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?

Denote the $k$-vector space in question by $V$. As $V$ is finite dimensional, $V$ and $V^* := L(V,k)$ (the dual space) have the same dimension. Define a map $\Theta_g \colon V \to V^*$ by $\Theta_g a = g(a, \cdot)$. As $g$ is non-degenerated, $\Theta_g$ is one-to-one: If $a \in \ker \Theta_g$, we have $g(a,b) = \Theta_ga(b) = 0$ for all $b \in V$, hence $a = 0$. As $\dim V = \dim V^*$, $\Theta_g$ is an isomorphism. Now define $T_1 \colon V \to V$ by 
$$ T_1 a := \Theta_g^{-1}\bigl(f(a, \cdot)\bigr) $$
Then, for all $a, b \in V$, we have 
\begin{align*}
  g(T_1 a, b) &= \Theta_g(T_1 a)b\\
              &= \Theta_g \Theta_g^{-1}\bigl(f(a,\cdot)\bigr)b\\
              &= f(a,\cdot)(b)\\
              &= f(a,b)
\end{align*}
The existence of $T_2$ is proved along the same lines.

Addendum: $T_1$ is unique, as if $T$ is any operator with $f(a,b) = g(Ta, b)$ for every $a, b \in V$, we have 
$$ f(a,\cdot) = g(Ta, \cdot) = \Theta_g(Ta) \iff Ta = \Theta_g^{-1}\bigl(f(a,\cdot)\bigr) = T_1 a $$
Uniqueness of $T_2$ follows along the same lines.
If $g$ is denenerate, for example $g = 0$. Then $g(T_1 a, b) = 0$ for any $T_1 \colon V \to V$, that is, if $f \ne 0$, such a $T_1$ will not exist.
