Consider the set of all matrices with spectral radius $<1$, denoted $\mathcal C=\{A\in \mathbb{C}^{n\times n}:\rho(A)<1\}$. This is an interesting set, partly because $A\in \mathcal C\iff \lim\limits_{k\to\infty} A^k= 0$. It seems difficult to describe the geometry of $\mathcal C$, but let's ask about its size in terms of the Lebesgue measure $\mu$ on $\mathbb{C}^{n\times n}$.

Question: What is the asymptotic behavior of $\mu(\mathcal C\cap \{A:\|A\|\le r\})$ as $r\to\infty$? (That is, is it some power of $r$, or $\log r$, etc?)


  1. It does not matter what matrix norm is used here, since they are all equivalent.
  2. Let's exclude the trivial case $n=1$.
  3. $\mu(\mathcal C)=\infty$, which can be shown as follows. Let $\epsilon>0$ be such that any $n\times n$ matrix with all entries satisfying $|a_{ij}|<\epsilon$ has spectral radius less than $1$. Denote by $\mathcal E$ the set of matrices with all entries between $\epsilon/2$ and $\epsilon$ in absolute value. Let $D$ be the diagonal matrix with diagonal entries $(2,1,\dots,1)$. Then the sets $D^k\mathcal E D^{-k}$, $k\in\mathbb Z$, are disjoint, have the same (positive) Lebesgue measure, and are contained in $\mathcal C$. The claim follows.
  4. The argument in item 3 shows that $\mu(\mathcal C\cap \{A:\|A\|\le r\}) \gtrsim \log r$.
  • $\begingroup$ I don't quite understand your argumentation in 3. Say, $n=2$ and you have a matrix $E\in\mathcal E$, then$$DED^{-1} = \begin{pmatrix}e_{11} & 2e_{12}\\\frac 1 2e_{11} & e_{22}\end{pmatrix}.$$If, e.g., $|e_{12}| = \frac 2 3\varepsilon$, then you might run out of $\mathcal C$. Or do I miss something? $\endgroup$ Mar 16 '17 at 0:00
  • 1
    $\begingroup$ Similar matrices have the same spectral radius (since they have the same spectrum). So, conjugation by $D$ preserves the set $\mathcal C$. It does move $\mathcal E$ to a set disjoint from $\mathcal E$, which is the goal: creating a sequence of disjoint sets within $\mathcal C$. $\endgroup$
    – user357151
    Mar 16 '17 at 0:16
  • $\begingroup$ Thanks! That was what I missed. $\endgroup$ Mar 16 '17 at 0:19

For a (maybe not very good) upper bound, note that $\rho(A) \ge |\text{trace}(A)|/n$. If we use the Frobenius norm, with $A$ chosen uniformly from the ball $B_R$ of radius $R$, $\text{trace}(A)$ is the dot product of the vectorized $A$ with a vector of norm $\sqrt{n}$, so with $V_{m}(r) = \pi^{m/2} r^{m}/\Gamma(1+m/2)$ the $m$-dimensional volume of an $r$-ball in $\mathbb R^m$,
$$ \eqalign{\mu(B_R \cap \{\rho(A) \le 1\}) \le \mu(B_R \cap \{|\text{trace}(A)| \le n\}) &= \int_{-\sqrt{n}}^{\sqrt{n}} dt\; V_{n^2-1}(\sqrt{R^2 - t^2})\cr &\sim c(n) R^{n^2-1} }$$ where $c(n)$ is a positive constant depending on $n$.

  • $\begingroup$ $\rho(A) < 1$ also implies $|\det(A)| < 1$. Can you do a similar thing with this? At least for $n=2$? $\endgroup$ Mar 16 '17 at 0:25
  • 2
    $\begingroup$ In the setting of complex matrices (as the question is posed) this estimate becomes $O(R^{2n^2-2})$. A shorter way to describe it: placing a bound on a linear form, such as the trace, leaves us with the product of a bounded set with a subspace of codimension 1, hence the result. $\endgroup$
    – user357151
    Mar 16 '17 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy