A lower bound on the second largest eigenvalue of a $4\times 4$ matrix. I have a $4\times4$ non-symmetric symbolic matrix with the following characteristics:


*

*there are positive, negative, and zero entries;

*the determinant and trace are both positive;

*under relevant conditions, there are two positive and two negative eigenvalues (using Routh's criterion).


Problem: I have to show analytically that the positive eigenvalues are both larger than some $x>0$ (which is true for their sum). I can show this numerically, for a wide range of parameters of interest, and analytically, for one special case of interest in which one element of the matrix is zero. But I cannot show it for the general case. (Analytical computation of the eigenvalues does not help, as the resulting expressions are just too complex.)
Any suggestions are welcome!
 A: My two cents: your assertion is true if the spectral radius of the matrix is bounded. In other words, we have the following:

Proposition. Suppose $A$ is a $4\times4$ matrix such that
  
  
*
  
*$A$ has two positive and two negative eigenvalues.
  
*$\det A$ and $\operatorname{tr} A$ are positive and bounded away from zero.
  
*$\rho(A)$ is bounded.
  
  
  Then the second highest (positive) eigenvalue of $A$ is bounded away from zero.

The proof should be easy and it is omitted here.
Clearly, one may replace the third assumption in the above by some stronger ones, e.g. the entries of $A$ are bounded in magnitude or $\|A\|$ is bounded. At any rate, without this third assumption, your assertion is not true in general. For a counterexample, consider a family of triangular matrices $\{A_n\}_{n\ge2}$ such that the diagonal entries of $A_n$ are $\{n,\,\frac1n,-\frac n2,-\frac 2n\}$. The determinant of each $A_n$ is $1$, which is bounded below. As $\operatorname{tr}A_n=\frac n2-\frac1n$ is strictly increasing in $n$, it is bounded below by $\operatorname{tr}A_2=\frac12$. Yet the second highest eigenvalue of $A_n$ (i.e. $\frac1n$) approaches $0$ as $n\to\infty$.
