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SqueeSpleen

  • 3 years ago

Finding a closed form for a matrix power.

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  1. SqueeSpleen
    • 3 years ago
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    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
    • 3 years ago
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    If it is to difficult to read I can use some LaTeX, feel free to ask me for it.

  3. SqueeSpleen
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    although i have studied matrices....but closed form....ill check

  6. SqueeSpleen
    • 3 years ago
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    http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form I'll close it.

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