# Isomorphism between dual space and bilinear forms

Studying from Roman’s Advanced Linear Algebra, I want to prove that $$U^* \otimes V^* \cong (U \otimes V)^* \cong \hom(U,V;\mathbb{F})$$ The author proves that $U^* \otimes V^* \cong (U \otimes V)^*$ by showing that there is an unique linear transformation

$$\theta:U^* \otimes V^* \to (U \otimes V)^*$$

defined by $\theta(f \otimes g)=f \odot g$ where $(f \odot g)(u \otimes v)=f(u)g(v)$

I want to prove that $U^* \otimes V^* \cong \hom(U,V;\mathbb{F})$. I tried to use the universal property by fixing $f \in U^*, g\in V^*$ and defining $S: U^*\times V^* \to \hom(U,V;\mathbb{F})$ as $S(f,g)=F_{f,g}$, where $F_{f,g}(u,v)=f(u)g(v)$, but I got stuck.

#### Solutions Collecting From Web of "Isomorphism between dual space and bilinear forms"

It is much easier to prove that $(U\otimes V)^{\ast}\cong \mathrm{hom}(U,V;\mathbb{F})$. This is because any bilinear map $F:U\times V\to\mathbb{F}$ induces a unique homomorphism $F’:U\otimes V\to\mathbb{F}$ and vice versa. Notice that $F’\in (U\otimes V)^{\ast}$; the universal property gives us the bijection.