TITLE: Basis for tensor products
QUESTION [19 upvotes]: Suppose $V_1$ and $V_2$ are $k$-vector spaces with bases $(e_i)_{i \in I}$ and $(f_j)_{j \in J}$, respectively. I've seen the claim that the collection of elements of the form $e_i \otimes f_j$ (with $\left(i,j\right) \in I \times J$) forms a basis for $V_1 \otimes V_2$. But I seem to get stuck with the proof.
My question: What's the easiest way to see that the above set is indeed linearly independent?

REPLY [8 votes]: First, take linear functions $\delta_i$ defined on $V_1$ and $\varphi_i$ defined on $V_2$ such that
$$\delta_i(e_i)=1$$
$$\delta_i(e_j)=0 \ \ i\neq j$$
$$\varphi_i(f_i)=1$$
$$\varphi_i(f_j)=0 \ \ i\neq j.$$
Now, consider the bilinear form $\phi_{ij}:V_1\times V_2\to \mathbb{R}$ given by $\phi_{ij}(v_1,v_2) = \delta_i(v_1)\cdot\varphi_j(v_2)$.
The universal property gives a linear function $\widetilde{\phi_{ij}}:V_1\otimes V_2\to \mathbb{R}$ that maps $e_i\otimes f_j$ to $1$ and any other $e_{i'}\otimes f_{j'}$ with $i\neq i'$ or $j\neq j'$ to $0$, then they are linearly independent.