Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    q=0 is not the only solution to pq=0....
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook