If I have $e_1 .. e_n$ - basis of vector space $V$ and $g_1 .. g_m$ - basis of $W$ and I know that system ${e_i \otimes g_j}$ - basis of $V \otimes W$, how I can prove, that if I fix another basis $e'_1 .. e'_n$ of the space $V$, then system ${e'_i \otimes g_j}$ also will be a basis of $V \otimes W$?
I don't find easy way to prove it. I tried to prove linear independence. $$\sum\limits_{i,j} \gamma_{ij}e'_i \otimes g_j = 0$$ and I find, that coefficient at term $e_i \otimes g_j$ is $\gamma_{ij} (\alpha_{i1} + .. + \alpha_{i1})$. And we know that $\gamma_{ij} (\alpha_{i1} + .. + \alpha_{i1}) = 0$, because ${e_i \otimes g_j}$ is basis. After this step I don't have idea what to do next? Please help me. Maybe you know how to finish this reasoning ir maybe you know textbooks, where I can find this proof.
It's simpler to go for proving the generating property: since $e_i'$ is a basis, we can express each $e_j$ by a linear combination of $e_i'$, so $e_j\otimes g$ is in the span of $e_i'\otimes g$ elements for every $g$, thus $e_i'\otimes g_k$ generates the tensor space, and contains the same number of elements as $e_j\otimes g_k$.