proving that symplectic lie algebra is a subalgebra of GL

Question

Suppose $S$ is n by n matrix over a field F. Define

$gl_S(n,F)=\{A \in gl(n,F): SA+A^TS= 0\}$

Show that this is a subalgebra of $gl(n,F)$

I get as far as:

$A \in gl_S(n,F)$ and $B \in gl_S(n,F)$

$S[AB] +[AB]^T S =SAB -SBA + (AB)^TS-(BA)^TS $ $=SAB + B^TA^TS - (SBA + A^TB^TS)$

I guess it's supposed to equal to zero somehow.

Answer 1

Since $SA=-A^TS$ and $SB=-B^TS$, $$ S[A, B]=SAB-SBA=-A^T SB+B^T SA=A^T B^T S-B^TA^TS=[A^T, B^T] S=-[A, B]^TS. $$