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I'm independently studying Stephen Roman's Advanced Linear Algebra, and I came across a line of reasoning that appears obvious but that I don't understand, and was hoping someone might help me clarify.

'$\textbf{The Rank of a Decomposable Tensor}$

If $\{u_i \; | \; i \in I\}$ is a basis for $U$ and $\{v_j \; | \; j \in J\}$ is a basis for $V$, then any decomposable vector has the form $$ u \otimes v = \sum_{i,j} r_i s_j (u_i \otimes v_j)$$ Hence, the rank of a decomposable vector is $1$, since the rank of a matrix whose $(i, j)$th entry is $r_i s_j$ is $1$.'

I do not understand how a matrix having $(i, j)$th entry $r_i s_j$ implies that it is rank $1$.

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The rank is one provided that $u\otimes v \neq 0$. In particular, $u, v \neq 0$.

Consider the matrix

$$A = \left[\begin{array}{cccc} r_1s_1 & r_1s_2 & \dots &r_1s_n\\ r_2s_1 & r_2s_2 & \dots & r_2s_n\\ \vdots & \vdots & \ddots&\vdots\\ r_ms_1 & r_ms_2 & \dots & r_ms_n \end{array}\right].$$

Note that

\begin{align*} \operatorname{rank}(A) &= \dim\operatorname{Col}A\\ &= \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1s_1\\ \vdots\\ r_ms_1\end{array}\right], \dots, \left[\begin{array}{c} r_1s_n\\ \vdots\\ r_ms_n\end{array}\right]\right\}\\ &= \dim\operatorname{span}\left\{s_1\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right], \dots, s_n\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\ &= \dim\operatorname{span}\left\{\left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right]\right\}\\ &= 1 \end{align*}

as $u = \sum r_iu_i$ where $u_i$ form a basis and $u \neq 0$, so $r_i \neq 0$ for some $i$.

Another way of seeing it uses the notion of outer product (see here and here). Let $v^T = [r_1\ \dots\ r_m]$ and $w^T = [s_1\ \dots\ s_n]$, then the outer product of $v$ and $w$ is

$$vw^T = \left[\begin{array}{c} r_1\\ \vdots\\ r_m\end{array}\right][s_1\ \dots\ s_n] = \left[\begin{array}{cccc} r_1s_1 & r_1s_2 & \dots &r_1s_n\\ r_2s_1 & r_2s_2 & \dots & r_2s_n\\ \vdots & \vdots & \ddots&\vdots\\ r_ms_1 & r_ms_2 & \dots & r_ms_n \end{array}\right] = A.$$

As $A$ can be written as the sum of one outer product and no fewer (because $A$ is not the zero matrix), $A$ has rank one.

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Let $r$ be the column vector with entries $r_i$ and let $s$ be the column vector with entries $s_j$. Then if $M=rs^T$, we have $M_{i,j}=r_is_j$. Clearly $M$ has rank one, since its column are all multiples of $r$.

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