TITLE: Rank of the Outer Product of two Vectors
QUESTION [16 upvotes]: I have come across the statement that the rank of the outerproduct of two vectors is always one - but why is that? 
Thanks

REPLY [7 votes]: 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 $2$.