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Let $a(M)$ be the maximum absolute value of entries of matrix $M\in\mathsf{GL}_k(\mathbb Z)$.

$M^{-1}\in\mathsf{GL}_k(\mathbb Z)$ holds.

What is a good upper bound for $|a(M)-a(M^{-1})|$?

I am thinking whether the dependence could be a little smaller than fully exponential in $k$ for $a(M)\cdot a(M^{-1})$ which will reflect upper bound for $|a(M)-a(M^{-1})|$.

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    $\begingroup$ A good upper bound for $|a(M)-a(M^{-1})|$ is $|a(M)-a(M^{-1})|$ itself :) you probably want an upper bound not depending too much on $M$, for instance, $a_{k,n}=\sup_{|a(M)|\le n}|a(M)-a(M^{-1})|$: estimate $a_{k,n}$? (for $k$ fixed) $\endgroup$
    – YCor
    Commented May 31, 2020 at 23:09
  • $\begingroup$ I think $a(M)\cdot a(M^{-1})\leq(\max(a(M),a(M)^{-1}))^k$ might be possible and can that $k$ be $o(k)$ or just even $k-\alpha$ with $\alpha\gg1$ and not just $k$? $\endgroup$
    – VS.
    Commented Jun 1, 2020 at 1:50
  • $\begingroup$ A trivial upper bound is $a(M^{-1})\le (k-1)!a(M)^{k-1}$. For a Jordan matrix $M_n=I+nJ$ in size $n$ one has $a(M_n)=n$ and $a(M_n^{-1})=n^{k-1}$, so in this case $a(M_n)a(M_n^{-1})=n^k$. $\endgroup$
    – YCor
    Commented Jun 1, 2020 at 9:27
  • $\begingroup$ [You wrote $a(M)a(M^{-1}\le \max(a(M),a(M^{-1}))^k$, but this is trivial with $k=2$: $ab\le \max(a,b)^2$... this might be a typo] $\endgroup$
    – YCor
    Commented Jun 1, 2020 at 9:29

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For $k=2$, the upper bound is zero.

For $k>2$, there is no upper bound. E.g., let $$M=\pmatrix{1&1&1\cr9&10&11\cr-n&n&3n-1\cr}$$ Then $$M^{-1}=\pmatrix{10-19n&2n-1&-1\cr38n-9&1-4n&2\cr-19n&2n&-1\cr}$$ and $a(M)-a(M^{-1})=(3n-1)-(38n-9)=-(35n-8)$ is unbounded.

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    $\begingroup$ actually the matrix $M_n=\begin{pmatrix}1 & n & 0\\ 0 & 1 & n\\0 & 0 & 1\end{pmatrix}$ with $a(M_n)=n$ even has $a(M_n^{-1})=n^2$, since $M_n^{-1}=\begin{pmatrix}1 & -n & n^2\\ 0 & 1 & -n\\0 & 0 & 1\end{pmatrix}$. $\endgroup$
    – YCor
    Commented May 31, 2020 at 23:14
  • $\begingroup$ So the gap is $(a(m))^{k-1}$? $\endgroup$
    – VS.
    Commented Jun 1, 2020 at 2:04
  • $\begingroup$ If the elements of $M$ are bounded in absolute value by $b$, then the elements of $M^{-1}$ can't be any bigger than $c_kb^{k-1}$ for some constant $c_k$ depending only on $k$. $\endgroup$
    – Gerry Myerson
    Commented Jun 1, 2020 at 3:33

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