I'm studying for a upcoming exam, and I found the following problem:
Let $V$ and $W$ be finite-dimensional vector spaces of equal dimension and $\phi: V \times W \rightarrow \mathbb{K}$ a bilinear, symmetric and non-degenerated form. Given a basis $\{e_1,…, e_n\}$ of $V$, show that exists a basis $\{f_1,…,f_n\}$ of $W$ such that $\phi(e_i,f_j) = \delta_{ij}$.
My attempt was:
Let $B = \{e_1,…,e_n\}$ be a basis of $V$. Since $V$ and $W$ have the same dimension, they are isomorphic, and therefore $W$ and $V^*$(the dual space of $V$) are isomorphic. Let $\psi: V^* \rightarrow W$ be that isomorphism.
Let $B^* = \{e_1^*,…,e_n^*\}$ be the dual basis of $B$. The set $B_o = \{\psi(e_1^*),…,\psi(e_n^*)\}$ is a basis of $W$. Define $f_j = \psi(e_j^*)$, and I believe that the basis $B_o$ will do the job, but I can't quite prove it.
Any help would be deeply appreciated.
Best Answer
Consider the elements $\phi(e_1,-),\ldots,\phi(e_n,-) \in W^\ast$. Now develop an isomorphism $\Psi: W^\ast \overset{\cong}{\to} W$, and let $f_j$ be the image of $\phi(e_j,-)$ under $\Psi$.
(This should be a comment but I don't have the reputation yet)