4
$\begingroup$

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.

$\endgroup$
3
  • 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. $\endgroup$ Apr 18, 2014 at 7:28
  • $\begingroup$ 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. $\endgroup$ Apr 18, 2014 at 15:08
  • $\begingroup$ @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 Rivin
    Apr 18, 2014 at 15:21

1 Answer 1

1
$\begingroup$

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 $1$.

$\endgroup$
2
  • $\begingroup$ 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). $\endgroup$
    – Igor Rivin
    Apr 18, 2014 at 18:18
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.