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Rotation try again.

  1. Jul 13, 2006 #1
    This is another attempt at an earlier question that maybe wasn't done in a readable format.

    This was in Strang page 230. Could someone derive this?

    if K = \left[\begin{array} {cc} 0&-1\\1&0\end{array}\right]

    then \quad e^{Kt} = \left[\begin{array} {cc} cos t & -sin t\\sin t & cos t\end{array}\right]
    Last edited by a moderator: Jul 13, 2006
  2. jcsd
  3. Jul 13, 2006 #2

    matt grime

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    By working out what exp{Kt} looks like: K has a nice form, so do the powers of K, hence it is possible to simply work out what the sum defining exp{Kt} is.
  4. Jul 13, 2006 #3


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    The eigenvalues of K are i and -i so, over the complex numbers, K is equivalent to the diagonal matrix with i and -i on the diagonal. In fact, since the corresponding eigenvectors are multiples of [1, -i] and [1, i] respectively, we have:
    [tex]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}1 & 1 \\-i & i \end{array}\right] \left[\begin{array}{cc}i & 0 \\0 & -i\end{array}\right]\left[\begin{array}{cc}\frac{i}{2} & -\frac{1}{2} \\-\frac{i}{2} & -\frac{1}{2}\end{array}\right][/tex]

    The point of "diagonalizing" like that is that if [itex]A= CDC^{-1}[/itex] then [itex]A^2= (CDC^{-1})(CDC^{-1})= CD^2C^{-1}[/itex] and similarly for higher powers. Then, using the Taylor's series for ex:[itex]e^x= 1+ x+ \frac{1}{2}x^2+ \cdot\cdot\cdot + \frac{1}{n!}x^n+ \cdot\cdot\cdot[/itex], we have
    [tex]e^{CDC^{-1}}= I+ CDC^{-1}+ \frac{1}{2}CD^2C^{-1}+ \cdot\cdot\cdot+ \frac{1}{n!}CD^nC^{-1}+ \cdot\cdot\cdot[/tex]
    Since I= CC-1, that is
    [tex]C(I+ D+ \frac{1}{2}D^2+ \cdot\cdot\cdot + \frac{1}{n!}D^n+ \cdot\cdot\cdot = Ce^DC^{-1}[/tex]
    For a diagonal matrix, powers just give powers on the diagonal and adding just adds the diagonal values, for diagonal matrix D, eD is just the diagonal matrix with exponentials on the diagonal. In this case
    [tex]e^{Kt}= \left[\begin{array}{cc}1 & 1 \\-i & i \end{array}\right]\left[ \begin{array}{cc}e^{it} & 0 \\ 0 & e^{-it}\end{array}\right]\left[\begin{array}{cc}\frac{i}{2} & -\frac{1}{2} \\-\frac{i}{2} & -\frac{1}{2}\end{array}\right][/tex]
    Use [itex]e^{it}= cos(t)+ isin(t)[/itex] and multiply it out.
  5. Jul 14, 2006 #4
    Wow! Thanks

    Wow! Such a quick and clear answer. Thanks!
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