= Separability of a tensor
{title2=$t=v\otimes w$}
In a two-factor <tensor product> over a <field>, a <pure tensor> is also called a separable or decomposable <tensor>: it can be expressed as one product $v\otimes w$. Every <tensor> is a finite sum of <pure tensors>, but a sum need not itself be a <pure tensor>.<Tensor Products I by Keith Conrad>[§3, printed p. 9: elementary, simple, decomposable, and pure tensors; Example 2.1, printed pp. 4–5: outer products do not fill the matrix space]{ref}
For <finite-dimensional vector spaces>, the <tensor-product coefficient matrix> gives a decisive test: a nonzero <tensor> is a <pure tensor> exactly when its <matrix rank> is one. Indeed, every column of $vw^T$ is a <scalar multiple> of $v$. Conversely, if all columns of a nonzero <matrix> are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=vw^T$. The zero <tensor> is also a <pure tensor>, for example $0\otimes w$, but its <matrix rank> is zero.
In $\mathbb R^2\otimes_{\mathbb R}\mathbb R^2$, the <tensor>
$$
t=e_1\otimes f_1+e_2\otimes f_2,\qquad C(t)=\begin{pmatrix}1&0\\0&1\end{pmatrix},
$$
is nonseparable: its <matrix rank> is two and its <determinant> is one, whereas $\det(vw^T)=0$. Directly, trying $t=(ae_1+be_2)\otimes(cf_1+df_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four <scalars> nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes(f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.
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