# Cayley - Hamilton Theorem on higher powers

1. Nov 14, 2013

### KMjuniormint5

1. The problem statement, all variables and given/known data
Given the following matrix A = [3 -1; -1 3] Find C = (0.5*A - I)100

2. Relevant equations
Using the knowledge that the Cayley - Hamilton Theorem must satisfy its own characteristic polynomial.

3. The attempt at a solution

Here the characteristic polynomial is λ2 - 6*λ + 8. When we plug in A = λ we can verify that the solution A2 - 6*A + 8 = 0 which it does. If I solve by hand/brute force way, I get C = [0.5 -0.5; -0.5 0.5] regardless of what the exponential is (here the exponential is 100). As far as applying the Cayley-Hamilton theorem, I am not connecting the two. I understand how to solve for A^3 or A^4, √A, or exp(A) but not for this problem. Any help in the right direction will be greatly appreciated.

2. Nov 14, 2013

### KMjuniormint5

Looking at it at a different angle take w = 0.5*A - I and find the characteristic equation which is λ2 -λ = 0. We see that λ2 = λ. Plug in w for λ Now for w3 we can do w2*w but that is w*w and that is w. So for w100=w99*w= ... = w*w = w. In the end we see that w = C = [0.5 -0.5; 0.5 -0.5]. Was that the right approach?

3. Nov 14, 2013

### Ray Vickson

A much, much simpler way is to note that if the eigenvalues of A are distinct (which they are in this case) then for any polynomial function f(x) we have
$$f(A) = E_1 f(\lambda_1) + E_2 f(\lambda_2),$$
with the matrices $E_1, E_2$ being the same for any function f. You can find E1 and E2 in this case by using, for example,
$$f(x) = x^0 = 1 \Longrightarrow f(A) = I = E_1 \lambda_1^0 + e_2 \lambda_2^0 = E_1 + E_2$$
and $$f(x) = x \Longrightarrow f(A) = A = E_1 \lambda_1 + E_2 \lambda_2$$
Now it is easy to compute f(A) for
$$f(x) = (0.5 x - 1)^{100}.$$