# zeros of sum of two (matrix-valued!) polynomials

• Apr 28th 2010, 06:05 AM
choovuck
zeros of sum of two (matrix-valued!) polynomials
hi,

I have a nontrivial problem. let f(z) and g(z) be two (possibly monic) matrix-valued polynomials (i.e. of the form $\sum_{j=0}^n A_j z^j$ with $A_j$ being $m\times m$ matrices, and $z\in\mathbb{C}$). assume that det(f(z)) and det(g(z)) has all real and negative roots. assume det(f(z)+g(z)) has all roots real.

prove (or disprove) that det(f(z)+g(z)) cannot have a positive zero.

for the case of dimension 1, it's easy since f(z) and g(z) are positive for z>0, and so f(z)+g(z) is positive for z>0, so all the roots are negative.

any suggestion/idea/reference would be highly appreciated...
• Apr 28th 2010, 07:18 AM
Laurent
Quote:

Originally Posted by choovuck
hi,

I have a nontrivial problem. let f(z) and g(z) be two (possibly monic) matrix-valued polynomials (i.e. of the form $\sum_{j=0}^n A_j z^j$ with $A_j$ being $m\times m$ matrices, and $z\in\mathbb{C}$). assume that det(f(z)) and det(g(z)) has all real and negative roots. assume det(f(z)+g(z)) has all roots real.

prove (or disprove) that det(f(z)+g(z)) cannot have a positive zero.

A piece of contribution, perhaps:
In dimension at least 2, you can easily find situations where $\det (f(z)+g(z))$ is identically zero. For instance, in dimension 2, choose $f(z)$ satisfying the assumptions, and define $g(z)$ by exchanging the columns of $f(z)$. Then the roots of their determinants are the same (hence negative), while $f(z)+g(z)$ has two identical columns hence is singular for any $z$. But maybe this is not allowed since all complex numbers are roots of $\det(f(z)+g(z))$...

You should specify what you mean by "monic" in your problem. There is no obvious definition, I think. In dimension 1, if you discard this hypothesis, the theorem becomes clearly false ( $(2z+1)+(-z-2)=z-1$ has a positive root).
• Apr 28th 2010, 07:48 PM
choovuck
Quote:

Originally Posted by Laurent
A piece of contribution, perhaps:
In dimension at least 2, you can easily find situations where $\det (f(z)+g(z))$ is identically zero. For instance, in dimension 2, choose $f(z)$ satisfying the assumptions, and define $g(z)$ by exchanging the columns of $f(z)$. Then the roots of their determinants are the same (hence negative), while $f(z)+g(z)$ has two identical columns hence is singular for any $z$. But maybe this is not allowed since all complex numbers are roots of $\det(f(z)+g(z))$...

You should specify what you mean by "monic" in your problem. There is no obvious definition, I think. In dimension 1, if you discard this hypothesis, the theorem becomes clearly false ( $(2z+1)+(-z-2)=z-1$ has a positive root).

by monic I mean $\sum_{j=0}^n A_j z^j$ with $A_n=I$, identity matrix. Then your counterexample doesn't really work, since after interchanging columns, we obtain $A_n$ not being identity anymore.

unfortunately I found another counterexample :( it's painfully simple. basically if the eigenvalues of two matrices are negative, then the eigenvalues of their sum might be ugly. the explicit example could be smth like this: take $f(z)=z-\left(\begin{array}{cc} -2 &0\\0& -1 \end{array}\right)$ having zeros -1 and -2. take $g(z)=z-\left(\begin{array}{cc} -c &-1-c^2\\1& c \end{array}\right) \left(\begin{array}{cc} -1 &0\\0& -2 \end{array}\right) \left(\begin{array}{cc} -c &-1-c^2\\1& c \end{array}\right)^{-1}$ having zeros -1,-2. Then their the sum is $2z-T$, where T is a matrix with determinant (using Mathematica) $8-c^2$. So taking $c=2\sqrt2$ will make 0 to be a root.

...that sucks :(

thanks anyway