Are non-degenerate bilinear forms equivalent to isomorphisms $V \cong V^*$?

$\newcommand{\Bilinear}{\operatorname{Bilinear}}$ 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?

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 $\phi(T) = g$.

1. The answer to this question is yes. The proof is analogous to that in this question.

Let $\varepsilon: V^* \times V \to \mathbb{F}$ be the evaluation map, which for all $f\in V^*$ and $v\ \in V$ is defined as: $$\varepsilon: (f,v) \mapsto f(v)\,. $$ $\varepsilon$ is clearly bilinear. Given a linear transformation $T \in \mathscr{L}(V,V^*)$, we define a bilinear form by: $$\Xi:T \mapsto \varepsilon(T(\cdot),\cdot) \,. $$ This function is linear, since given $s,t \in \mathbb{F}, S,T \in \mathscr{L}(V,V^*)$, by the bilinearity of $\varepsilon$: $$\varepsilon((sS+tT)(\cdot),\cdot )= \varepsilon(sS(\cdot)+tT(\cdot),\cdot)=s\varepsilon(S(\cdot),\cdot)+t\varepsilon(T(\cdot),\cdot) \,.$$ To conclude the proof, we show that $\Xi: \mathscr{L}(V,V^*) \to \Bilinear(V \times V, \mathbb{F})$ is bijective (1)(2).

Injectivity: This is essentially a computation using linearity. If you are unsure about any step please feel free to comment on this post. Assume that, for $S,T \in \mathscr{L}(V,V^*)$, $\Xi S = \Xi T$. Then: $$\Xi S = \Xi T \iff \varepsilon(S(\cdot),\cdot)=\varepsilon(T(\cdot),\cdot) \\ \iff 0=\varepsilon(S(\cdot),\cdot)-\varepsilon(T(\cdot),\cdot) =\varepsilon(S(\cdot)-T(\cdot),\cdot)=\varepsilon((S-T)(\cdot),\cdot) \,.$$

For any fixed $f\in V^*$, clearly, $\varepsilon(f, v)=f(v)=0$ for all $v \in V$ if and only if $f = 0$.

By the above, for any fixed $f \in V^*$, $$\varepsilon((S-T)f, v)=0$$ for all $v \in V$. It follows that $(S-T)f=0$ for any $f \in V^*$. Thus $S-T=0$ and $S=T$. $\square$

Now, if one knows already that $\mathscr{L}(V,V^*)$ and $\Bilinear(V\times V, \mathbb{F})$ have the same dimension, then the above actually suffices to show that $\Xi$ is bijective.

However, a direct proof of surjectivity of $\Xi$ not only shines light onto why this proposed isomorphism works, but is also no more difficult than a verification of the fact that the domain and target of $\Xi$ have the same dimension. Thus we will use this approach here.

Surjectivity: Let $\beta \in \Bilinear(V \times V, \mathbb{F})$. Fix a basis $v^1, \dots, v^n$ of $V$. Then for all $i =1,\dots,n$, $$ \beta(v^i,\cdot ) \in V^* \,.$$ Using the fact that any linear transformation is determined by its action on a basis, we define $T \in \mathscr{L}(V,V^*)$ by $\quad Tv^i = \beta(v^i, \cdot) \quad$ for all $i$.

Claim: $\beta = \varepsilon(T(\cdot),\cdot)$. Let $u=u_1v^1+\dots+u_nv^n, w \in V$ be arbitrary. Then: $$\varepsilon(Tu,w)=\varepsilon(T(u_1v^1+\dots+u_nv^n),w)=u_1\varepsilon(Tv^1,w)+ \dots + u_n\varepsilon(Tv^n,w) \\ = u_1 \varepsilon(\beta(v^1,\cdot),w)+\dots+u_n\varepsilon(\beta(v^n,\cdot),w) = u_1 \beta(v^1,w)+\dots+u_n \beta(v^n,w) \\ = \beta(u_1v^1+\dots+u_nv^n,w)=\beta(u,w)\,. \square $$

2. Using the first part, all we need to show is that: $$T \in \mathscr{L}(V,V^*)\text{ is invertible} \iff \Xi T =\beta \in \Bilinear(V \times V, \mathbb{F})\text{ is non-degenerate.}$$ Two observations make this easier to show.

(i) Since $\dim V = \dim V^*$, $T$ is invertible $\iff$ $T$ is injective.

(ii) For any $x \in V$ (i.e. not just $v^1, \dots, v^n$), $Tv=\beta(x,\cdot)$.

Proof of (ii): Let $x=x_1v^1+\dots+x_nv^n$. Then: $$Tx=T(x_1v^1+\dots+x_nv^n) =x_1Tv^1+\dots+x_nTv^n= x_1 \beta(v^1,\cdot)+\dots+x_n\beta(v^n,\cdot)=\beta(x_1 v^1+ \dots+x_nv^n,\cdot)=\beta(x,\cdot)\,.$$

$\beta$ is non-degenerate $\implies$ $T$ is injective: Assume $Tx_1 = Tx_2$. Then: $$0=Tx_1 - Tx_2 = T(x_1-x_2)=\beta(x_1-x_2,\cdot)=0 \implies x_1 - x_2=0 \iff x_1=x_2\,, $$ the implication following from the non-degeneracy of $\beta$. Thus $T$ is injective. $\square$

$T$ is injective $\implies$ $\beta$ is non-degenerate: Because $T$ is injective, $Tx=T(0)=0 \implies x=0 $. Since $Tx = \beta(x,\cdot)$, this means that $\beta(x,\cdot)=0 \implies x=0$. $\square$