Why $X^* X\in \mathfrak{g}(V)$.

Question

A finite dimensional vector space $V$ with a non-degenerate form (,) s.t. $(u,v)=\epsilon (v,u) \forall u,v\in V$ is called a quadratic space of type $\epsilon$.

Let $V$ be a quadratic space of type $\epsilon$ and $U$ be a quadratic space of type $-\epsilon$.

Let $X:V\rightarrow U$ be a linear map and the adjoint map $X^*:U\rightarrow V$ is defined by $(Xv,u)=(v,X^*u)$ for $v\in V, u\in U$.

Why $X^* X\in \mathfrak{g}(V)$(Lie algebra of V)?

Answer 1

$(X^*Xu,v)=\varepsilon(v, X^*Xu)=\varepsilon (Xv, Xu)=-\varepsilon^2(Xu, Xv)=-(u, X^*Xv)$. That's it