= Tensor-product coefficient matrix
{title2=$C(t)=(c_{ij})$}
After choosing <ordered bases>, identify a <tensor> $t=\sum_{i,j}c_{ij}e_i\otimes f_j$ with its <matrix> $C(t)=(c_{ij})$. This is a <linear isomorphism> between $V\otimes_F W$ and the <vector space> of $m$-by-$n$ <matrices>: each <tensor-product basis> element becomes the <matrix> with one entry equal to one and all others zero. In particular, a <pure tensor> has the <outer product> coefficient formula
$$
C(v\otimes w)=vw^T,\qquad C\bigl((2,1)\otimes(-1,3)\bigr)=\begin{pmatrix}-2&6\\-1&3\end{pmatrix}.
$$
Here $v,w$ on the right denote column <coordinate vectors>. The identification depends on the chosen <bases>, while the <tensor> itself does not. Over the <complex numbers>, the same formula uses the ordinary <transpose> $w^T$, with no <complex conjugation>; replacing it by the <conjugate transpose> $w^*$ would conjugate the <scalar> in the second input, violating <bilinearity>.
\Image[/tensor-product-coefficient-grids.png]
{title=A pure tensor has an outer-product coefficient grid, while the identity grid needs two pure tensors}
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