matrix stability criterion - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:25:33Z http://mathoverflow.net/feeds/question/95483 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95483/matrix-stability-criterion matrix stability criterion Felix Goldberg 2012-04-29T02:57:15Z 2012-05-08T11:52:20Z <p>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. </p> <p>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.</p> <p>So, to put it in a general way:</p> <blockquote> <p>Under which additional assumptions is a non-symmetric nonnegative P-matrix stable?</p> </blockquote> http://mathoverflow.net/questions/95483/matrix-stability-criterion/95518#95518 Answer by Bazin for matrix stability criterion Bazin 2012-04-29T19:46:53Z 2012-04-29T19:46:53Z <p>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.</p> <p>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 $$\frac{\partial p}{\partial t}(0,0)=0,\quad (\partial_t^2 p)(0,0) (\partial_X p)(0,0)&lt;0.$$ Bazin.</p> http://mathoverflow.net/questions/95483/matrix-stability-criterion/95583#95583 Answer by Felix Goldberg for matrix stability criterion Felix Goldberg 2012-04-30T15:17:14Z 2012-04-30T15:17:14Z <p>After some hard work showing sign-symmetry, I realized that the answer is much simpler:</p> <p>Being a $P$-matrix is equivalent to having the property that the real eigenvalues of all principal submatrices are positive.</p> <p>Since I had known that my matrix had real eigenvalues to begin with, I should have just QEDed it after showing $P$-matricity.</p>