Let f,g be bilinear forms on a finite dimensional vector space. Show that there exist unique linear operators Let $f,g$ be bilinear forms on a finite dimensional vector space.
(a) Suppose $g$ is non-degenerate. Show tha there exist unique linear operators $T_{1}, T_{2}$ on $V$ such that $f(a,b)=g(T_{1} a,b)=g(a,T_{2} b)$ for all $a,b$.
(b) Show that this result may not be true if $g$ is degenerate.
My attempt:
Although I am clueless about this question, I was wondering if it has something to do with the adjoint of a linear operator. If we have a linear operator $T$ on a hermitian space $V$ then $ \langle Tv,w \rangle= \langle v,T^{*}w \rangle$ and $ \langle T^{*} v,w \rangle= \langle v,Tw \rangle$ for all $v,w$ in $V$. But then how would you relate it to the other bilinear form and how would you use the fact that one of the bilinear forms is non-degenerate?

Hint: If $g$ is non-degenerate, then $g$ induces two isomorphisms $V \to V^*$:


*

*$\phi:V \to V^*$, given by $\phi(v)(y)=g(v,y)$

*$\psi:V \to V^*$, given by $\psi(v)(x)=g(x,v)$
