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I have come across the statement that the rank of the outer product of two vectors is always $1$, but why is that true?

Rodrigo de Azevedo
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2 Answers2

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Outer Product generates the matrix whose first row is $u_1(v_1,v_2,..,v_n)$ and the ith row is $u_i(v_1,v_2,..,v_n)$. So the rows are the vector $(v_1,v_2,..,v_n)$ multiplied by scalars. So this itself is the basis.Hence dimension is 1.

happymath
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  • What if ${\bf u} = {\bf 0}_n$? – Rodrigo de Azevedo Jun 28 '23 at 14:55
  • I guess that it is worth to note that the correct statement should be:

    Let $u \in \mathbb{R}^{m \times 1}$, $v \in \mathbb{R}^{n \times 1}$, then $$\operatorname{rk}(uv^T) = \begin{cases} 1, ~ u \ne \boldsymbol{0}_m \text{ and } v \ne \boldsymbol{0}_n \ 0, \text{ otherwise} \end{cases}$$

     – perepelart Jun 06 '25 at 07:52
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To put it into a more philosophical frame: The rank $r$ of a matrix $A$ is the minimal number of dyadic products $u_kv_k^\top$ required to express the matrix as

$$A=\sum_{k=1}^r u_kv_k^\top.$$

Obviously, such representations exist as $A=\sum_{k=1}^n a_ke_k^\top$ where $a_k$ are the columns of $A$ and $e_k$ the canonical basis vectors of length $n$.

So if your matrix is constructed as a dyadic product, it obviously has a representation as a sum of one dyadic product and thus rank 1.

The outer product in its usual meaning is the anti-symmetric tensor product or wedge product $u\wedge v=\frac12(u\otimes v-v\otimes u)$. This obviously is either zero if $u\sim v$ or has rank $2$.

Lutz Lehmann
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  • What if ${\bf u} = {\bf 0}_n$? – Rodrigo de Azevedo Jun 28 '23 at 14:56
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    Obvious trivial cases are obviously trivial. So the zero matrix obviously has a representation with zero dyadic products. @RodrigodeAzevedo – Lutz Lehmann Jun 28 '23 at 17:09
  • Trivial case is enough to render the statement in the question incorrect, sorry. – Rodrigo de Azevedo Jun 28 '23 at 17:13
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    The main statement remains correct and the examples after that are (as usual) intended to be non-trivial. @RodrigodeAzevedo – Lutz Lehmann Jun 28 '23 at 17:15
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    Logic does not bend to your wishes. The correct statement is that the rank is $\leq 1$. Perhaps you are decades past the annoying rigorous stage of one's mathematical education, but your relaxed attitude sets a bad example for the young yahoos first encountering the pedantic rigorous way of doing things – Rodrigo de Azevedo Jun 28 '23 at 17:18