Rank of Matrix A with Kronecker Symbol and Sum Condition

[zed123]

helloo
while working on a combinatorics problem I have found the following result:

let A=(a_{ij})_{1\leq i,j\leq2n+1} where n is a positive integer , be a real Matrix such that :
i) a_{ij}^2=1-\delta_{ij} where \delta is the kronecker symbol
ii) \forall i \displaystyle{ \sum_{j=1}^{2n+1}a_{ij}=0}
then rankA=2n
any idea ?

 

[Hurkyl]

Er, what are you asking? Did you mean that you have observed it in some cases, and are wondering if it's true in general?

Can you describe qualitatively what such a matrix looks like?


I feel like induction is the most likely way to go about it, if it is true. How many particular examples have you tested, and of what sizes? Do you have a conjecture for how things behave if the dimension is even instead of odd?

(Or, maybe you could explain the combinatorics problem you were solving; maybe it's easier to do that problem than it is to work with this matrix)

 

[HallsofIvy]

For n= 1, that is saying that
A= \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 0\end{bmatrix}
What is the rank of that matrix?

 

[micromass]

For n= 1, that is saying that
A= \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 0\end{bmatrix}
What is the rank of that matrix?

Not exactly For n=1, it's a matrix that looks like this

A= \begin{bmatrix}0 & 1 & -1 \\ 1 & 0 & -1\\ -1 & 1 & 0\end{bmatrix}

So the entries on the diagonal must be 0, and all the other entries are 1 and -1. But the sum of every row must be 0.

It is very easy to see that such a matrix cannot have full rank (the sum of all the columns is 0, so the columns cannot be linear independent). So the rank is at most 2n. That it's exactly 2n is a bit harder...