Tensor product of dual spaces

Question

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?

Answer 1

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 $(1)$.