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If $A∈M(n;\mathbb{R})$, let $A^t$ denote its transpose. A matrix $S\in M(n;\mathbb{R})$ is said to be skew-symmetric if $S^t = −S$. Pick out the true statements:

a. If S ∈ $M(n;\mathbb{R})$ is skew-symmetric and non-singular, then $n$ is even.

b. Let $G = \{T ∈ GL(n;\mathbb{R})\mid T^t ST = S$, for all skew-symmetric $S ∈ M(n;\mathbb{R}$}. Then $G$ is a subgroup of $GL(n;\mathbb{R})$.

c. Let $I_n$ and $O_n$ denote the $n \times n$ identity and null matrices respectively. Let $S$ be the $2n \times 2n$ matrix given in block form by $\left[\matrix{O_n&I_n\cr -I_n&O_n}\right]$.

If $X$ is a $2n×2n$ matrix such that $X^t S+SX = 0$, then the trace of $X$ is zero.

Please help anyone to solve the problem. My thinking as far:- (a) is true as every skew symmetric matrix of odd order is singular. For(b) & (c) no idea. Thanks.

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Hints:
(a) Note Aaron's hint (that $\det(A^t)=\det(A)$) and remember that $\det(\alpha A)=\alpha^n \det(A)$ for all $\alpha\in\mathbb{R}$.
(b) Remember that for $T\in\rm{GL}_n(\mathbb{R})$ we have $\left(T^{-1}\right)^t=\left(T^t\right)^{-1}$ and that for all $A,B\in \rm{M}_n(\mathbb{R})$, $B^tA^t=(AB)^t$. Now check the group axioms.
(c) What is $S^{-1}$? You can find it by checking smaller examples. What is $SXS^{-1}$? Compute this using small examples as well.

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  • $\begingroup$ Could you give more hints for $c$... I see that $S^{-1}=\begin{bmatrix} 0&-I_n\\I_n&0 \end{bmatrix}$ and $SXS^{-1}=-X^t$.. but then it does not say anything except that Trace of $X$ is same as that of $SXS^{-1}$... could you please extend your hint $\endgroup$ – user87543 Jan 8 '14 at 3:26
  • $\begingroup$ @Praphulla Koushik: It means that $\operatorname{trace}(X)=\operatorname{trace}(SXS^{-1})=\operatorname{trace}(-X^{t})=-\operatorname{trace}(X)$. Then... $\endgroup$ – Dennis Gulko Jan 8 '14 at 12:39
  • $\begingroup$ but i do not have $X=SXS^{-1}$ so.. i do not understand the concussion $\operatorname{trace}(X)=\operatorname{trace}(SXS^{-1})$ remaining this are fine... $\endgroup$ – user87543 Jan 8 '14 at 12:45
  • $\begingroup$ I now understand why $\operatorname{trace}(X)=\operatorname{trace}(SXS^{-1})$ thank you :) $\endgroup$ – user87543 Jan 9 '14 at 2:12

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