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Unitary matrix of a hermitian form matrix

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Given the matrix H=
    \begin{array}{cc}
    4 & 2+2i & 1-i \\
    2-2i & 6 & -2i \\
    1+i & 2i & 3 \\
    \end{array}

    Find a unitary matrix U such that U*HU is diagonal
    (U* is the conjugate transpose of U, and U* = U-1)

    3. The attempt at a solution

    I find the eigenvalues
    λ1 = 9
    λ2 = 2
    λ3 = 2
    and the corresponding eigenvectors
    v1 = [ 1-i , -2i , 1 ]
    v2 = [ -1+i , 0 , 2 ]
    v3 = [ -1-i , 1 , 0 ]
    Normalizing them and using them as column vectors for U I get
    \begin{array}{cc}
    (1-i)/\sqrt{7} & (-1+i)/\sqrt{6} & (-1-i)/\sqrt{3} \\
    -2i/\sqrt{7} & 0 & 1/\sqrt{3} \\
    1/\sqrt{7} & 2/\sqrt{6} & 0 \\
    \end{array}

    but this doesn't work. I try U*U and I try U*HU but neither come out right.
    What am I missing?

    *edit* I've just noticed that U isn't in hermitian form because of v1. Is there a way of transforming v1 so I get a real number for u1,1? Like multiplying it my the conjugate of 1-i? I'm sure it has something to do with the repeated eigenvalue, but I don't know what to do
     
    Last edited: Nov 24, 2011
  2. jcsd
  3. Nov 25, 2011 #2

    I like Serena

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    Hi Locoism! :smile:

    Your v2 and v3 form a basis for the eigenspace of eigenvalue 2.
    However, they are not orthogonal.
     
  4. Nov 25, 2011 #3
    Ok but then do I construct an orthogonal basis from v2 and v3 (using gram-schmidt)?
     
  5. Nov 25, 2011 #4

    I like Serena

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