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Show that minimal poly for a sq matrix and its transpose is the same

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data
    show that minimal poly for a sq matrix and its transpose is the same


    2. Relevant equations




    3. The attempt at a solution
    no clue.
     
    Last edited: Jan 30, 2014
  2. jcsd
  3. Jan 30, 2014 #2

    pasmith

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    Let [itex]\lambda[/itex] be an eigenvalue of [itex]A[/itex] such that
    [tex](A - \lambda I)^n = 0[/tex]
    but
    [tex](A - \lambda I)^{m} \neq 0[/tex]
    for every positive integer [itex]m < n[/itex].

    Given that, can you show that [itex](A^T - \lambda I)^n = 0[/itex] and that there does not exist a positive integer [itex]m < n[/itex] such that [itex](A^T - \lambda I)^m = 0[/itex]?
     
    Last edited: Jan 30, 2014
  4. Jan 30, 2014 #3

    Ray Vickson

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    The polynomial ##p(x) = x^m + a_1 x^{m-1} + \cdots + a_{m-1} x + a_m ## is a minimal polynomial for matrix ##A## if and only if ##p(A) x = 0## for all column vectors ##x##, but this is not true for any polynomial of degree < m.

    In other words, the vectors ##x, Ax, A^2 x, \ldots, A^{m-1}x## are linearly independent, but ##A^m x## is a linear combination of ##x, Ax, A^2 x, \ldots, A^{m-1}x##; furthermore, this same m and this same linear combination holds for all ##x##.

    Basically, this is how some computer algebra packages find minimal polynomials, without finding the eigenvalues first. In fact, if we restrict the field of scalars to the reals a real matrix ##A## may not have (real) eigenvalues at all, but it will always have a real minimal polynomial.
     
    Last edited: Jan 30, 2014
  5. Jan 30, 2014 #4
    i'm not sure where we are going with this.

    i assume this is a property
    [tex](A - \lambda I)^n = 0[/tex]
    but
    [tex](A - \lambda I)^{m} \neq 0[/tex]
    for every positive integer [itex]m < n[/itex].

    but i don't get how showing the next part will help with minimal poly problem.
     
  6. Jan 30, 2014 #5

    pasmith

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    Hint: [itex](A^n)^T = (A^T)^n[/itex]
     
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