Show that $(X^*)^*=\epsilon\epsilon 'X$

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 $v\in V, u\in U$.

Show that $(X^*)^*=\epsilon\epsilon 'X$

Answer 1

First, note that $(X^{*})^*$ means that it satisfies the equation: $$ (u,(X^{*})^*v) = (X^*u,v) $$ for $u \in U$, $v \in V$, but $$ (X^* u,v) = \epsilon'(v, X^{*}u) = \epsilon' (Xv,u) = \epsilon' \epsilon(u,Xv) $$ by definition of "quadratic space $\epsilon'$", by definition of $X^*$, and by definition of "quadratic space $\epsilon$", respectively. Therefore, $$ (u,(X^{*})^*v) = \epsilon' \epsilon(u,Xv) $$ for any $u \in U$ and $v \in V$, which implies $(X^{*})^* = \epsilon \epsilon' X$ since the form is non-degenerate.