Minimal polynomial of $T(A) = A^\top - A$

Question

As said in the title , I need to find the minimal polynomial of the linear transformation $$T(A)=A^\top-A.$$ The matrices are $M_n(\mathbb{C})$. I've figured out that $T^2 = 2A - 2A^t$ , so a polynomial $p(t) = t^2 + 2t$ works so $p(T) = 0$. Now $p(t)$ breaks to $t(t+2)$ but non of them kills T. Therefore $p(t)$ is the minimal polynomial.

I'm having trouble with this, because I guessed $p(t)$, and Im not sure on how to actually find the polynomial. For example, I have no idea how to find a matrix, because of that transpose. Is there another way to do this?

Answer 1

I do not know if i can say anything better than what you have done...

You have seen what $T^2$ would be... this is what you actually have to do..

see what would $T,T^2,T^3\cdots$ be and check for a liner combination that would result zero map ..

You have seen the very first non trivial power of $T$ namely $T^2$ and realized it as $-2T$

So, You have $T^2=-2T$ and remaining thing i want to say is not any better than yours..

So, What you have done is natural for me..

P.S : All this is just for your statement I guessed $p(t)$ and I'm not sure on how to actually find the polynomial

Answer 2

Computing powers of $T$ and seeing what happens is a good approach here, because of the simple form that $T$ is given in. Another approach would be to view $T$ as a linear map on $M_n(\Bbb{C})$ given by $$T:\ M_n(\Bbb{C})\ \longrightarrow\ M_n(\Bbb{C}):\ (a_{ij})_{1\leq i,j\leq n}\ \longmapsto\ (a_{ji}-a_{ij})_{1\leq i,j\leq n},$$ with respect to the natural basis. For all $1\leq i,j\leq n$ the subspaces $$\operatorname{span}(E_{ii})\qquad\text{ and }\qquad\operatorname{span}(E_{ij}, E_{ji}),$$ are invariant under $T$, so $T$ is represented by a block diagonal matrix, with blocks of the form $$\begin{pmatrix}0\end{pmatrix}\qquad\text{ and }\qquad\begin{pmatrix}-1&1\\1&-1\end{pmatrix},$$ respectively, and the characteristic polynomial of the latter is $$\det\left(\begin{pmatrix}-1&1\\1&-1\end{pmatrix}-tI\right)=(-1-t)^2-1^2=t^2+2t$$ Hence this is its minimal polynomial of every $2\times2$-block, and therefore it is also the minimal polynomial of $T$.