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Homework Help: Matrices in power diagonal question

  1. Nov 2, 2011 #1
    find B matrices so [tex]B^{3}=A=\left(\begin{array}{cc}14 & 13\\13 & 14\end{array}\right)[/tex]

    ,the diagonal form of A is [tex]D=\left(\begin{array}{cc}a & 0\\0 & b\end{array}\right)[/tex]

    i got weird numbers so for convinience the eigenvalues are a,b

    so there is U for which

    [tex]U^{-1}AU=\left(\begin{array}{cc}a^{\frac{1}{3}} & 0\\0 & b^{\frac{1}{3}}\end{array}\right)^{3}[/tex]

    i can find U
    because A is simetric so U is consists of the orthonormal eigenvectors of A
    what to do next,how to find B

  2. jcsd
  3. Nov 2, 2011 #2


    Staff: Mentor

    You have U-1AU = D, where D is a diagonal matrix whose entries are the eigenvalues of A.

    It turns out that U-1A1/3U is a cube root of D (or D1/3), where D1/3 is also a diagonal matrix whose entries are the cube roots of the entries of D. In short, U-1A1/3U = D1/3.

    To verify that U-1A1/3U is a cube root of D,
    notice that (U-1A1/3U)3
    = (U-1A1/3U)(U-1A1/3U)(U-1A1/3U)
    = U-1A1/3UU-1A1/3UU-1A1/3U
    = U-1A1/3A1/3A1/3U
    = U-1AU = D.

    In other words, if we cube U-1A1/3U, we get D, so U-1A1/3U is a cube root of D.

    To get B, which is A1/3, multiply both sides of the equation U-1A1/3U = D1/3 on the left by U, and multiply both sides on the right by U-1.
  4. Nov 2, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper

    Why do you say the eigenvalues of A are 'weird'? They don't look weird to me. In fact they are both integer cubes. What did you get? I think you can do this whole exercise with easy numbers.
    Last edited: Nov 2, 2011
  5. Nov 3, 2011 #4
    [tex]\left(\begin{array}{cc}14 & 13\\13 & 14\end{array}\right)[/tex]

    the caracterstic polinomial


    t=1 t=27

    for t=1 i get (-1,1) and for t=27 i got the same (-1,1)

    so the transformation matrices is not invertible which is wrong

    where is my mistake?
  6. Nov 3, 2011 #5


    User Avatar
    Science Advisor
    Homework Helper

    Your mistake is in what you didn't show. (-1,1) can't have an eigenvalue of both 1 and 27. Which is it?
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