Linear Forms induce Bilinear Forms

Here I make use of the Einstein convention to cut down some keyboard work.

Let $V$ be a vector space over the field $K$ and $V^*$ is the dual space of it. The map $F\in\mathrm{Hom}(V,V^*)$ induces a bilinear form $\left<v,w\right>_F:=F(v)(w)$, a more or less trivial fact. 

$B:=(b^i)$ ... Basis sequence of $V$.

Every vectors $v,w\in V$ can be unambiguously expressed by coordinates $v_i,w_i$ regarding the above basis. $$v=v_ib^i ~~~; ~~~ w=w_ib^i$$ By making use of the linearity and the commutativity of the underlying field we get $$F(v)(w)=v_iw_jF(b^i)(b^j)=v_iF(b^i)(b^j)w_j$$ It can be directly seen that the field elements $F(b^i)(b^j)=\left<b^i,b^j\right>_F$ shape a matrix - the so called Gramian Matrix. Further we realize that it depends on the basis vector's of $v$ and $w$. In this case we used the same basis for both vectors but that is not essential.

Back to the beginning: $F$ induces a bilinear form, but can every bilinear form be induced by a mapping $V\rightarrow V^*$? 

Let $B_V$ be the set of all bilinear forms $V\times V\rightarrow K$. One way to start working on our question is to show that $\mathrm{Hom}(V,V^*)$ and $B_V$ are equipotent. This can be done by constructing a bijective map.

So we focus on the map $H:\mathrm{Hom}(V,V^*)\rightarrow B_V:F\mapsto \left<.,.\right>_F$.

First we show that $H$ is surjective, that is for every $\beta\in B_V$ there exists a $F\in\mathrm{Hom}(V,V^*)$ such that $H(F)=\beta$. We just set $F(b^i):=\beta(b^i,.)$ for all $i$, which forms a linear map $V\rightarrow V^*$ since $\beta$ is linear in each parameter and a linear map is uniquely defined by defining the image of basis vectors, which we just did. Now we need to show that $H(F)=\left<.,.\right>_F=\beta$. $$\beta(v,w)=v_i\beta(b^i,w)=v_iF(b^i)(w)=v_i\left<b^i,w\right>_F=v_iH(F)(b^j,w)=H(F)(v,w)$$

Second we need to prove that $H$ is injective. That is for every $F,G\in\mathrm{Hom}(V,V^*)$ $$F=G\Longleftrightarrow H(F)=H(G)$$ Let $F\neq G$, then there exists at least a single index $k$ such that $F(b^k)\neq G(b^k)$, because otherwise the mappings would be equal (uniquely defined by the images of basis vectors). Further if $H$ is injective there have to exist vectors $v,w\in V$ such that $H(F)(v,w)\neq H(G)(v,w)$. Suppose  $H(F)(v,w)= H(G)(v,w) \forall v,w\in V$ then we get $$\left<v,w\right>_F=v_i\left<b_i,w\right>_F= v_i\left<b_i,w\right>_G=  \left<v,w\right>_G$$ and if we set all $v_i=1_K$ for all $i$ (we can do so since the vectors can be arbitrarily chosen): $$\left<b^i,w\right>_F=\left<b^i,w\right>_G\Leftrightarrow F(b^i)(w)=G(b^i)(w)~~\forall i~\forall w\in V$$ Particularly it is $$F(b^k)(w)=G(b^k)(w)~~ \forall w\in V$$ in contradiction to our assumption $F\neq G$. So the map $H$ is injective and therefore bijective. 

Conclusion: $\left|\mathrm{Hom}_K(V,V^*)\right| = \left| B_V\right|$ for finite dimensional vector spaces $V$ where $B_V$ is the set of all bilinear forms on $V$.

Note that this also suggests that every bilinear form is uniquely defined by the respective Gramian matrix!