= Two-factor tensor rank
{title2=$\operatorname{rank}_{\otimes}(t)=\operatorname{rank}C(t)$}
The <two-factor tensor rank> is the smallest number of <pure tensors> whose sum equals $t$, with zero allowed as the empty sum. For two <finite-dimensional vector spaces> over the same <field>, it equals the <matrix rank> of the <tensor-product coefficient matrix>. A sum of $r$ <outer products> has <column space> contained in the span of their $r$ first factors, so its <matrix rank> is at most $r$. Conversely, choose a <basis> $u_1,\ldots,u_r$ for the <column space>, express each column in that <basis>, and collect the coefficients into row <vectors> $w_k^T$. This yields $C=\sum_{k=1}^ru_kw_k^T$, hence a sum of $r$ <pure tensors>.
The <pure tensor> $(2,1)\otimes(-1,3)$ therefore has <two-factor tensor rank> one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has <two-factor tensor rank> two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the <real numbers>, the <singular value decomposition> supplies a useful decomposition $C=\sum_k\sigma_ku_kv_k^T$, with one nonzero term per <matrix rank>. The two-factor rank equality does not extend to three or more factors by merely choosing one <matrix> layout.
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