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Math Help - Diagonalize unitary matrix

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

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

    <br />
V=\frac{1}{\sqrt{3}}\left[\begin{array}[pos]{cc}<br />
    1    & 1-i \\<br />
    1+i  & -1 \\<br />
\end{array}\right]<br />

    Pathetic attempt:
    The trace of \textbf{V} is zero and since it is unitary it has eigenvalues with absolute value one. This mean that \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 \textbf{V} to reach \textbf{V}=\textbf{U}\Lambda\textbf{U}^*.

    <br />
V=\frac{1}{\sqrt{3}}\left[\begin{array}[pos]{cc}<br />
    1    & 1-i \\<br />
    1+i  & -1 \\<br />
\end{array}\right]<br />

    Pathetic attempt:
    The trace of \textbf{V} is zero and since it is unitary it has eigenvalues with absolute value one. This mean that \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 \lambda=1, do you solve this system by elimination?

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

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

    Then from row 1:
    <br />
-\frac{2}{\sqrt{3}(1+i)}y + \frac{2}{\sqrt{3}(1+i)}y = 0<br />

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

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