure that $1,(-1)$ are the only eigenvalues of $T_P$ (again, with minimal computational effort)? Any suggestions?
 A: The minimal polynomial divides any polynomial that anahilates the matrix. 
This means that if you already recognized such a polynomial you only have a few options for the minimal polynomial, since in your case it's easy to factor that specific polynomial
A: It's given as a hint that $P^2+3I = 0$, so:
$$
\begin{align}
&P^2 = -3I \\
&P\cdot-\frac{1}{3}P = I \\
\therefore \quad &P^{-1} = -\frac{1}{3}P
\end{align}
$$
So, $T_P$ is actually
$$
T_P(X)=-\frac{1}{3}PXP
$$
Note that $P^2=-3I$ and we have two $P$s in the above expression for $T_P$, so it might be useful to calculate $T_P^2$:
$$
\begin{align}
T_P^2(X) &= T_P(T_P(X)) \\
&= T_P(-\frac{1}{3}PXP) \\
&= -\frac{1}{3}P(-\frac{1}{3}PXP)P =\\
&= \frac{1}{9}P^2XP^2 \\
&= \frac{1}{9}(-3I)X(-3I) \\
&= \frac{1}{9}\cdot 9\cdot IX =X \\
\end{align}
$$
So we got $T^2_P=I$, which means $T^2_P-I=O$, so the following polynomial satisfies $m(T_P)=0$:
$$
m(t)=t^2-1=(t+1)(t-1)
$$
Since $T_P$ is not scalar, it's trivial to prove that indeed $m(t)$ is the minimal polynomial of $T_P$.
