I have a $5 \times 5$ parametric nonnegative matrix and want to show that it's stable (in the sense that all eigenvalues are positive). It is not symmetric, but I do know in advance that it has 5 real eigenvalues. 


I have also shown that it's a $P$-matrix (i.e. all principal minors are positive). So now I can try to check if it's sign-symmetric and if yes, I'll have stability by Carlson's theorem. However, the sign-symmetry involves checking lots of cases and I wonder if there is an alternative approach that I'm missing.


So, to put it in a general way:



> 
> Under which additional assumptions is
>  a non-symmetric nonnegative P-matrix stable?
> 
> 
> 



matrix stability criterion
Just a remark. Following your assumptions, your matrix is hyperbolic in the sense that you know that all eigenvalues are real-valued. I understand that it depends on some (real) parameters, then from a theorem due to Bronshtein, the characteristic roots are Lipschitz-continuous functions of the parameters.


Of course, if the roots are all positive at some value of the parameter (assume that there is one real parameter $t$ near the distinguished value $0$), then they will stay positive for $t$ near $0$. So the only case to look at is a root $\lambda(t)$ such that $\lambda(0)=0$. If it is a simple root, it is smooth and you just have to check for instance that
$$
\dot \lambda (0)=0,\quad\ddot\lambda(0)>0
\tag 1
$$
which will ensure that $\lambda$ will stay positive near $t=0$. The characteristic polynomial
$$
p(t,X)=\prod\_{0\le j\le 4}(X-\lambda\_j(t))
$$
and you may assume that you know
$\frac{\partial p}{\partial X}(0,0)\not=0:$
then $$
p(t,X)=e(t,X)(X-\lambda\_0(t)),\qquad p(t,\lambda\_0(t))\equiv 0
$$
and
$
\partial\_t p +\partial\_X p\dot \lambda\_0=0,\quad
\partial\_t^2 p+2\partial\_t \partial\_X p \dot \lambda\_0 +
\partial\_X^2 p\dot \lambda\_0+\partial\_X p \ddot \lambda\_0 =0.
$
As a result to ensure (1), it is enough to check
$$
\frac{\partial p}{\partial t}(0,0)=0,\quad
(\partial\_t^2 p)(0,0) (\partial\_X p)(0,0)<0.
$$
Bazin.


