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Homework Help: Diagonalizing Martrix (S) question

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    First off, this is not for a course, I'm reviewing material. This also *should* be straightforward! I think I'm forgetting something simple, so if someone could point it out to me, I would be able to sleep easy tonight :)
    OK! the question:

    Given a matrix [tex]T = \begin{pmatrix}
    cosx & -sinx\\
    sinx & cosx\\
    \end{pmatrix}$[/tex]

    we want to find the inverse of [tex]{\bf S}^{-1}={\bf S}[/tex] and then take [tex]STS^{-1}[/tex].

    2. Relevant equations
    So first I find the eigenvalues which are cosx +/- isinx
    Next I calculated the eigenvectors and got a(1) = (1,-i) [column vector]
    and a(2) = (1,i) [column vector]
    If you normalize the two eigenvectors you get a constant 1/sqrt[2] for both.
    That gives me
    [tex]S^{-1} = \frac{1}{\sqrt{2}}\begin{pmatrix}
    1 & 1\\
    -i & i\\
    \end{pmatrix}$[/tex]

    and when I solve I get [tex]S = \frac{1}{\sqrt{2}}\begin{pmatrix}
    1/2 & i/2\\
    1/2 & -i/2\\
    \end{pmatrix}$[/tex]

    3. The attempt at a solution
    If you look above, you should see that I did most everything correctly (I believe, let me know if I made an error)! However, clearly [tex]STS^{-1}[/tex] should give me back a matrix with my eigenvalues on the diagonal. However, I get an extra coefficient in front of the matrix which should cancel out. Where is the mistake? Also, along those lines, I read somewhere that [tex]S=(S^{-1})^{\dag}[/tex] is this true? I was always under the impression that S is simply the inverse and you do not need to take the adjoint?
     
  2. jcsd
  3. Jan 7, 2010 #2
    I get:

    [tex]
    S=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\-i&i\end{array}\right)\rightarrow S^{-1}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)
    [/tex]

    The latter term is where it seems you encounter your extra factor. My result does lead to the expected solution:

    [tex]
    S^{-1}TS=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\cdot\left(\begin{array}{cc}\cos(x)&-\sin(x)\\ \sin(x)&\cos(x)\end{array}\right)\cdot\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\-i&i\end{array}\right)=\left(\begin{array}{cc}\cos(x)+i\sin(x)&0\\0&\cos(x)-i\sin(x)\end{array}\right)
    [/tex]
     
  4. Jan 8, 2010 #3
    Yes, thank you. I made an error in the analysis of [tex]S[/tex] and that lead to the errors :)
     
    Last edited: Jan 8, 2010
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