# Rank of the outer product of two vectors

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

• This depends on the context. And what is the outer product, the wedge product $a\wedge b$ or the dyadic product $a\cdot b^\top$? Feb 25, 2014 at 10:49
• It would be the dyadic product. The context was just a remark on the slides we received.
– user66280
Feb 26, 2014 at 8:20
• @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. Jun 28, 2023 at 15:08
• @CameronWilliams 9 people disagree with you Jun 28, 2023 at 15:13

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.

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

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 $2$.

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