Is There a General Formula for Finding the Nth Power of a 2x2 Matrix?

[pbandjay]

I am trying to find the N^th 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.

 

[tiny-tim]

To find eigenvectors …

Hi pbandjay!

(nice LaTeX, btw )

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).

 

[pbandjay]

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!

 

[Spartan Math]

Not every 2 x 2 matrix is diagonalizable.

 

[Hurkyl]

The only solution I can see for this is (x,y) = (0,0)

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