Basis for tensor products 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?
 A: Construct a set $\{\phi_{i,j}\}$ of linear forms on the tensor product which is a dual basis to the family $\{e_i \otimes f_j\}$ (in the sense that for each pair $\left(i, j\right) \in I \times J$, the map $\phi_{i,j}$ sends $e_i \otimes f_j$ to $1$ while sending all $e_{i'} \otimes f_{j'}$ with $\left(i', j'\right) \neq \left(i, j\right)$ to $0$). That will immediately imply linear independence.
If $\{\psi_i\}$ is a dual basis to your basis $\{e_i\}$ of $V_1$ and $\{\rho_j\}$ is a dual basis to your basis $\{f_j\}$ of $V_2$, then you can consider the map $\phi_{i,j} = \psi_i\otimes \rho_j:V_1\otimes V_2\to k\otimes k\cong k$ for each pair $\left(i, j\right) \in I \times J$. This map $\phi_{i,j}$ sends $e_i \otimes f_j$ to $1$ while sending all $e_{i'} \otimes f_{j'}$ with $\left(i', j'\right) \neq \left(i, j\right)$ to $0$.
A: 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.
A: I'll only show that it spans $V_1\otimes V_2$.
Consider $U=\operatorname{span}\{e_i\otimes f_j:i\in I,j\in J\}$, with $\{e_i:i\in I\}$ basis for $V_1$ and $\{f_j:j\in J\}$ basis for $V_2$.
We have trivially that $U$ is a subspace of $V_1\otimes V_2$.
Let $h:V\times W \to V\otimes W$ be the bilinear map associated to the tensor product via its universal property. That is, $h(v,w)=v\otimes w$.
It is clear that $\operatorname{im}(h)=\operatorname{span}\{h(e_i,f_j)=e_i\otimes f_j: i\in I,j\in J\}=U$. But, in this case, $(U,h)$ also satisfies the universal property of tensor product. Thus, $U\cong V_1\otimes V_2$. Then, $U=V_1\otimes V_2$.
Thus, $V_1\otimes V_2$ is generated by $\{e_i\otimes f_j:i\in I,j\in J\}$.
