= Tensor-product basis
{title2=$\{e_i\otimes f_j\},\quad\dim(V\otimes_F W)=\dim V\dim W$}
If $(e_1,\ldots,e_m)$ and $(f_1,\ldots,f_n)$ are <bases> of <finite-dimensional vector spaces> $V,W$ over $F$, then the $mn$ products $e_i\otimes f_j$ form a <basis> of $V\otimes_F W$. There is one independent <coordinate> for every pair of indices.<Linear Algebra Done Right by Sheldon Axler (4th edition)>[§9D, Theorems 9.72 and 9.74, printed pp. 372–373: dimension and tensor-product basis]{ref}
For a proof outline, <bilinearity> expands each <pure tensor> as
$$
\left(\sum_i v_ie_i\right)\otimes\left(\sum_j w_jf_j\right)=\sum_{i,j}v_iw_j(e_i\otimes f_j),
$$
so these products form a <spanning set>. For <linear independence>, take the <coordinate functionals> $e^k,f^\ell$ from the two <dual bases>. The <bilinear map> $(v,w)\mapsto e^k(v)f^\ell(w)$ induces a <linear functional> $\lambda_{k\ell}$ on the <tensor product>, satisfying $\lambda_{k\ell}(e_i\otimes f_j)=\delta_{ki}\delta_{\ell j}$. Applying it to a proposed zero <linear combination> selects just $c_{k\ell}$, forcing every coefficient to vanish. Thus <dimensions> multiply, rather than add. This finite-dimensional count includes a zero factor, whose <tensor product> has <dimension> zero.
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