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 $1$.


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.


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 $r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.