I have come across the statement that the rank of the outerproduct of two vectors is always one - but why is that?


  • 1
    $\begingroup$ 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

2 Answers 2


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.


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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