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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?

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2 Answers 2

up vote 2 down vote accepted

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.

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Bazin, thanks for the interesting answer. –  Felix Goldberg Apr 30 '12 at 9:44

After some hard work showing sign-symmetry, I realized that the answer is much simpler:

Being a $P$-matrix is equivalent to having the property that the real eigenvalues of all principal submatrices are positive.

Since I had known that my matrix had real eigenvalues to begin with, I should have just QEDed it after showing $P$-matricity.

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However, I'm glad that Brazin gave his answer - it contains a useful idea that might help in other situations. –  Felix Goldberg Apr 30 '12 at 15:17

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