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Spectral decomposition of a diagonal matrix

  1. Oct 1, 2012 #1

    syj

    User Avatar

    1. The problem statement, all variables and given/known data
    I have
    [tex]
    J=\begin{bmatrix}
    \frac{\pi}{2}&0&0\\
    1&\frac{\pi}{2}&0\\
    0&1&\frac{\pi}{2}\\
    \end{bmatrix}
    [/tex]

    I need to find [tex] \sin(J) \text{ and } \cos(J) \text{ and show that } \sin^{2}(J)+\cos^{2}(J)=I
    [/tex]


    2. Relevant equations



    3. The attempt at a solution
    I have the following:

    [tex]
    \sin(J)=
    \begin{bmatrix}
    1&0&0\\
    0&1&0\\
    0&0&1\\
    \end{bmatrix}
    [/tex]

    and
    [tex]
    \cos(J)=
    \begin{bmatrix}
    0 & 0 & 0\\
    0 & 0 & 0\\
    0 & 0 & 0\\
    \end{bmatrix}
    [/tex]

    I don't know if this is correct. All the questions I have examples of have more than one eigenvalue, this one only has one eigenvalue.
     
  2. jcsd
  3. Oct 1, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    J is basically a Jordan form, except its 1's are below the diagonal instead of above; that is, it is a transpose of a Jordan matrix. So, you need to know what is the form of
    [tex] \begin{pmatrix} \lambda & 0 & 0 \\
    1 & \lambda & 0 \\
    0 & 1 & \lambda
    \end{pmatrix}^n [/tex]

    See, eg., http://en.wikipedia.org/wiki/Jordan_normal_form to see why your final answers are incorrect.

    RGV
     
  4. Oct 1, 2012 #3

    syj

    User Avatar

    The textbook I have defines a Jordan matrix to be one where the 1s are below the diagonal. So the matrix given is already a Jordan matrix.
     
  5. Oct 1, 2012 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    OK, fine. But did you read the rest of the message?

    RGV
     
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