# 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

• what do you mean by $gl(n,F)$?
– QED
Commented Sep 8, 2020 at 12:17
• the lie algebra, oops sorry i'll add that Commented Sep 8, 2020 at 12:18
• Oops, edited my question, sorry that was wrong. $\mathbb{R}$ was the field, $gl(n,\mathbb{R})$ is the lie algebra of concern Commented Sep 8, 2020 at 12:24

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.

• what does the $!$ above the $=$ mean?? Commented Sep 9, 2020 at 17:17
• What I meant with that is that this equation "must" be true for $x$ to be in $gl_S$, i.e. it is the condition we "have to check". It's maybe not too standard notation, sorry if it was confusing. Commented Sep 9, 2020 at 18:31
• Wouldn't it have to be $(Sx)^t=-Sx$? Commented Sep 9, 2020 at 19:21
• In principle yes, but in characteristic $2$ we have $A=-A$ for every matrix $A$. Commented Sep 9, 2020 at 21:00