Thread: New to MHF / Cayley Hamilton 2x2 matrix to power n

1. New to MHF / Cayley Hamilton 2x2 matrix to power n

A 2x2 matrix has (in this case) 2 distinct eigenvalues. The matrix to the power n can be expressed as a coefficient times the matrix plus a second coefficient times the identity matrix. Can someone point me to a simple derivation of a formula that computes the two coefficients? That is, find the two coefficients in terms of the two eigenvalues and n. Yes, I know the problem can also be soved by diagonalizing the matrix and then raising the diagonal elements to the power n, but I want to avoid computing and then multiplying by the pre and post-matrices that are required for that method. There are (at least) two ways to skin this cat, and I'm asking for help on the method that does not involve explicit diagonalization. Many thanks.

2. Re: New to MHF / Cayley Hamilton 2x2 matrix to power n

we DO have a linear algebra forum. this post should be there.

if you KNOW the eigenvalues, then you know that A satisfies the polynomial equation:

$x^2 - (\lambda_1 + \lambda_2) + \lambda_1\lambda_2 = 0$

that is:

$A^2 = (\lambda_1+\lambda_2)A - (\lambda_1\lambda_2)I$

for convenience, let's write this as:

$A^2 = cA + dI$

i suspect you can work out some formula, but it's going to get messy, for example:

$A^5 = (c^4 + 3c^2d + d^2)A + (c^3d + 2cd^2)I$

3. Re: New to MHF / Cayley Hamilton 2x2 matrix to power n

I will move this post to the correct forum. Thanks.