Checkboard matrix, brand new or old?

Ok so what I found was a square matrix of order $n×n$ where $n$ follows $2m+1$ and $m$ is a natural number

the pattern these matrices follow is as follows:

for a $3×3$ matrix: $$ A = \left( \begin{array}{ccc} b & a & c \\ a & a+b+c & a \\ c & a & b \end{array} \right)$$ and the cool thing is that $$ A^x = \left( \begin{array}{ccc} q & p & r \\ p & p+q+r & p \\ r & p & q \end{array} \right)$$ where $ x $ is any natural number

Now for $5×5$ it goes like $$ A = \left( \begin{array}{ccc} b & b & a & c & c\\ b & b & a & c & c\\ a & a & 2b+a+2c & a & a\\ c & c & a & b & b\\ c & c & a & b & b\end{array} \right)$$ and again $$ A^x = \left( \begin{array}{ccc} q & q & p & r & r\\ q & q & p & r & r\\ p & p & 2q+p+2r & p & p\\ r & r & p & q & q\\ r & r & p & q & q\end{array} \right)$$

Once more for a $7×7$ matrix we have $$ A = \left( \begin{array}{ccc} b & b & b & a & c & c & c\\ b & b & b & a & c & c & c\\ b & b & b & a & c & c & c\\ a & a & a & 3b+a+3c & a & a & a\\ c & c & c & a & b & b & b\\ c & c & c & a & b & b & b\\ c & c & c & a & b & b & b\\ \end{array} \right)$$ and then $$ A^x = \left( \begin{array}{ccc} q & q & q & p & r & r & r\\ q & q & q & p & r & r & r\\ q & q & q & p & r & r & r\\ p & p & p & 3q+p+3r & p & p & p\\ r & r & r & p & q & q & q\\ r & r & r & p & q & q & q\\ r & r & r & p & q & q & q\\ \end{array} \right)$$

and so on and so forth, I didn't find whether this has already been observed and neither a name for this so I am calling this checkboard matrix, because it looks like that.

I just have one question, has this been found already, if yes please give details.

Addendum: Thanks to Robert Israel for pointing out the error, I have made some changes, please check into it. The pattern for the middle element is $ m (b+c)+a $

Solution 1:

For what it's worth, the set of matrices of this form form a subgroup of the additive group of the matrix ring. As Robert points out, they also form a subring for $n=3$, which is interesting.

As to whether or not anyone has ever constructed such a matrix and found this property: it's possible that they have, but as Pedro said it is not worth publishing a paper entirely about these matrices simply because they look pretty and behave well under exponentiation. Maybe if you could do some more neat tricks with them you could get it published in a recreational mathematics magazine. Published papers in journals tend to have results that are applicable to something, even if that thing is pure math. This is truly recreational mathematics because there is no purpose other than observing something interesting.

Solution 2:

For the $3 \times 3$ case, you have the commutative ring (with identity) generated by $\pmatrix{0 & 1 & 0\cr 1 & 0 & 1\cr 0 & 1 & 0\cr}$. For the $2n+1 \times 2n+1$ case with $n > 1$, you could take the commutative ring (without identity) generated by the block matrices $\pmatrix{0_{n,n} & 0_{n,1} & 1_{n,n}\cr 0_{1,n} & n & 0_{1,n}\cr 1_{n,n} & 0_{n,1} & 0_{n,n}\cr}$ and $\pmatrix{0_{n,n} & 1_{n,1} & 0_{n,n}\cr 1_{1,n} & 0 & 1_{1,n}\cr 0_{n,n} & 1_{n,1} & 0_{n,n}\cr}$ (where the blocks are of sizes $n$, $1$ and $n$, and $0_{i,j}$ or $1_{i,j}$ denotes an $i \times j$ block filled with $0$'s or $1$'s).

EDIT: This gives you matrices of the form $$\pmatrix{a \; 1_{n,n} & b \; 1_{n,1} & c\; 1_{n,n} \cr b \; 1_{1,n} & n(a+b) & b \; 1_{1,n}\cr c\; 1_{n,n} & b \; 1_{n,1} & a\; 1_{n,n}\cr}$$

Solution 3:

Here's my attempt to explain the pattern with stuff already known:

We can express the matrix as $$ A = \pmatrix{ bX & ax & cX\\ ax^T & (a+b+c) & ax^T\\ cX & ax & bX} $$ Where $x = (1,\dots,1)^T \in \Bbb R^n$, and $X = xx^T$. Note that $Xx = \|x\|^2x = nx$.

Extend $u_1 = x/\sqrt{n}$ into an orthonormal basis $u_1,\dots,u_n$. Let $U$ be the $n \times n$ matrix with columns $u_i$.

Apply the change of basis $S^{-1}AS = S^TAS$ where $$ S = \pmatrix{U\\&1\\&&U} $$ We find $$ S^{-1}AS = \pmatrix{ nbE_{11} & \sqrt n a e_1 & cn E_{11}\\ a\sqrt n e_1^T & (a+b+c) & a\sqrt n e_1^T\\ cnE_{11} & a \sqrt n e_1 & b n E_{11} } $$ where $e_1 = (1,0,\dots,0)^T$ is the first standard basis vector, and $E_{11} = e_1e_1^T$.

You should find that this new matrix has the same properties, and quite a bit less of the mystery.

So, in the $5 \times 5$ case, we're left with $$ S^{-1}AS = \left( \begin{array}{ccc} 2b & 0 & \sqrt 2 a & 2c & 0\\ 0 & 0 & 0 & 0 & 0\\ \sqrt 2 a & 0 & a+b+c & \sqrt 2a & 0\\ 2c & 0 & \sqrt 2a & 2b & 0\\ 0 & 0 & 0 & 0 & 0\end{array} \right) $$