Since it seems this question is getting a lot of attention, but at the same time it is quite difficult to solve in its generality, let us try to tackle a special case.
So, let $A$ be a $3\times 3$ random matrix where every entry is i.i.d. and uniformly distributed on $[0,1]$. What is the probability that $A$ has only real eigenvalues? Or if that is not possible to compute exactly, do we have lower/upper bounds?
From simulations with 10 million points, the correct answer seems to be around $0.708...$, so one could use it as a benchmark to test the validity for proposed upper/lower bounds.
Some initial bounds:
- Take the diagonal matrix $A$ with entries $0,1/2,1$. Consider all matrices $B\in [0,1]^{3\times 3}$ such that $\max_{i,j}|A_{i,j}-B_{i,j}|\le 1/12$. By perturbation theory, $B$ still has real eigenvalues, since $\|A-B\|\le 1/4$. The space where you can take $B$ has probability $2(1/12)^9$ and since you can permute the elements on the diagonal of $A$, we can multiply that by 6 and obtain a lower bound of $(1/12)^8$
- A similar reasoning can be done starting from $\begin{pmatrix}1&1&0\\1&0&1\\1&0&0\end{pmatrix}$ that has complex eigenvalues and considering all matrices $B\in [0,1]^{3\times 3}$ such that $\max_{i,j}|A_{i,j}-B_{i,j}|\le 1/8$. By Bauer-Fike theorem, $B$ will still have imaginary eigenvalues, and the same holds if we permute the columns/rows of $B$ in all the 6 possible ways, so we get an upper bound of $1-6(1/8)^9$
I realize these are very weak bounds (they are at most $10^{-8}$ better than the obvious bounds $[0,1]$). In theory one could work with the discriminant of the degree 3 characteristic polynomial, but it's a degree 6 polynomials in the 9 variable entries. Are there any better ways to get bounds?
