Does there exist a matrix $S$ such that $gl_S(n,\mathbb{R})$ is equal to the set of all diagonal matrices in $\mathbb{R}$? Let $S$ be an $n \times n$ matrix with entries in field the $\mathbb{R}$ and define:
$gl_S(n,F)=\{x \in gl(n,F) : x^tS = -Sx \}$
Does there exist a matrix $S$ such that the lie subalgebra $gl_S(n,\mathbb{R})$ is equal to the set of all diagonal matrices in the lie algebra $gl(n,\mathbb{R})$?
I've tried
$$ A = \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right) $$
$$ B = \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right) $$
and a few other things but can't seem to find anything that forces $S$ to be the set of all diagonal matrices. Help would be appreciated here! Thanks
 A: Just writing out the equation with a general $x=diag(x_1, ... x_n)$ gives
$$S_{ij}x_i=-S_{ij}x_j$$
for all $x_i, x_j$, and this forces $S=0$: namely, for $i\neq j$, we can choose $x_i\neq -x_j$, and for $i=j$, we can choose $x_i\neq -x_i$.
But for $S=0$, all $n \times n$-matrices are in $gl_0(n,F)$, and for $n>1$ many of them are not diagonal. So for $n\ge 2$, such an $S$ does not exist.

Note that this argument works over any field of characteristic $\neq 2$. For characteristic $2$, I made an erroneous claim in an earlier version (sorry), now I believe that again for $n \ge 2$, no such $S$ exists.
Namely, for $i \neq j$ the above still works, so at least we know $S$ must be diagonal. Assume we have such an $S$. Now using $S^t=S$ and the characteristic, the condition on $x$ to be in $gl_S(n, F)$ becomes
$$x^t S^t = (Sx)^t \stackrel{!}=Sx$$
i.e. $Sx$ is symmetric. For $n=1$ this works trivially, but as soon as $n \ge 2$, we find non-diagonal $x$ which satisfy this. E.g. if $S_{ii}=0$ for some $i$, we can choose as $x$ any matrix with non-zero entries in its $i$-th column, zeros elsewhere; if, on the other hand, all $S_{ii} \neq 0$, then $S$ is invertible, we can choose any non-diagonal symmetric matrix $y$ and set $x:=S^{-1}y$ which still is non-diagonal.
