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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?

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    $\begingroup$ This depends on the context. And what is the outer product, the wedge product $a\wedge b$ or the dyadic product $a\cdot b^\top$? $\endgroup$ Feb 25, 2014 at 10:49
  • $\begingroup$ It would be the dyadic product. The context was just a remark on the slides we received. $\endgroup$
    – user66280
    Feb 26, 2014 at 8:20
  • $\begingroup$ @RodrigodeAzevedo The massive amount of edits you have been doing, including creating (?) the rank one matrix tag is flooding the front page and the tag is frankly unnecessary. $\endgroup$ Jun 28, 2023 at 15:08
  • $\begingroup$ @CameronWilliams 9 people disagree with you $\endgroup$ Jun 28, 2023 at 15:13

2 Answers 2

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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.

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  • $\begingroup$ What if ${\bf u} = {\bf 0}_n$? $\endgroup$ Jun 28, 2023 at 14:55
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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 $2$.

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  • $\begingroup$ What if ${\bf u} = {\bf 0}_n$? $\endgroup$ Jun 28, 2023 at 14:56
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    $\begingroup$ Obvious trivial cases are obviously trivial. So the zero matrix obviously has a representation with zero dyadic products. @RodrigodeAzevedo $\endgroup$ Jun 28, 2023 at 17:09
  • $\begingroup$ Trivial case is enough to render the statement in the question incorrect, sorry. $\endgroup$ Jun 28, 2023 at 17:13
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    $\begingroup$ The main statement remains correct and the examples after that are (as usual) intended to be non-trivial. @RodrigodeAzevedo $\endgroup$ Jun 28, 2023 at 17:15
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    $\begingroup$ 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 $\endgroup$ Jun 28, 2023 at 17:18

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