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?


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

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    $\begingroup$ I know that. I still believe that searching for a polynomial which is annihilated by $T_P$ is not very effective. Again, there must be an elegant solution to this question. $\endgroup$ – user265336 Aug 29 '15 at 15:44

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 $T_P$.


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