# Tensor product of dual spaces

Let $$V$$ and $$W$$ be vector spaces over the field $$k$$, $$V^*$$ be the dual space of $$V$$ etc. and $$BLF(V,W)$$ the vector space of bilinear forms $$V\times W \rightarrow k$$. Then the bilinear map $$V^*\times W^* \rightarrow BLF(V,W)$$ defined by $$(\phi, \psi) \mapsto \left ((v,w) \mapsto \phi(v)\psi(w) \right )$$ yields a linear map $$\Phi: V^*\otimes W^* \rightarrow BLF(V,W)$$ via the universal property of the tensor product. Apparently $$\Phi$$ is always injective (even if the vector spaces are not finite dimensional), why is that?

We prove that the kernel of $$\Phi$$ is trivial.

$$(1)$$ At first we prove that, for every $$\phi \in V^*$$ and every $$\psi \in W^*$$, if $$\Phi(\phi\otimes \psi)$$ is null, then $$\phi \otimes \psi$$ is null as well.

Let $$\phi \in V^*$$ and $$\psi \in W^*$$ be such that $$\Phi(\phi\otimes\psi)$$ is the null element of $$BLF(V,W)$$, hence $$(\Phi(\phi \otimes \psi))(v,w)=0$$ for every $$v \in V$$ and every $$w \in W$$. As $$\Phi \circ \otimes$$ is the bilinear map $$V^* \times W^* \to BLF(V,W)$$ that you defined, it is the case that $$(\Phi(\phi \otimes \psi))(v,w)=\phi(v)\psi(w)$$ for every $$v \in V$$ and every $$w \in W$$. Hence it is the case that $$\phi(v)\psi(w)=0$$ for every $$v \in V$$ and $$w \in W$$.

If $$\phi$$ is the null element of $$V^*$$ then $$\phi \otimes \psi$$ is the null element of $$V^* \otimes W^*$$ and we are done. Otherwise, if $$\phi$$ is not the null element of $$V^*$$ then there is $$v' \in V$$ such that $$\phi(v')\neq 0$$. Then for every $$w \in W$$ it is the case that $$\phi(v')\psi(w)=0$$, that is, $$\psi(w)=0$$. Therefore $$\psi$$ is the null element of $$W^*$$ and in particular $$\phi \otimes \psi$$ is the null element of $$V^* \otimes W^*$$ and again we are done.

$$(2)$$ Secondly, we prove that, for every $$\phi,\phi' \in V^*$$ and $$\psi,\psi' \in W^*$$, if $$\Phi(\phi \otimes \psi + \phi'\otimes \psi')$$ is null, then $$\phi \otimes \psi + \phi'\otimes \psi'$$ is null as well. After that, it's clear that you can conclude by induction over the number of addends composing a given element of $$V^* \otimes W^*$$ (remind that every element of $$V^* \otimes W^*$$ is a finite sum of elements of the form $$\alpha \otimes \beta$$, being $$\alpha \in V^*$$ and $$\beta \in W^*$$).

So, let us assume that $$\Phi(\phi \otimes \psi + \phi'\otimes \psi')$$ is null for some $$\phi,\phi' \in V^*$$ and $$\psi,\psi' \in W^*$$. Hence, for every $$v \in V$$ and $$w \in W$$, it is the case that $$\phi(v)\psi(w)+\phi'(v)\psi'(w)=0$$.

If $$\phi'$$ is null, then $$\phi \otimes \psi +\phi'\otimes \psi'=\phi \otimes \psi$$ and we are done by $$(1)$$. Otherwise, if $$\phi'$$ is not the null element of $$V^*$$, then there is $$v' \in V$$ such that $$\phi'(v')\neq 0$$. Hence, for every $$w \in W$$, it is the case that $$\psi'(w)=\lambda\psi(w)$$, being $$\lambda:=-\phi(v')/\phi'(v')$$. Then $$\psi'=\lambda \psi$$ and in particular $$\phi \otimes \psi +\phi'\otimes \psi'=(\phi + \lambda \phi')\otimes\psi$$. Again we are done by $$(1)$$.