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.