Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$?

I have a couple of questions about tensor products:

Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$?

Why is an element of $V^{*\otimes m}\otimes V^{\otimes n}$ the same thing as a multilinear map $V^m \to V^{\otimes n}$?

What is the general formulation of this principle?

Solutions Collecting From Web of "Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$?"

The result is generally wrong for infinite-dimensional spaces: see this question.

For finite dimensional space $V$, let’s build an isomorphism $f : V^* \otimes W \to \hom(V,W)$ by defining

$$f(\phi \otimes w)(v) = \phi(v) w$$

This clearly defines a linear map $V^* \otimes W \to \hom(V,W)$ (it’s bilinear in $V^* \times W$). Reciprocally, take a basis $(e_i)$ of $V$, then define $g : \hom(V,W) \to V^* \otimes W$ by:

$$g(u) = \sum_i e_i^* \otimes u(e_i)$$

Where $(e_i^*)$ is the dual basis to $(e_i)$ (I will use a few of its properties in $\color{red}{red}$ below). This is well-defined because $V$ is finite-dimensional (the sum is finite). Let’s check that $f$ and $g$ are inverse to each other:

  • For $u : V \to W$, $$f(g(u))(v) = \sum e_i^*(v) u(e_i) = u \left( \sum e_i^*(v) e_i \right) \color{red}{=} u(v)$$ and so $f(g(u)) = u$.

  • For $\phi \otimes w \in V^* \otimes W$, $$g(f(\phi \otimes w)) = \sum e_i^* \otimes f(\phi \otimes w)(e_i) = \sum e_i^* \otimes \phi(e_i) w = \sum \phi(e_i) e_i^* \otimes w \color{red}{=} \phi \otimes w$$

And so $f$ and $g$ are isomorphisms, inverse to each other.

It is known that for finite dimensional $V$, then $(V^*)^{\otimes m} = (V^{\otimes m})^*$. Then an element of $V^{* \otimes m} \otimes V^{\otimes n}$ is an element of $(V^{\otimes m})^* \otimes V^{\otimes n} = \hom(V^{\otimes m}, V^{\otimes n})$. So by definition / universal property of the tensor product, it’s a multilinear map $V^m \to V^{\otimes n}$.