Basis for tensor products

Solution 1:

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 $0$.

Solution 2:

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.