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Thread: Matrix Polynomials Question

  1. #1
    Junior Member
    Jun 2010

    Matrix Polynomials Question

    I'm studying for a PhD preliminary exam, and it seems I have some holes in my linear algebra knowledge. I need help with the following problem:

    Let A and B be operators V -> V, where V is a finite-dimensional vector space over the complex numbers. Let p be any polynomial such that p(AB) = 0. Then (a) Prove that that q(x) = xp(x) satisfies q(BA) = 0.

    Interestingly, I have shown that this problem is equivalent to proving that the minimal polynomial of AB is equal to the minimal polynomial of BA. Therefore, a direct solution to this problem would be to directly show that q(BA) = 0 whereas, given what I've already shown, it would suffice to simply prove that the minimal polynomials of BA and AB are equal. I am interested to a solution of either kind (ideally both!)

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  2. #2
    MHF Contributor
    Oct 2009

    Re: Matrix Polynomials Question

    I think the fact that this problem is about operators over finite-dimensional vector space is irrelevant. This statement can be proved about any associative algebra over a ring.
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  3. #3
    MHF Contributor

    Mar 2011

    Re: Matrix Polynomials Question

    suppose p(x) = c0 + c1x + ...+ cmxm, so that

    q(x) = c0x + c1x2 +....+ cmxm+1.

    then q(BA) = c0BA + c1(BA)2 +...+ cm(BA)m+1.

    note that (BA)k = B(AB)k-1A, so we have:

    q(BA) = B(c0I + c1(AB) +.... + cm(AB)m)A

    = B(p(AB))A = B(0)A = 0.

    your assertion that AB and BA must have the same minimal polynomial is false. suppose A =

    [1 -1]
    [1 -1], and that B =

    [0 1]
    [0 1]

    then AB = 0, while BA = A.

    clearly, the minimal polynomial for AB is x, while the minimal polynomial for BA is x2. note this agrees with the result above.
    Last edited by Deveno; Jun 19th 2012 at 09:43 AM.
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