 Solution

Writing matrices in $2\times 2$ blocks, the defining relation givesThe trace is automatically zero. The relation is linear, and if $as=-s{a}^T$, $bs=-s{b}^T$, thenThus $L$ is closed under the commutator and is a Lie subalgebra of ${\mathfrak{sl}}_4(F)$.

Its diagonal subalgebra isPut ${\epsilon }_1(D(a,b))=a$, ${\epsilon }_2(D(a,b))=b$. Since $[D,E_{ij}]=(D_{ii}-D_{jj})E_{ij}$, the required block decomposition isThe four lines have respective characters ${\epsilon }_1-{\epsilon }_2$, $-{\epsilon }_1+{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $-{\epsilon }_1-{\epsilon }_2$. They are distinct and nonzero because $\mathrm{char}F\ne 2$. The two-dimensional summand $H$ has two trivial composition factors. These five submodules therefore have exactly the factor-grouping properties of part (b). Their dimensions sum to $6$, the dimension obtained from four free entries in $A$ and one each in the skew-symmetric blocks $B,C$.