Find elements $f,g$ of the orthogonal Lie_Algebra with $f \circ g \neq 0$ Let $K$ be a field with $char(K) \neq 2$. Let $0 \neq V$ be a $K$-vectorspace with $dim(V)< \infty$ and let $B: V \times V \to K$ be a nondegenerate bilinearform. Consider the subspace $W = \{f \in End(V) : B(f(x),y))+B(x,f(y))=0$ for all $x,y \in V \} \subset End(V)$.
My question is if there are always $f,g \in W$ with $f \circ g \neq 0$ (an idempotent $f \neq 0$ would therefore also be enough). This property seems so general that it should be obvious , but except for $0$ I don't know how a single element of $W$ looks like.
 A: I assume we are talking about a symmetric bilinear form $B$ here.
Let $n := \mathrm{dim}(V)$. For $n=1$, the space $W$ is reduced to $\{0\}$, but for $n \ge 2$, indeed there exist $f \in W$ with $f\circ f \neq 0$. Much stronger statements can be made with using scalar extension and classification of semisimple Lie algebras (at least if $\mathrm{char}(K)=0$), but here is a rather elementary proof for the claim:
First, it is well known that one can "diagonalise" the symmetric bilinear form (reference for linear algebra books that teach reverse Hermite method for symmetric matrices, Bilinear Form Diagonalisation, How to diagonalize $f(x,y,z)=xy+yz+xz$). That means there is a basis $v_1, ..., v_n$ of $V$ such that $B(v_i, v_j) =0$ for $i\neq j$ and further, if the form is non-degenerate, that all $a_i :=B(v_i, v_i)\neq 0$.
Now define $f\in \mathrm{End}(V)$ as follows:
$$f(v_i):= \begin{cases} v_2 \qquad \; \;\text{ if } i=1\\ \frac{-a_2}{a_1}v_1 \quad \text{ if } i=2 \\0 \qquad \quad\text{ if } i \ge 3\end{cases}$$ 
Check that $f$ is in $W$, and obviously $(f\circ f) (v_1) = -\frac{a_2}{a_1} v_1 \neq 0$.
Note that in the special case that $B$ is the standard "scalar product" and $v_i$ an orthonormal basis, $W$ consists of all skew-symmetric matrices and the above $f$ corresponds to the matrix
$$\pmatrix{0 &-1& &\\\
1&0& &\\
&&&\Large{0}\\}$$
which was my motivating example for the general construction.
