Understanding tensor products of vector spaces In the lecture we defined the tensor product of two finite dimensional vector spaces over a field by the universal property.
Now we claimed:

Let $\phi: U \times V \to W$ be bilinear, $(e_i)_{1 \leq i \leq m}$
  basis of $U$, $(f_j)_{1 \leq j \leq n}$ basis of $V$. 
Then $(W, \phi)$ is a tensor product of $U$ and $V$ iff $(\phi(e_i,
f_j))_{1 \leq i \leq m, 1 \leq j \leq n}$ is a basis of $W$.

Can anyone give me a proof of this? Thank you in advance.
 A: First assume that $\left(\phi(e_i,f_j)\right)_{1 \leq i \leq m, 1 \leq j \leq n}$ is a basis for $W$. To prove that $(W, \phi)$ is a tensor product, take a vector space $Z$ and a bilinear operator $h : U \times V \to Z$. Assume $h$ is defined as $h(e_i, f_j) = z_{ij} \in Z$: this uniquely determines the behaviour of $h$ on any $(u, v) \in U\times V$.
We have to show that there exists a unique linear operator $\hat{h} : W \to Z$ such that $\hat{h}\circ\phi = h$.
Note that if such $\hat{h}$ exists, $\hat{h}\left(\phi(e_i,f_j)\right) = h(e_i, f_j) = z_{ij}$ must hold. But, since $\left(\phi(e_i,f_j)\right)_{1 \leq i \leq m, 1 \leq j \leq n}$ is a basis for $W$, this uniquely defines a linear operator $\hat{h} : W \to Z$.
Let's show that $\hat{h}$ is the map we are looking for:
take $(u, v) \in U\times V$, $u = \sum_{i=1}^m \alpha_i e_i$, $v = \sum_{j=1}^n \beta_j f_j$.
Using the bilinearity of $h$ and $\phi$, and linearity of $\hat{h}$:
\begin{align}\hat{h}(\phi(u, v)) &= \hat{h}\left(\phi\left(\sum_{i=1}^m \alpha_i e_i, \sum_{j=1}^n\beta_j f_j\right)\right)\\ &= \hat{h}\left(\sum_{i=1}^m\sum_{j=1}^n\alpha_i\beta_j\phi(e_i, f_j)\right) \\ &= \sum_{i=1}^m\sum_{j=1}^n\alpha_i\beta_j\hat{h}\left(\phi(e_i, f_j)\right) \\ &= \sum_{i=1}^m\sum_{j=1}^n\alpha_i\beta_j h(e_i, f_j) \\ &= h\left(\sum_{i=1}^m \alpha_i e_i, \sum_{j=1}^n\beta_j f_j\right) \\ &= h(u, v)\end{align}
This proves that $(W, \phi)$ is indeed a tensor product of $U$ and $V$.
Now note that $\dim W = \dim U \cdot \dim V = mn$. Since all tensor products are isomorphic (because it is defined with a universal property), it follows that $\dim U \cdot \dim V$ is the dimension of all tensor products of $U$ and $V$.
For the reverse implication, assume that $W$ is a tensor product of $U$ and $W$. To prove that $\left(\phi(e_i,f_j)\right)_{1 \leq i \leq m, 1 \leq j \leq n}$ is its basis, it suffices to prove linear independence.
Let $({e_i}^*)_{1\leq i\leq m}$ be the dual basis for $U$ and $({f_j}^*)_{1\leq j\leq n}$ be the dual basis of $V$. Now define bilinear functionals $\psi_{ij} : U \times V \to \mathbb{F}$ as $\psi_{ij}(u, v) = {e_i}^*(u){f_j}^*(v)$. Notice that $\psi_{ij}(e_k, f_l) = \delta_{ik} \delta_{jl}$. Because $W$ is a tensor product of $U$ and $W$, there exist unique $\hat{\psi}_{ij} : W \to \mathbb{F}$ such that the universal property holds.
Now assume $0 = \sum_{i=1}^m\sum_{j=1}^n\lambda_{ij}\phi(e_i, f_j)$ for some $\lambda_{ij} \in \mathbb{F}$. Apply $\hat{\psi}_{kl}$ to both sides of the equality:
$$0 = \hat{\psi}_{kl}\left(\sum_{i=1}^m\sum_{j=1}^n\lambda_{ij}\phi(e_i, f_j)\right) = \sum_{i=1}^m\sum_{j=1}^n\lambda_{ij}\hat{\psi}_{kl}\left(\phi(e_i, f_j)\right) = \lambda_{kl}$$
So $\lambda_{kl} = 0$, for arbitrary $1\leq k\leq m, 1\leq l\leq n$, which proves linear independence.
Thus, $\left(\phi(e_i,f_j)\right)_{1 \leq i \leq m, 1 \leq j \leq n}$ is a basis for $W$.
