 Tensor-product basis

If $(e_1,\dots ,e_m)$ and $(f_1,\dots ,f_n)$ 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.^[ref]

For a proof outline, bilinearity expands each pure tensor asso 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.