Question on bilinear forms and invertible matrix

Question

Let $k$ be a field and $V$,$W$ be two $k$-vectorspaces of same dimension. Now if $V \times W \longrightarrow k$ is a bilinear pairing, how can I see that this bilinear pairing is nonsingular if and only if for any basis $\{v_1,...,v_n \}$ for $V$ and $\{w_1,...,w_n \}$ for $W$ the matrix $b_{ij}= \langle v_i,w_j \rangle$ is invertible.

The bilinear pairing is nonsingular if the maps $V \to W^{\ast}$ and $W \to V^{\ast}$ are isomorphisms.

Answer 1

I'm assuming the first map in question, $V\to W^*$, is given by mapping $v$ to the functional $w\mapsto \langle v,w\rangle$. And let $B$ be the matrix with entries $b_{ij}=\langle v_i, w_j\rangle$. Then we have the following chain of equivalences:

\begin{align*} \text{$V\to W^*$ is an isomorphism} &\iff \text{It has trivial kernel} \\ &\iff \text{The only vector $v$ satisfying $\langle v,w \rangle =0 $ for all $w$ is $v=0$} \\ &\iff \text{The only vector $v$ satisfying $\langle v,w_j\rangle=0$ for all $j$ is $v=0$} \\ &\iff \text{The only $\alpha_i$ satisfying $\langle \sum \alpha_i v_i, w_j \rangle=0$ for all $j$ are $\alpha_i=0$.}\\ &\iff \text{The only $\alpha_i$ satisfying $\sum \alpha_i b_{ij}=0$ for all $j$ are $\alpha_i=0$.}\\ &\iff \text{The matrix $B$ has trivial left null space.}\\ &\iff \text{The matrix $B$ is invertible.} \end{align*} By a symmetric argument, we can also conclude that the map $W\to V^*$ is an isomorphism if and only if $B$ is invertible.

Answer 2

If you compute with the bases, you'll see the matrix of the mappring $V \to W^{\ast},\,$ $\, W^{\ast}$ equipped with the dual basis $(w_1^\ast,\dots w_n^{\ast})$, is but the transpose of $(b_{ij})$. Hence the mapping $V \to W^{\ast}\,$ is an isomorphism if and only if $(b_{ij})$ is invertible.