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If I have that a matrix $A \in \mathbb{R}^{n \times n}$ is centered and m-banded i.e.

$\textit{m-banded:}$

there is an index $l$ such that $$a_{i,j}=0 \hspace{4mm} if \hspace{2mm}j \notin[i-l,i-l+m]$$

$\textit{centered}$: $$m=2s, \hspace{4mm}s \in \mathbb{Z}, \hspace{3mm}l=\frac{m}{2}$$

how to prove that $A^k$ is still centered and $km-$banded?

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  • $\begingroup$ Can you solve this problem for a matrix which is tridiagonal, i.e. $m=1$? Have you worked with the adjacency graph of a matrix before? $\endgroup$ – Carl Christian Jan 18 at 10:28
  • $\begingroup$ no, it's the first time I see this definition $\endgroup$ – James Arten Jan 18 at 10:33
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    $\begingroup$ Then I recommend that you first study the case of a tridiagonal matrix. Take a small value of $n$, set the nonzero entries to 1, and multiply the matrix with itself, 1, 2 and 3 times. You will get a very good idea of what is happening and why. $\endgroup$ – Carl Christian Jan 18 at 10:46
  • $\begingroup$ thank you, I tried with a 2-banded matrix 4x4 (hence triadiagonal) and I actually get the result (4-banded)..I think to obtain it just by doing an induction $\endgroup$ – James Arten Jan 18 at 10:48
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    $\begingroup$ Indeed, this is a good way to do it. Through experimentation, we can quite often find the pattern and understand how the induction step will work , but a formal proof is necessary to complete the analysis. $\endgroup$ – Carl Christian Jan 18 at 10:54

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