Dual isomorphisms induced by a nondegenerate pairing ($U\cong V^{\ast },V\cong U^{\ast }$)

A bilinear form $b:U\times V\to K$ with zero left and right radicals induces injective linear maps $U\to V^{\ast }$ and $V\to U^{\ast }$. Finite dimensions force $\dim U=\dim V$, so both maps are isomorphisms. For any other bilinear form $c$, composing its maps to the dual spaces with these inverse isomorphisms gives unique endomorphisms $S,T$ with $c(x,y)=b(Sx,y)=b(x,Ty)$. In bases with invertible pairing matrix $B$ and coefficient matrix $C$, $S=B^{-T}{C}^T$ and $T=B^{-1}C$.