Prove or disprove that trace of matrix $X$ is zero I was trying to solve a question  from a competitive exam paper.
This is a part of that question.

Let $I_n$ and $O_n$ be $n\times n$ identity and null matrices respectively.Let $S$ be $2n\times 2n$ matrix given in block form by $$S=\begin{bmatrix} O_n & I_n \\ -I_n & O_n\end{bmatrix}$$
If $X$ is a $2n\times 2n$ matrix such that $X^tS+SX=O_{2n}$ then determine wheather trace of $X$ is zero or not.

From the given information in the question I just managed to find that $S^t=-S$(i.e $S$ is skew symmetric) and det($S$)=$1$. Then $SX=-X^tS=X^tS^t=(SX)^t\Rightarrow SX$ is symmetric. Also det($SX$)=det($X$).
Am I right upto this?and I do not know whether these are necessary or not to solve the problem. I can not proceed further,completely stuck.
Please help.Thnx in advance.
 A: $X^\top=-SXS^{-1}$, so $\operatorname{tr}X=\operatorname{tr}X^\top = -\operatorname{tr}(SXS^{-1})=-\operatorname{tr} X$.
A: Let denote
$$X=\begin{bmatrix} X_1 & X_2 \\ X_3 & X_4\end{bmatrix}$$
so
$$X^tS+SX=\begin{bmatrix} X^t_1 & X^t_3 \\ X^t_2 & X^t_4\end{bmatrix}\begin{bmatrix} O_n & I_n \\ -I_n & O_n\end{bmatrix}+\begin{bmatrix} O_n & I_n \\ -I_n & O_n\end{bmatrix}\begin{bmatrix} X_1 & X_2 \\ X_3 & X_4\end{bmatrix}\\
=\begin{bmatrix} -X^t_3 & X^t_1 \\ -X^t_4 & X^t_2\end{bmatrix}+\begin{bmatrix} X_3 & X_4 \\ -X_1 & -X_2\end{bmatrix}=\begin{bmatrix} -X_3^t+X_3 & X^t_1+X_4 \\ -X_1-X_4^t & X_2^t-X_2\end{bmatrix}=O_{2n}$$
hence we see that $X_1=-X^t_4$ and then YES, $\operatorname{tr}(X)=\operatorname{tr}(X_1)+\operatorname{tr}(X_4)=\operatorname{tr}(X_1)-\operatorname{tr}(X_1)=0$.
A: Present X as a block matrix with $n$ x $n$ blocks $X(1, 1)=A$, $X(1, 2)=B$, $X(2, 1) = C$, $X(2, 2) = D$. Then $-C^t = A, A^t = B, -D^t = C, B^t = D$ and desired value is $tr X = tr A + tr D$. However $tr C = -tr A = tr B = tr D = -tr C$ and hence $tr C = 0$ and $tr D = 0$ and $tr A = 0$. So $tr X = 0$. 
