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[Linear Algebra] Closed formula for recursive sequence

  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data
    qsEKf8z.png

    2. Relevant equations
    a) the one given
    b) det(A-λI) = 0
    find λ values using A
    c)use λ values to find eigenvectors

    3. The attempt at a solution
    This wasn't explained well enough so I can understand it in class. So far, I made the matrix being multiplied to A have the following elements from top to bottom: x0 = 1, x1 = 1, x2 = 2. The one on the right has the following elements: x1 = 1, x2= 2, x3=7. Is this correct thinking? Do I just find a matrix A that maps the vector on the left to the vector on the right?
    I know how to do b) and c), I just don't know how to get A correctly.
     
  2. jcsd
  3. Jul 26, 2016 #2

    Stephen Tashi

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    The usual way to describe an element of a matrix is to give it two indices, such as A[1][1] , A[1][2] or ## A_{1,1}, A_{1,2} ## etc. Which elements are you trying to describe with the phrase "from top to bottom" ?

    You need a matrix A such that:

    eq. 1: [itex] A_{1,1} x_k + A_{1,2} x_{k+1} + A_{1,3} x_{k+2} = x_{k+1} [/itex]
    eq 2: [itex] A_{2,1} x_k + A_{2,2} x_{k+1} + A_{2,3} x_{k+2} = x_{k+2} [/itex]
    eq 3: [itex] A_{3,1} x_k + A_{3,2} x_{k+1} + A_{3,3} x_{k+2} = x_{k+3} [/itex]

    An solution to eq. 1 is [itex] A_{1,1} = 0, A_{1,2} = 1, A_{1,3} = 0 [/itex].
     
  4. Jul 26, 2016 #3
    Sorry, I completely forgot that A was a 3x3 matrix.
    Anyways, I see how you are getting the solutions.
    For equation 2, I found these solutions:
    A21 = 0 A22 =0 A23= 1
    For equation 3, I found:
    A31 = 0 A32 = 1 A33 =3

    Are those correct? Does what they give for xk+3 have anything to do with the last equation?
     
  5. Jul 26, 2016 #4
    So A can be more than 1 matrix?
     
  6. Jul 26, 2016 #5

    Ray Vickson

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    You can (and should) check that for yourself. When you substitute in your values for ##A_{31}, A_{32}## and ##A_{33}## do you get the correct recursion?

    And, for your next question in post #4, why do you think there is more than one matrix? The matrix ##A## has 9 elements called ##A_{11}, A_{12} \ldots, A_{33}##.
     
  7. Jul 26, 2016 #6
    Yes, I do. I just changed the bottom row of A to be the constants of the recursive formula given, but I kept the first 2 rows I stated the same. I verified and got the eigenvalues in b.
    I thought there was more than one matrix because in the equations, there are a combination of values for each element in A that can produce the desired elements. But it seems like there will be only one because the eigenvalues have to be verified, correct?
     
  8. Jul 26, 2016 #7

    Ray Vickson

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    I have no idea what you are saying.

    While it is the case that IF you apply the matrix to some smal number of initial values like ##x_0=1,x_1=, x_2 =2## (and a few more obtained from the recursion) you will get numbers on the left and numbers on the right, so will have equations like
    $$ A_{31}\: \text{number one} + A_{32}\: \text{number two} + A_{33} \:\text{number three} = \text{number four} $$
    with actual numbers. Of course you can have many possible values of ##A_{31}, A_{32}, A_{33}## that "work". However, that is irrelevant: you are supposed to find a matrix that works for all values of ##x_k, x_{k+1}, x_{k+2}, x_{k+3}## that you will ever encounter in the calculations (say for ##x_{27}, x_{28}, x_{29}, x_{30}## or for ##x_{1000}, x_{1001}, x_{1002}, x_{1003}##, etc.) and when you use that fact there can be only one possible choice for the three numbers ##A_{31}, A_{32}, A_{33}##. There is only one matrix ##A##.
     
    Last edited: Jul 26, 2016
  9. Jul 26, 2016 #8

    Stephen Tashi

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    You solution for eq. 3 doesn't work. Your solution gives [itex] x_{k+2} = x_{k+3} [/itex].

    Yes !
     
  10. Jul 26, 2016 #9

    Stephen Tashi

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