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Thread: Diagonalize unitary matrix

  1. #1
    Member Mollier's Avatar
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    Diagonalize unitary matrix

    Problem statement:
    Diagonalize this unitary matrix $\displaystyle \textbf{V}$ to reach $\displaystyle \textbf{V}=\textbf{U}\Lambda\textbf{U}^*$.

    $\displaystyle
    V=\frac{1}{\sqrt{3}}\left[\begin{array}[pos]{cc}
    1 & 1-i \\
    1+i & -1 \\
    \end{array}\right]
    $

    Pathetic attempt:
    The trace of $\displaystyle \textbf{V}$ is zero and since it is unitary it has eigenvalues with absolute value one. This mean that $\displaystyle \lambda = 1,-1$

    What is confusing me is the fact that I am supposed to make a matrix with orthonormal column vectors out of a unitary matrix (which has orthonormal column vectors)...

    Any hints are greatly appreciated, thanks!
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  2. #2
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    Quote Originally Posted by Mollier View Post
    Problem statement:
    Diagonalize this unitary matrix $\displaystyle \textbf{V}$ to reach $\displaystyle \textbf{V}=\textbf{U}\Lambda\textbf{U}^*$.

    $\displaystyle
    V=\frac{1}{\sqrt{3}}\left[\begin{array}[pos]{cc}
    1 & 1-i \\
    1+i & -1 \\
    \end{array}\right]
    $

    Pathetic attempt:
    The trace of $\displaystyle \textbf{V}$ is zero and since it is unitary it has eigenvalues with absolute value one. This mean that $\displaystyle \lambda = 1,-1$

    What is confusing me is the fact that I am supposed to make a matrix with orthonormal column vectors out of a unitary matrix (which has orthonormal column vectors)...

    Any hints are greatly appreciated, thanks!
    Find the eigenvectors corresponding to eigenvalues 1 and -1. Because those are distinct eigenvalues, the eigenvectors will be orthogonal. Make their lengths 1 and the matrix having those unit length eigenvectors as columns will diagonalize the matrix.
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  3. #3
    Member Mollier's Avatar
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    To find the eigenvectors for $\displaystyle \lambda=1$, do you solve this system by elimination?

    $\displaystyle
    \left[\begin{array}[pos]{cc}
    \frac{1}{\sqrt{3}}-1 & \frac{1-i}{\sqrt{3}} \\
    \frac{1+i}{\sqrt{3}} & -\frac{1}{\sqrt{3}}-1 \\
    \end{array}\right] \textbf{x} = \textbf{0}
    $

    I write from row 2:
    $\displaystyle x = \frac{1+\sqrt{3}}{1+i}y$

    Then from row 1:
    $\displaystyle
    -\frac{2}{\sqrt{3}(1+i)}y + \frac{2}{\sqrt{3}(1+i)}y = 0
    $

    Basically I can choose y to be whatever I want, but I guess I would get the "prettiest" solution if I choose $\displaystyle y = 1+i $.

    Is this the right way to go about it?
    Thanks.
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