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...

 

[blue_raver22]

Based on the information provided, it seems like the matrix A has some special properties. The first condition given, a_{ij}^2=1-\delta_{ij}, indicates that the matrix is likely symmetric and has a diagonal of all 1's except for the main diagonal, which is filled with -1's. This can also be seen as a Hadamard matrix, which has a well-known property of having a rank equal to its order. In this case, the order of A is 2n+1, so the rank would be 2n+1.

The second condition states that the sum of each row of A is equal to 0. This means that the rows of A are linearly dependent, which would decrease the rank. However, since the matrix is symmetric and has a special structure, it is possible to show that the rank is still equal to 2n.

One approach to proving this would be to use the properties of the Kronecker symbol and the sum condition to construct a basis for the null space of A. This basis would consist of vectors that satisfy both conditions and can be used to show that the rank of A is 2n.

In conclusion, the rank of matrix A with the given conditions is 2n, which is equal to its order. This is a interesting result and could potentially have applications in combinatorics and linear algebra. Further investigation and analysis of this matrix could lead to a better understanding of its properties and potential uses.