$A\in M_{3}(\mathbb{R})$ is a matrix such that $\mbox{ Tr }(A-A^t)^{2n}=0$

Question

If $A\in M_{3}(\mathbb{R})$ is a matrix such that $\mbox{ Tr }(A-A^t)^{2n}=0$ for some integer $n>0$. Show that $A=A^t$

I have tried like this: Let $X=A-A^t$. Then $X$ is a skew symmetric matrix with all its eigenvalues are zero. This follows from the trace condition. From this how I can proceed?

Answer 1

Hint for the number-crunching route: Show that for $n>1$ and an arbitrary skew-symmetric matrix, we have $$ \begin{bmatrix} 0&a&b\\-a&0&c\\-b&-c&0 \end{bmatrix}^{2n} = (-1)^n(a^2+b^2+c^2)^{n-1}\begin{bmatrix}a^2 + b^2 & bc&-ac\\bc & a^2 + c^2 & ab\\-ac&ab&b^2+c^2\end{bmatrix} $$

Answer 2

$X=A-A^t$, being a $3\times 3$ skew-symmetric real matrix, have eigenvalues $0,\pm ic$ for some $c\geq 0$. So $\operatorname{Tr} X^{2n}=2(-1)^nc^{2n}$

Answer 3

Let $X=A-A^t$. Note that $X$ is a skew symmetric matrix. Now $X$ being a $3\times3$ skew symmetric real matrix, its eigenvalues are $0,\pm i c$ for some $c\geq0$. Then $$\mbox{ Tr }X^{2n}=0\implies2(-1)^{n}c^{2n}=0\implies c=0$$ Hence all the eigenvalues of $X$ are zero. Now consider $$\mbox{ Tr }(XX^t)=\mbox{ Tr }(X(-X))=-\mbox{ Tr }X^2=0.$$ Consequently we have $X=0\implies A=A^t.$