# Rank of outer sum

Suppose we have two vectors.

$u=\begin{bmatrix}u_1\\\vdots\\u_n\end{bmatrix}$ and $v=\begin{bmatrix}v_1&\dots&v_n\end{bmatrix}$ then we know rank of $uv$ is $1$.

1. What is the rank of $\begin{bmatrix}u_1+v_1&\dots&u_1+v_n\\\vdots&\ddots&\vdots\\u_n+v_1&\dots&u_n+v_n\end{bmatrix}$ for the cases where $u_i,v_j\in\Bbb R$ and $u_i,v_j\in\Bbb F_q$?

2. Suppose $u_i,v_j\in\Bbb Z$ then let $a_{ij}=u_i+v_j\bmod 2$ where $a_{ij}$ is $0/1$. Consider $\begin{bmatrix}a_{11}&\dots&a_{1n}\\\vdots&\ddots&\vdots\\a_{n1}&\dots&a_{nn}\end{bmatrix}$ for the case where $a_{ij}$ is considered to be in $\Bbb Z$ and for the case $a_{ij}$ is in $\Bbb F_2$?

The crucial part for me is where $0/1$ is in $\Bbb Z$ for 2. Is rank (rank over rational numbers) here also bound by $2$?

1. In any field: take away the first row from every row, and get a matrix with rows 2 to n all the same. Thus, the rank is at most 2. It can be 2, e.g. in case $n=2$ the determinant is:

$$(u_1+v_1)(u_2+v_2)-(u_1+v_2)(u_2+v_1)=-(u_1-u_2)(v_1-v_2)$$

which is $\ne 0$ if $u_1\ne u_2$ and $v_1\ne v_2$. The rank can also be 1 (imagine $u\ne 0, v=0$) or 0 ($u=v=0$).

This applies to any field, including $\mathbb F_2$.

1. Take the entries $u_i+v_j\pmod 2$ but keep them in $\mathbb Z$ and then calculate rank in $\mathbb Q$... Note that, in the final matrix, we can only have up to two different rows:

$$[v_1\pmod 2,\ldots,v_n\pmod 2]\;\text{where}\;u_i\;\text{is even}$$

or

$$[1-v_1\pmod 2,\ldots, 1-v_n\pmod 2]\;\text{where}\;u_i\;\text{is odd}$$

thus the rank in this case is also up to 2. It can be 2 for $\begin{pmatrix}0+0\pmod 2&0+1\pmod 2\\1+0\pmod 2&1+1\pmod 2\end{pmatrix}$, it could be 1 if $v=0$ and at least one $u_i\not\equiv 0\pmod 2$ and it could be 0 for $u=v=0$.

• you missed the crucial part in 2. where $0/1$ is in $\Bbb Z$. – T.... Jan 10 '18 at 10:19
• Ah, thanks for clarification. So you take mod 2, but keep the entries in $\mathbb Z$, and you want the rank over $\mathbb Q$? It was not clear from your post. I will think about it and either update or delete my answer soon. – user491874 Jan 10 '18 at 10:52
• Yes......................... – T.... Jan 10 '18 at 10:53
• I've updated it, hope to have covered all your questions. – user491874 Jan 10 '18 at 11:07
• I also found a proof similar to yours but a little less succinctly done. – T.... Jan 10 '18 at 11:35