How to show using the universal property that $V\otimes V^{\ast} \cong \mathcal{L}(V,V)$?

Question

Let $V$ a vector space of finite dimension and $V^{\ast}$ its dual space.

How to use the universal property to show that $V\otimes V^{\ast} \cong \mathcal{L}(V,V)?$

I just know that I can construct the function $ g : V\times V^{\ast} \to \mathcal{L}(V,V)$ as: $$g(v,f)w := vf(w).$$

Bow how to use universal property to conclude the exercise? I think that the problem is that I did understand very well what the universal property tells.

Answer 1

By the universal property you have a set theoretic map $l : V \times V^* \to V \otimes V^*$, $l(v,f) = v \otimes f$, and a linear map $\phi : V \otimes V^* \to \mathcal{L}(V,V)$ such that $\phi \circ l = g$. Hence $\phi$ takes $v\otimes f \mapsto vf(\cdot)$. Show $\phi$ is an isomorphism.

Answer 2

The universal property of the tensor product applied here says that a $k$-bilnear map $f:V\times V^*\rightarrow W$ factors by unique linear map $V\otimes V^*$.

Let $(e_1,..,e_n)$ be a basis of $V$, you have a map $p: V\times V^*\rightarrow L(V,V)$ defined by $p(e_i\otimes e_j^*)(e_k)=e_j^*(e_k)e_i$. You can define $g(f):L(V,V)\rightarrow W$ as follows: , for every $h\in L(V,V)$, $g(f)(h)=f(\sum h(e_i)\otimes e_i^*)$. You have $g(f)\circ p=f$. Then shows that $g(f)$ is unique.