Biliniear form to inner product

Question

Let $f:V\times V\rightarrow F$ be a bilinear form in a finite inner product space V.

If $F=R$, how can I prove that there exists a single linear transformation $T:V \rightarrow V$ so that for each $v,u\in V$, $f(u,v) = (Tu, v)$ ?

Would that sill be the case if $F=C$?

Answer 1

Pick an orthonormal basis $\{e_1,\dots,e_n\}$ for $V$, and write $u$ and $v$ in this basis as $u = \sum_i u_i e_i$ and $v = \sum_i v_i e_i$, where $u_i$ and $v_i$ are scalar coefficients. Then use the bilinearity of $f$:

$$ \begin{eqnarray} f(u,v) &=& \left( \sum_i u_i e_i, \sum_j v_j e_j \right) \\ &=& \sum_{i,j} u_i v_j \, f(e_i,e_j) \end{eqnarray} $$

If $G$ is the Gram matrix of $f$ in this basis (the matrix with $f(e_i,e_j)$ as entries), then the above calculation shows that $f(u,v) = (Gu,v)$.