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

  1. #1
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    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...
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  2. #2
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    Quote Originally Posted by choovuck View Post
    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).
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  3. #3
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    Quote Originally Posted by Laurent View Post
    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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