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

**choovuck** · April 28th 2010, 05:05 AM

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...

**Laurent** · April 28th 2010, 06:18 AM

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).

**choovuck** · April 28th 2010, 06:48 PM

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

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