Relation between non-degenerate bilinear forms and linear transformations

Question

Let $V$ be a finite vector space over $\Bbb R$, let $q$ be an non-degenerate bilinear form. Let $T: V \to V$ be a linear transformation.

Prove that exists one and only one $S: V \to V$ so that $S$ is a linear transfomration satisfying:

$\forall v,w \in V \hspace{1.5mm} q(T(v), w) = q(v, S(w))$

My initial thought was to choose a Matrix $M$ that represents $q$ in some basis $C$ of $V$, and prove:

$[T(v)^t]_CM[w]_C \hspace{1.5mm} - [v^t]_CM[S(w)^t]_C = 0$

Obviously, if $M$ is positive-definite then we can take $S$ to be $T^*$ and we achieve the desired result since $q$ is an inner product. However, if $M$ is not definite-positive, I'm not sure how to tackle this - mainly unsure how to use the fact that $M$ is invertible. Any help is appreciated very much.

Answer 1

Using your approach we obtain, thinking of $T$ and $S$ as square matrices, $$v^t T^t M w = v^t M S w \quad \textrm{for all }v, \, w,$$ so we must have $T^t M = M S$. This implies, using the fact that $M$ is invertible, $$S = M^{-1} T^t M.$$

Answer 2

Here is a proof that is coordinate free.

Existence Since $q$ is nondegenerate, given $w\in V$ there is a unique $u\in V$ such that for every $v\in V$, $q(T(v),w) = q(v,u)$. I leave this for you to check.

Thus, we can define a map $S:V\to V$ by $S(w) = u$. Combining the fact that $T$ is linear and $q$ is bilinear, you can prove that $S$ is linear: For all $v\in V$,

$$q(v,S(\alpha w_1 + w_2)) = q(T(v),\alpha w_1 + w_2) = q(T(v),\alpha w_1) + q(T(v),w_2) $$ $$ = q(\alpha T(v), w_1) + q(v,S(w_2)) = q(T(\alpha v), w_1) + q(v,S(w_2)) = q(\alpha v, S(w_1)) + q(v,S(w_2))$$ $$= q(v,\alpha S(w_1)) + q(v,S(w_2)) = q(v,\alpha S(w_1) + S(w_2)).$$ Thus, $$S(\alpha w_1 + w_2) = \alpha S(w_1) + S(w_2)$$ by nondegeneracy of $q$, so $S$ is linear.

Uniqueness This is easy. Supppose you have two such transformations $S$ and $S'$ and use nondegeneracy of $q$ to prove they are equal (as transformations).