Take your favorite matrix norm $\|\bullet\|$ (my favorite is the Frobenius norm $\|A\| = \sqrt{\operatorname{tr} A A^t}$). Now consider the set $S_x$ of matrices $A,$ such that $\|A\| < x$ and $\|A^{-1}\| < x.$ Is there some other way to characterize $S_x$? Presumably a closely related question (which does not use $S_x$) is to have some alternate characterization of matrices such that $\|A^{-1}\| \leq C \|A\|,$ for some fixed $C.$ This sort of thing must come up a lot.
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1$\begingroup$ $S_x$ is an empty set for $x < 1$ and $S_1$ is a subgroup of the general linear group. One alternate characterization of $S_x$ is that it is the set of all matrices (n-by-n) whose condition number is $\leq x^2$. I know these observations are trivial. $\endgroup$– Mustafa SaidApr 18, 2014 at 7:28
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$\begingroup$ It seems to me to be more natural to ask this of a submultiplicative norm, which the Frobenius norm is not. For example, one has in the submultiplicative case that $S_{x}S_{y} \subseteq S_{xy}.$ In general, it may be more revealing to consider families of $S_{x}$'s rather than a single one. Once can probably say more about such subsets within discrete subgroups. $\endgroup$– Geoff RobinsonApr 18, 2014 at 15:08
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$\begingroup$ @GeoffRobinson In fact, I REALLY care about $SL(n, \mathbb{Z})$ so if you can say something about that case, please do :) Submultiplicative norm (e.g. operator norm, I guess) works fine for me... $\endgroup$– Igor RivinApr 18, 2014 at 15:21
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I will write a few things down for the case of ${\rm SL}(n,\mathbb{Z}),$ and I will use the operator norm (with respect to Euclidean norm on vectors). None of these is deep: Each $S_{x}$ is finite and admits a two sided action by ${\rm O}(n,\mathbb{Z}) = (\mathbb{Z}/2\mathbb{Z}) \wr {\rm Sym}_{n}.$ The stabilizer of $u$ is $\{(a,u^{-1}au) \}$ as $u$ runs through ${\rm O}(n,\mathbb{Z})$, so the orbit size is easy to calculate for any given $u.$ In fact, $S_{1} = {\rm SO}(n,\mathbb{Z}),$ if, as appears to be the case in the question, attention is restricted to matrices of determinant $1$.
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$\begingroup$ Great, waiting with bated breath! Re finiteness, the set of elements in $SL(n, \mathbb{Z})$ of norm (any norm) bounded by $x$ is asymptotic to $x^{n^2-n}$ -- this is due to Morris Newman for $n=2$ and Duke, Rudnick, Sarnak in general (so $S_x$ is strictly smaller). $\endgroup$ Apr 18, 2014 at 18:18
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$\begingroup$ Well, the finiteness is obvious by a compactness argument, though I would not have known the asymptotic behaviour $\endgroup$ Apr 18, 2014 at 19:16