I have a sequence of matrices $M_n$ with a certain pattern. All the entries of these matrices are 0's and 1's with each row containing a maximum of three 1's. I know they are invertible. Is there any way to prove that the row sums of $\{M_n^{-1}\}$ are bounded uniformly for all $n?$
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What are the dimensions of $M_n$? $k \times k$ for a fixed $k$? – Caleb Stanford Jul 30 '16 at 08:38
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It is 5n+3 for each n – XYZ Jul 31 '16 at 17:06
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$5n+3\times 5n+3.$ – XYZ Jul 31 '16 at 17:12
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Sure, they are bounded uniformly, although many of your specifications are not necessary for this result. There are finitely many possible $k \times k$ $(0,1)$-matrices; in particular there are finitely many invertible ones, and for each of those invertible matrices, the inverse has a finite operator norm (by the bounded inverse theorem, see also this question). Thus there is a maximum operator norm, $C$, and we have $$ \|M_n^{-1}\| \le C \quad \forall n $$ which implies that, for $u$ the vector of all $1$s, $$ \|M_n^{-1} u\| \le C\|u\|. $$ Now the vector $v_n = M_n^{-1} u$ is the vector of row-sums, and its entries are bounded above by $\|v_n\|$, which is in turn bounded above by $C\|u\|$. So the row-sums are uniformly bounded.
Caleb Stanford
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Sorry I forgot to mention that the size of $M_n$ is not fixed k: $M_n$ are of size $5n+3\times 5n+3.$ – XYZ Jul 31 '16 at 17:08
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@XYZ I figured, that the problem was too easy this way :) Okay. – Caleb Stanford Aug 01 '16 at 01:18