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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2This depends on the context. And what is the outer product, the wedge product $a\wedge b$ or the dyadic product $a\cdot b^\top$? – Lutz Lehmann Feb 25 '14 at 10:49
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It would be the dyadic product. The context was just a remark on the slides we received. – Feb 26 '14 at 08:20
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@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. – Cameron Williams Jun 28 '23 at 15:08
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@CameronWilliams 9 people disagree with you – Rodrigo de Azevedo Jun 28 '23 at 15:13
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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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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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1Obvious 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
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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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1The 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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1Logic 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