$\newcommand{\Bilinear}{\operatorname{Bilinear}}$ Let $V$ be a finite-dimensional vector space over the field $\mathbb{F}$, and let $V^*$ denote its dual. Denote by $\mathscr{L}(V,V^*)$ by the space of all linear transformations from $V$ to $V^*$. I define a bilinear form to be a bilinear function $V \times V \to \mathbb{F}$. A bilinear form $\beta$ is said to be non-degenerate if for any $x\in V$, $\beta(x, \cdot) \equiv 0 \implies x =0$.
1. Are the spaces $\Bilinear(V\times V, \mathbb{F})$ and $\mathscr{L}(V,V^*)$ linearly isomorphic? In other words, does every bilinear form uniquely determine a homomorphism $V \to V^*$, and vice versa, in such a manner that respects the linear structure of both spaces?
2. Does every non-degenerate bilinear form uniquely determine an isomorphism $V \to V^*$, and does every isomorphism $V \to V^*$ uniquely determine a non-degenerate bilinear form, in such a way that respects the linear structure of both spaces?
Best Answer
Let me offer a short proof using dual bases. Let $v_1, \dots, v_n$ be some basis of $V$ and $v^1, \dots, v^n$ the associated dual basis of $V^{*}$. Define the map $\phi \colon \mathcal{L}(V,V^{*}) \rightarrow \operatorname{Bi}(V \times V, \mathbb{F})$ by $\phi(T) := g_T$ where $g_T(u,v) = T(u)(v) = \varepsilon(T(u), v)$. It is readily verified that this map is linear so it is enough to show that $\phi$ is one-to-one and onto. Since $$ T(v_i) = \sum_{i=1}^n \varepsilon(T(v_i), v_j) v^j $$
we see that the map $T$ is determined uniquely by the $n \times n$ scalars $\varepsilon(T(v_i), v_j)$ (since the matrix $(\varepsilon(T(v_i),v_j))$ is the matrix representing $T$ with respect to the bases $v_1,\dots,v_n$ of $V$ and $v^1,\dots,v^n$ of $V^{*}$). Hence, if $\phi(T) = 0$ then all the scalars $\varepsilon(T(v_i), v_j)$ are zero and $T = 0$. This shows the injectivity.
To show surjectivity, let $g$ be a bilinear form on $V$ and define a linear map $T$ by requiring that
$$ T(v_i) = \sum_{i=1}^n g(v_i, v_j)v^j $$
for all $1 \leq i \leq n$. Then a direct calculation shows that $\phi(T) = g$.