Non-degenerate Bilinear Forms and Counterexample

Question

Let $f$ and $g$ bilinear forms in a finite-dimensional space $V$. Assume $g$ is non-degenerate. Show that there are only operators ${T_1}:V \to V$ and ${T_2}:V \to V$ such that $$f\left( {u,v} \right) = g\left( {{T_1}\left( u \right),v} \right) = g\left( {u,{T_2}\left( v \right)} \right),\quad \forall u,v \in V.$$ Display a counterexample to the above statement in which $f$ is non-degenerate, $g$ is degenerate and there is no $T_1$ satisfying $f\left( {u,v} \right) = g\left( {{T_1}\left( u \right),v} \right)$ for all $u,v\in V$.

Answer 1

We’re in finite dimension. Take a basis and consider the two symmetric matrices $F$ and $G$ such that $f(u,v)=U^TFV$ and $g(u,v)=U^TGV$ where $U$ and $V$ are the columns of coordinates of $u$ and $v$.

$g$ Is non degenerate if and only if $G$ is inversible

We have

$$\begin{align}U^TFV=& \left(G^{-1}FU\right)^TGV\\ =&U^TG\left(G^{-1}FV\right) \end{align} $$

And we’re done with $T_1=T_2=G^{-1}F$