The size of the set of matrices with spectral radius less than 1
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?)
Remarks
- It does not matter what matrix norm is used here, since they are all equivalent.
- Let's exclude the trivial case $n=1$.
- $\mu(\mathcal C)=\infty$, 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.
- The argument in item 3 shows that $\mu(\mathcal C\cap \{A:\|A\|\le r\}) \gtrsim \log r$.
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 $n$.