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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 $

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    $\begingroup$ This result seems pretty nice, in order to complety this fact, I'd prove this behaviour for arbitrary $x$ and $m$ and parameters $a,\,b,\,c$. After this comes a problem: if this result has no practical applications, then it remains nothing but a amuzing fact. If it does have such applications (e.g., for some numerical method) and if this is a new result, then I suggest publishing it somewhere - and, of course, name this type of matrix as you wish=) $\endgroup$ – TZakrevskiy Jan 22 '15 at 16:06
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    $\begingroup$ This looks cool me to me. I agree with @TZakrevskiy. If you can generalized it and can find some interesting properties, then at least you can uploaded it in arXive (History & Overview section). There may be someone can advice you about it. $\endgroup$ – Krish Jan 22 '15 at 16:15
  • $\begingroup$ Why are you switching from a b c to p q r when showing $A^x$? Is it to make me work harder? $\endgroup$ – dustin Jan 22 '15 at 16:19
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    $\begingroup$ no thats to show that the elements in $ A $ and in $ A^x $ are not same $\endgroup$ – Rijul Gupta Jan 22 '15 at 16:20
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    $\begingroup$ It's not true for $5\times 5$. For example, try $a=b=c=1$. The top row of $A^2$ is $[5,5,7,5,5]$, and the $(3,3)$ entry is $13 \ne 5 + 7 + 5$. $\endgroup$ – Robert Israel Jan 22 '15 at 16:29
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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.

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    $\begingroup$ Not just of the additive group. It's actually a subring in the case $n=3$. Unfortunately it doesn't work for higher $n$. $\endgroup$ – Robert Israel Jan 22 '15 at 16:25
  • $\begingroup$ Can't I publish it as it stands? and maybe later if I found some application publish that too but at the same time allowing others to work on it to find some application. $\endgroup$ – Rijul Gupta Jan 27 '15 at 20:11
  • $\begingroup$ @rijulgupta In my opinion it's not likely to get published without further application. If you disagree then of course you are free to try. I may be wrong. $\endgroup$ – Matt Samuel Jan 27 '15 at 20:27
  • $\begingroup$ @MattSamuel: thanks for the input, I'll try both finding application and publishing and update when something new happens. $\endgroup$ – Rijul Gupta Jan 28 '15 at 4:14
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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 $$ S^{-1}AS = \left( \begin{array}{ccc} 2b & 0 & \sqrt 2 a & 2c & 0\\ 0 & 0 & 0 & 0 & 0\\ \sqrt 2 a & 0 & a+b+c & \sqrt 2a & 0\\ 2c & 0 & \sqrt 2a & 2b & 0\\ 0 & 0 & 0 & 0 & 0\end{array} \right) $$

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  • $\begingroup$ I am sorry I dont really understand some of the stuff including how is this related to my question, my question was just that whether that particular type of matrix has been already found. please elaborate and explain a little more. $\endgroup$ – Rijul Gupta Jan 22 '15 at 16:40
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    $\begingroup$ This doesn't answer your question at all, in retrospect. However, it's a useful way to see why your matrix should do what it does. While your matrix is cool, I don't think it tells us anything mathematical that we didn't already know. $\endgroup$ – Omnomnomnom Jan 22 '15 at 16:44
  • $\begingroup$ oh yes, that's true. I just found it pretty amusing so I thought I should adk about it, thanks for the input $\endgroup$ – Rijul Gupta Jan 22 '15 at 16:53
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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 $$\pmatrix{a \; 1_{n,n} & b \; 1_{n,1} & c\; 1_{n,n} \cr b \; 1_{1,n} & n(a+b) & b \; 1_{1,n}\cr c\; 1_{n,n} & b \; 1_{n,1} & a\; 1_{n,n}\cr}$$

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