# Bilinear forms and basis.

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.

• By the way, I'm not super happy about the tittle of the question. If someone comes up with a better one, please share. Jul 9 '19 at 20:17
• It's meaningless to say $\phi$ is symmetric, unless $V=W$. Jul 9 '19 at 20:39

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 $$\Psi$$.