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!