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.

  • 1
    $\begingroup$ Cultural remark: $S$ is called a symplectic matrix. $\endgroup$ – Potato Jun 8 '14 at 16:46
  • $\begingroup$ @Potato Thank you very much for the remark. $\endgroup$ – usermath Jun 8 '14 at 17:08

$X^\top=-SXS^{-1}$, so $\operatorname{tr}X=\operatorname{tr}X^\top = -\operatorname{tr}(SXS^{-1})=-\operatorname{tr} X$.

  • $\begingroup$ @@Ted Shifrin) Actually the given $S$ is NOT needed at all.Am I right ? $\endgroup$ – Empty Nov 8 '15 at 6:15
  • $\begingroup$ @S.Panja-1729: You're correct. It holds for any invertible $S$. But this question arises naturally from a certain Lie algebra. $\endgroup$ – Ted Shifrin Nov 8 '15 at 6:24

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, $\operatorname{tr}(X)=\operatorname{tr}(X_1)+\operatorname{tr}(X_4)=\operatorname{tr}(X_1)-\operatorname{tr}(X_1)=0$.

  • 3
    $\begingroup$ You seem to be assuming that the blocks in $X$ are symmetric. Also, your conclusion would be that $X_1=-X_4$. $\endgroup$ – Martin Argerami Jun 8 '14 at 16:24
  • $\begingroup$ @user63181 thank you very much for your help.Also as Martin said please see that $X_1=-X_4$ $\endgroup$ – usermath Jun 8 '14 at 16:28

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 $tr X = 0$.


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