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Unitary matrix - linear alg

  1. May 10, 2009 #1
    I am going over old practice exams and came across this question:

    Find the unitary matrix which diagonalizes the matrix

    .......i 1 0
    A = (-1 i 0 )
    ...... 0 0 -i


    First off, can someone explain to me about unitary matrices and get me started on this question? I do not know where to begin solving it. Thank you!
     
  2. jcsd
  3. May 10, 2009 #2

    Cyosis

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    A unitary matrix is a matrix for which [itex]U^{\dagger}U^{-1}=U^{-1}U^\dagger=I[/itex]. So [itex]U^\dagger=U^{-1}[/itex]. A matrix B is diagonalizable if there exists a matrix C such that [itex]C^{-1}BC[/itex] is diagonal.
     
  4. May 10, 2009 #3
    so i found the conjugate transpose of the matrix and got

    (-i -1 0)
    (1 -i 0)
    (0 0 i)

    is that a start?
     
  5. May 10, 2009 #4

    Cyosis

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    No, the idea is that you find [itex]U^\dagger A U[/itex]. A need not be unitary. The first step is to find the eigenvalues of A and their respective eigenvectors. You can however use [itex]A^\dagger[/itex] to check if A is unitarily diagonalizable. A matrix A is unitarily diagonalizable if and only if [itex]A^\dagger A=A A^\dagger[/itex]. Perhaps it's nice to check this for A, although of course A is unitarily diagonalizable.

    After you have the eigenvectors check if they form an orthonormal set. If they don't you have to make an orthonormal set by the Gram-Schmidt process. Now make a matrix with its columns being the eigenvectors.
     
    Last edited: May 10, 2009
  6. May 10, 2009 #5
    so i came up with the equation:
    -h(h^2+1)+h^2 i

    so the eigenvalues are 0, -i, and something else but i dont know how to find that. 0 = ih^2

    the eigenvectors are (1,- i,0) , (0,0,1), and something..
     
  7. May 10, 2009 #6

    Cyosis

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    That equation doesn't look right to me. I got [itex](-i-\lambda)[(i-\lambda)(i-\lambda)+1]=0[/itex]
     
  8. May 10, 2009 #7
    so its
    [tex]
    \left|\begin{array}{ccc}i-\lambda & 1 & 0 \\ -1 & i-\lambda & 0 \\ 0 & 0 & -i-\lambda\end{array}\right|
    [/tex]

    so then
    [itex]
    [(i-\lambda)(i-\lambda)(-i-\lambda)]-[(-1)(1)(-i-\lambda)]=0
    [/itex]
    oo.. i forgot the lambda for the last (-i) before.. ok so
    [itex]
    (i-\lambda)(i-\lambda)(-i-\lambda)-(i+\lambda)=0
    [/itex]
    and so
    [itex]
    (-i-\lambda)[(i-\lambda)(i-\lambda)+1]=0
    [/itex]
    is right

    so eigenvalues are 0, -i, 2i
    with eigenvectors: (1,- i,0), (0,0,1), (1, i,0)

    ok... so as for orthonormal set:
    (1,- i,0), (0,0,1) are perpendicular
    (0,0,1), (1, i,0) are perpendicular

    then? lost from there
     
  9. May 10, 2009 #8

    Cyosis

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    Well this set of eigenvectors is orthogonal and we want an orthonormal set. So you need to normalize them. After that you can simply form a matrix with the eigenvectors as columns. Then test if it's unitary and diagonalize the matrix A.

    A tip when it comes to calculating determinants. You don't always have to start from the top left and work to the top right when you write a 3x3 determinant as the sum of 2x2 determinants. Like in this case the fastest way is:

    [tex]
    \left|\begin{array}{ccc}i-\lambda & 1 & 0 \\ -1 & i-\lambda & 0 \\ 0 & 0 & -i-\lambda\end{array}\right|=(-i-\lambda)\left|\begin{array}{cc}i-\lambda & 1 \\ -1 & i-\lambda & \end{array}\right|
    [/tex]

    Start at the bottom right, because that row/column has the most zeros.
     
  10. May 10, 2009 #9
    aren't they already normalized?
    u1 = (1,- i,0)
    u2 = (0,0,1) - (0+0+0)/... doesnt matter
    u3 = (1, i,0) - (0+0+0)/... doesnt matter

    so then i would divide by ||u||
    u1: squareroot(1^2+(-i)^2+0) = 0 --well then itd be undef if u divide by 0
    u2: 1
    u3: 0
     
  11. May 10, 2009 #10

    Cyosis

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    A normalized vector has length 1. So [itex]||u_1||=\sqrt{(u_1,u_1)}=\sqrt{2}[/itex]. Normalizing u1 gives [itex]1/\sqrt{2}(1,-i,0)[/itex]. It works the same for the others.
     
  12. May 10, 2009 #11
    ok. i ended up with the vectors
    [itex]
    (1/\sqrt{2}, -i/\sqrt{2}, 0), (0, 0, 1), (1/\sqrt{2}, i/\sqrt{2}, 0)
    [/itex]

    which gives me the matrix
    [tex]
    \left|\begin{array}{ccc}1/\sqrt{2} & 0 & 1/\sqrt{2} \\ -i/\sqrt{2} & 0 & i/\sqrt{2} \\ 0 & 1 & 0\end{array}\right|
    [/tex]

    and the conjugate transpose is
    [tex]
    \left|\begin{array}{ccc}1/\sqrt{2} & i/\sqrt{2} & 0 \\ 0 & 0 & 1 \\ 1/\sqrt{2} & -i/\sqrt{2} & 0\end{array}\right|
    [/tex]

    then
    [tex]
    \left|\begin{array}{ccc}1/\sqrt{2} & 0 & 1/\sqrt{2} \\ -i/\sqrt{2} & 0 & i/\sqrt{2} \\ 0 & 1 & 0\end{array}\right|
    [/tex] * [tex]
    \left|\begin{array}{ccc}1/\sqrt{2} & i/\sqrt{2} & 0 \\ 0 & 0 & 1 \\ 1/\sqrt{2} & -i/\sqrt{2} & 0\end{array}\right|
    [/tex] = [tex]
    \left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|
    [/tex]
    which is already diagonalized

    so then the unitary matrix/answer is:
    [tex]

    \left|\begin{array}{ccc}1/\sqrt{2} & i/\sqrt{2} & 0 \\ 0 & 0 & 1 \\ 1/\sqrt{2} & -i/\sqrt{2} & 0\end{array}\right|

    [/tex]
    ?
     
  13. May 10, 2009 #12

    Cyosis

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    Yep that is correct. You've shown that [itex]U^\dagger U=I[/itex]. Therefore U is unitary. You can test if U indeed diagonalizes A by calculating [itex]U^\dagger A U[/itex].
     
  14. May 10, 2009 #13
    ok. thank you!
     
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