# Existence of dual basis for an algebra with a nondegenerate bilinear pairing

Let $A$ be an algebra over a field $k$. Suppose that we have a non degenerate bilinear pairing $\beta:A \otimes A \to k$.

Let $\{a_i\}$ be a basis of $A$.

I would like to show that there exist a basis $\{a'_i\}$ of $A$ such that we have $\beta(a_i, a'_j)=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta.

I thought I could use Gram-Schmidt process. But the problem is not to construct an orthonormal basis (the basis $\{a_i\}$ is fixed).

Also is it always true that $\beta(a_i,a_i)\neq0$?

I appreciate any help.

This has nothing to do with algebras in particular and is just a construction on finite-dimensional vector spaces. If $V$ is a finite-dimensional vector space over a field $k$, equipped with a nondegenerate bilinear pairing $\beta$, then the natural map $\iota: V \to V^*$ sending $w \mapsto (v \mapsto \beta(v,w))$ is injective (by nondegeneracy of $\beta$) and therefore surjective as well (since $\dim V^* = \dim V$ and both are finite-dimensional). So $\iota$ gives an isomorphism between $V$ and $V^*$.
Hence, given a basis $\{a_i\}$, we may, for each $j$ ($1 \le j \le n$), construct an element of $V^*$ which sends $a_i \to \delta_{ij}$ (i.e. $a_j$ to 1 and all other $a$'s to 0). The inverse image of this element under $\iota$ is the $a'_j$ you are looking for.
It is not true, in general, that $\beta(a_i, a_i) \ne 0$ for $a_i \ne 0$. For example, over any field $k$, we could take the nondegenerate inner product on the vector space $k^n$ defined by $\beta(x,y) = \sum x_i y_i$. Then $\beta(x,x) = \sum x_i^2$ and the right side could be 0 even if $x \ne 0$, for many fields $k$.