Here's the problem:
Let $f$ be a symmetric bilinear form over an n-dimensional vector space $X$, and $f$ is non-degenerate (for all nonzero $x \in X$, there exists $y \in X$ so $f(x,y) \neq 0$). Prove that $L_f: X \to X'$ defined as $x \mapsto f(x, -)$ is an isomorphism.
I can easily prove $L_f$ is one-to-one, and argue that it's an isomorphism since we are dealing with finite-dimensional spaces. (If $L_f(x) = L_f(y)$ then $f(x-y, z) = 0$ for all $z$. However, since $f$ is non-degenerate, if $x-y \neq 0$ then there would exist a $z$ that would make $f$ nonzero. Therefore, $x=y$)
However, I am stuck trying to prove that it is onto, to maybe extend this result to infinite dimensional spaces. Is it even possible?