Let $A$ be an $n \times n$ random matrix, such that the entries $a_{ij}$ are i.i.d. from the standard normal distribution.
I'm curious on the probability that $A$ is Schur stable. That is, $$P(\rho(A) < 1),$$ where $\rho(A)=\max\{|\lambda|:\lambda \in spec(A)\}$.
Are there practical lower bounds (i.e. something other than insanely tiny values)? The best I was able to figure out was $P(\rho(A)<1)=\sum_{j=1}^nP(|\lambda_j(A)|<1)$, where $\lambda_j$ is the $j$th eigenvalue of $A$.
We form the (real symmetric) matrix $$ B=AA^T\sim\operatorname{Wishart}_n(I_n,n) $$ so Weyl's inequality $\lvert\lambda_1\rvert\leq\sigma_1$ for $A$ gives $$ \mathbb{P}(\rho(B)\leq 1)\leq\mathbb{P}(\rho(A)<1). $$
But we know the spectral density of $B$ is given by $$ f(\lambda_1,\dots,\lambda_n)=c_n\exp(-\tfrac12\sum_i\lambda_i)\prod_i \lambda_i^{-1/2}\prod_{i<j}\lvert\lambda_i-\lambda_j\rvert\quad(\lambda_i>0\forall i) $$ where the normalisation constant $$ c_n=\frac{\pi^{n/2}}{2^{n^2/2}[\prod_{i=1}^n\Gamma(\frac12(n-i+1))]^2}. $$ and so the distribution of the largest eigenvalue $\lambda_1=\lambda_1(B)$ is given by the CDF $$ F_{\lambda_1}(x)=\mathbb{P}(\lambda_1\leq x)=c_n\cdot\operatorname{Pf}(M). $$ where $\operatorname{Pf}(M)$ is the Pffafian of the real skew-symmetric $(2\lceil n/2\rceil)\times(2\lceil n/2\rceil)$ matrix $M$ $$ M_{ij}=\begin{cases} \int_0^x\int_0^x\operatorname{sgn}(\alpha-\beta) \alpha^{i-\frac32}e^{-\alpha/2}\cdot\beta^{j-\frac32}e^{-\beta/2}\,\mathrm{d}\alpha\,\mathrm{d}\beta & 1\leq i,j\leq n\\ \int_0^x \alpha^{i-\frac32}e^{-\alpha/2}\,\mathrm{d}\alpha&i\leq n,j=n+1\text{ and }2\nmid n\\ -\int_0^x\beta^{j-\frac32}e^{-\beta/2}\,\mathrm{d}\beta&j\leq n,i=n+1\text{ and }2\nmid n\\ 0 & i=j=n+1\text{ and }2\nmid n \end{cases}. $$