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Nth Power of 2x2 Matrix

  1. Dec 23, 2009 #1
    I am trying to find the Nth power of a general 2x2 real matrix. This seemed simple at first, but I am running into trouble of finding general eigenvectors and cannot figure out where to go.

    [tex]A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mbox{ with } a,b,c,d \in \mathbb{R}[/tex]

    For my purposes, it is an element of SL(2,R), therefore det(A) = ad - bc = 1. I am trying to find An using An = PDnP-1. To find the eigenvalues:

    [tex]\det(A - \lambda{I_2}) = \left| \begin{array}{cc} a - \lambda & b \\ c & d - \lambda \end{array} \right| = (a-\lambda)(d-\lambda) - bc = \lambda^2 - (a + d)\lambda + ad - bc = \lambda^2 - (a + d)\lambda + 1 = 0[/tex]

    [tex]\lambda_{1,2} = \frac{a+d \pm \sqrt{(a+d)^2 - 4}}{2}[/tex]

    To find eigenvectors:

    [tex]\left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \lambda \left( \begin{array}{c} x \\ y \end{array} \right)[/tex]

    [tex]ax + by = \lambda x[/tex]
    [tex]cx + dy = \lambda y[/tex]

    Solving the first for y and inserting y into the second equation:

    [tex]y = \frac{x(\lambda - a)}{b}[/tex]

    [tex]cx + \frac{dx(\lambda - a)}{b} = \frac{\lambda x(\lambda - a)}{b}[/tex]

    The only solution I can see for this is (x,y) = (0,0), whether I use for first or second eigenvalue, which doesn't make sense to me. I would think that there would have to be some way to find a general formula since it is easy to use this method to find numerical examples of diagonalization and such. Or maybe I am missing something. My knowledge of linear algebra isn't very strong.
    Last edited: Dec 23, 2009
  2. jcsd
  3. Dec 23, 2009 #2


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    Hi pbandjay! :smile:

    (nice LaTeX, btw :wink:)

    For the eigenvectors, all you need is the ratio y/x,

    so just go back to your y = x(λ - a)/b (or y = xc/(λ - d), which is the same thing). :smile:
  4. Dec 23, 2009 #3
    Ah, of course. That helps a lot, thank you!

    I finally found the formula for An, but I'm afraid it isn't telling me what I expected. I may have think of some other ways to solve my question. But thank you!
    Last edited: Dec 23, 2009
  5. Dec 25, 2009 #4
    Not every 2 x 2 matrix is diagonalizable.
  6. Dec 25, 2009 #5


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    q=0 is not the only solution to pq=0....
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