1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cayley - Hamilton Theorem on higher powers

  1. Nov 14, 2013 #1
    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. jcsd
  3. Nov 14, 2013 #2
    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?
     
  4. Nov 14, 2013 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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
    [tex] f(A) = E_1 f(\lambda_1) + E_2 f(\lambda_2), [/tex]
    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,
    [tex] f(x) = x^0 = 1 \Longrightarrow f(A) = I = E_1 \lambda_1^0 + e_2 \lambda_2^0 = E_1 + E_2[/tex]
    and [tex] f(x) = x \Longrightarrow f(A) = A = E_1 \lambda_1 + E_2 \lambda_2[/tex]
    Now it is easy to compute f(A) for
    [tex] f(x) = (0.5 x - 1)^{100}. [/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted