Rank of the outer product of two vectors

Question

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

Answer 1

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.

Answer 2

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.

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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 $2$.