- #1
pbandjay
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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.
A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mbox{ with } a,b,c,d \in \mathbb{R}
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:
\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
\lambda_{1,2} = \frac{a+d \pm \sqrt{(a+d)^2 - 4}}{2}
To find eigenvectors:
\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)
ax + by = \lambda x
cx + dy = \lambda y
Solving the first for y and inserting y into the second equation:
y = \frac{x(\lambda - a)}{b}
cx + \frac{dx(\lambda - a)}{b} = \frac{\lambda x(\lambda - a)}{b}
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.
A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mbox{ with } a,b,c,d \in \mathbb{R}
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:
\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
\lambda_{1,2} = \frac{a+d \pm \sqrt{(a+d)^2 - 4}}{2}
To find eigenvectors:
\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)
ax + by = \lambda x
cx + dy = \lambda y
Solving the first for y and inserting y into the second equation:
y = \frac{x(\lambda - a)}{b}
cx + \frac{dx(\lambda - a)}{b} = \frac{\lambda x(\lambda - a)}{b}
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.
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