MHB Natural Isomorphism b/w Dual Spaces Tensor Prod & Multilinear Form Space

[caffeinemachine]

I am trying to prove the following.

Let $V_1, \ldots, V_k$ be finite dimensional vector spaces over a field $F$.
There is a natural isomorphism between $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^k(V_1, \ldots, V_k;\ F)$.

Define a map $A:V_1^*\times\cdots\times V_k^*\to \mathcal L^k(V_1, \ldots, V_k;\ F)$ as
\begin{equation*}
A(\omega_1, \ldots, \omega_k)(v_1, \ldots, v_k)=\omega(v_1)\cdots\omega_k(v_k)
\end{equation*}
for all $(\omega_1, \ldots, \omega_k)\in V_1^*\times\cdots\times V_k^*$.
It can be seen that $A$ is a multilinear map.
By the universal property of tensor product, there exists a unique linear map $\tilde A: V_1^*\otimes\cdots\otimes V_k^*\to \mathcal L^{k}(V_1, \ldots, V_k; \ F)$ such that $\tilde A\circ \pi=A$.

We also know that
\begin{equation*}
\dim V_1^*\otimes\cdots\otimes V_k^*=\dim \mathcal L^k(V_1, \ldots, V_k; \ F)
\end{equation*}
Thus we just need to show that $\ker \tilde A=0$ to show that $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^{k}(V_1, \ldots, V_k; \ F)$ are isomorphic.

My Problem: I want to show the triviality of the kernel in a basis free manner. But here I am stuck.

Can anybody help?

 

[Opalg]

I am trying to prove the following.

Let $V_1, \ldots, V_k$ be finite dimensional vector spaces over a field $F$.
There is a natural isomorphism between $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^k(V_1, \ldots, V_k;\ F)$.

Define a map $A:V_1^*\times\cdots\times V_k^*\to \mathcal L^k(V_1, \ldots, V_k;\ F)$ as
\begin{equation*}
A(\omega_1, \ldots, \omega_k)(v_1, \ldots, v_k)=\omega(v_1)\cdots\omega_k(v_k)
\end{equation*}
for all $(\omega_1, \ldots, \omega_k)\in V_1^*\times\cdots\times V_k^*$.
It can be seen that $A$ is a multilinear map.
By the universal property of tensor product, there exists a unique linear map $\tilde A: V_1^*\otimes\cdots\otimes V_k^*\to \mathcal L^{k}(V_1, \ldots, V_k; \ F)$ such that $\tilde A\circ \pi=A$.

We also know that
\begin{equation*}
\dim V_1^*\otimes\cdots\otimes V_k^*=\dim \mathcal L^k(V_1, \ldots, V_k; \ F)
\end{equation*}
Thus we just need to show that $\ker \tilde A=0$ to show that $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^{k}(V_1, \ldots, V_k; \ F)$ are isomorphic.

My Problem: I want to show the triviality of the kernel in a basis free manner. But here I am stuck.

Can anybody help?

Instead of trying to show that $\tilde A$ is injective, I think it would be easier to show that it is surjective. This should somehow be equivalent to the fact that a $k\times k$ matrix is a sum of rank $1$ matrices.

 

[caffeinemachine]

Instead of trying to show that $\tilde A$ is injective, I think it would be easier to show that it is surjective. This should somehow be equivalent to the fact that a $k\times k$ matrix is a sum of rank $1$ matrices.

Hello Opalg,

Sorry for the late reply. I somehow forgot about this post.

I can show that $A$ is surjective by choosing a basis. I am getting more and more convinced that this cannot be done without choosing a basis.

 

[Opalg]

I am getting more and more convinced that this cannot be done without choosing a basis.

I tend to agree. For one thing, I believe that the result is false if the spaces are infinite-dimensional. So you somehow need to make use of the fact that the spaces are finite-dimensional, and the obvious way is to choose bases for them.