Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

SqueeSpleen

  • one year ago

Finding a closed form for a matrix power.

  • This Question is Closed
  1. SqueeSpleen
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm trying to deal with the following problem: A={{0,1},{1,1}} X^2-X-1 is the charasteristic polynomial of A, so: A^2-A-I_2=0 I had a similar problem in the theory, but it was simplier. It was: A={{0,-1},{1,2}} And it charasteristic polynomial was: X^2-2X+1=(X-1)^2 By the division algorythm we have: P(X) \in R[X] and a_n, b_n in R such that: X^2=(X-1)^2.P(X)+a_n.X+b_n * Using X=1 we have: 1=0.P(X)+a_n.1+b_n then a_n+b_n=1 We derivate * and we have: n.X^(n-1)=2(X-1).P(X)+(X-1)^2.P'(X)+a_n We take X=1 and we have: n=a_n So we have b_n=1-n Then: X^n=(X-1)^2.P(X)+n.X+1-n So: A^n=n.A+(1-n).I_2={{1-n,-n},{n,n+1}} But with: A={{0,1},{1,1}} I can't solve this, because P'(X) and P(X) don't share any roots, so I can't isolate a_n or b_n, the best I achieved was: ((1+5^(1/2))/2)^n=((1+5^(1/2))/2).a_n+b_n and other equations, but I can never isolate a term. I know A^n.{{0,1}} is {{f_n},{f_(n+1)}} where f_n is the nth fibonnaci's number, but I'm trying to prove it in this way :p Sorry for my bad English :p I don't know a lot of mathematical terms in this language, neither I'm good speaking it.

  2. SqueeSpleen
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If it is to difficult to read I can use some LaTeX, feel free to ask me for it.

  3. SqueeSpleen
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I also have the link to the book where it comes from (it isn't piracy :P, they are a class notes), but it's in Spanish.

  4. SqueeSpleen
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I diagonalized the matrix, and by the fact that if A=C.B.C^(-1) then A^n = C.B^n.C^(-1) and a nth power of a diagonal matrix is the matrix which has the autovalors at the nth power, we have: ... {{-(1+5^(1/2))/2,(-1+5^(1/2))/2},{1,1}}.{{((1-5^(1/2))/2)^n,0},{0,((1+5^(1/2))/2)^n}}.{{-(1+5^(1/2))/2,(-1+5^(1/2))/2},{1,1}}^(-1) I suppose that my major mistake was to try to solve it in the same way instead of thinking what tools of the chapter were useful.

  5. nitz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    although i have studied matrices....but closed form....ill check

  6. SqueeSpleen
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form I'll close it.

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.