4
$\begingroup$

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.

$\endgroup$
  • 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
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @@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
4
$\begingroup$

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

$\endgroup$
  • 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
0
$\begingroup$

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 $tr X = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.