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Difference equation and diagonal block matrix

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data

    Compute ##A^j~\text{for} ~~j=1,2,....,n## for the block diagonal matrix##A=\begin{bmatrix}
    J_2(1)& \\
    &J_3(0)
    \end{bmatrix}##,

    And show that the difference equation ##x_{j+1}=Ax_{j}## has a solution satisfying ##|x_{j}|\rightarrow\infty~\text{as}~j\rightarrow\infty##
    2. Relevant equations


    3. The attempt at a solution


    So ##A^1=\begin{bmatrix}
    1&1&0&0&0 \\
    0&1&0&0&0 \\
    0&0&0&1&0 \\
    0&0&0&0&1 \\
    0&0&0&0&0
    \end{bmatrix}
    ,~~A^2=\begin{bmatrix}
    1&2&0&0&0 \\
    0&1&0&0&0 \\
    0&0&0&0&1 \\
    0&0&0&0&0 \\
    0&0&0&0&0
    \end{bmatrix}
    ,~~A^j=\begin{bmatrix}
    1&j&0&0&0 \\
    0&1&0&0&0 \\
    0&0&0&0&0 \\
    0&0&0&0&0 \\
    0&0&0&0&0
    \end{bmatrix}\forall ~~j\geq 3 ##



    I am certain that this isn't a difficult question, but I am not sure how to apply this to the difference equation. Which is probably due to my lack of experience with them. Help would be appreciated, thanks.
     
    Last edited by a moderator: Oct 14, 2015
  2. jcsd
  3. Oct 14, 2015 #2

    BvU

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    For my sake: apparently ##J_2## and ##J_3## have a special meaning. Could you explain ?
     
  4. Oct 14, 2015 #3
    Yes sorry about that, I should have put it in the question.

    ##J_2(1) = ~\text{the 2x2 jordan block matrix with 1 as eigenvalues}~=\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix}##
    And ## J_3(0) = ~\text{the 3x3 jordan block matrix with 0 as eigenvalues}~=\begin{bmatrix} 0&1&0 \\ 0&0&1 \\0&0&0 \end{bmatrix}##
     
  5. Oct 14, 2015 #4

    BvU

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    Well, I learned something !

    But now I can't be of much further help, because I can't imagine what they want to count as a 'solution' for ##
    x_{j+1}=Ax_{j}##, which to me is just a recipe to get ##
    x_{j+1}## from ##x_{j}##.

    I mean, is ##x_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}## a 'solution' ? (I ignore the other dimensions that go to 0 anyway)
    Because ##x_{j+1} = A x_j## is satisfied (always !?) and ##x_j = \begin{bmatrix} 1 + j \\ 1 \end{bmatrix}## so that
    ##|x_{j}|\rightarrow\infty~\text{as}~j\rightarrow\infty##

    --
     
  6. Oct 14, 2015 #5
    I am glad I was involved in helping someone learn something new!

    Mod note: Fixed typo in post #1
    I think I should point out that there is a typo in the original question, it should be:

    ##A^j=\begin{bmatrix}
    1&j&0&0&0 \\
    0&1&0&0&0 \\
    0&0&0&0&0 \\
    0&0&0&0&0 \\
    0&0&0&0&0
    \end{bmatrix}\forall ~~j\geq 3 ~\text{rather than}~
    A^j=\begin{bmatrix}
    1&n&0&0&0 \\
    0&1&0&0&0 \\
    0&0&0&0&0 \\
    0&0&0&0&0 \\
    0&0&0&0&0
    \end{bmatrix}\forall ~~j\geq 3##

    And yes what you say makes sense.. I think as long as ##x_0## has a positive value in the 2nd row it will be satisfied (since the second column in ##A^j## is unbounded in ##j##) and ##|x_ j| =\sqrt{j^2+1}## which tends to infinity as ##j## does. Do you agree ?
     
    Last edited by a moderator: Oct 14, 2015
  7. Oct 14, 2015 #6

    BvU

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    I agree with you, but that doesn't mean much (real knowledge trumps common sense). Would appreciate it very much if you would post if it turns out we are both wrong :smile:

    Here they call ## a_n = r^n## the solution for the difference equation ##a_n = r a_{n-1}## so in that jargon I have inadvertently posted the actual solution,

    which is in fact against PF rules. May I be forgiven by O2 and all the other good spirits that watch over us, because I did not know what I was doing
     
  8. Oct 14, 2015 #7

    Ray Vickson

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    Try writing out the first few terms ##x_2, x_3 x_4 \ldots##, to see how they relate to ##x_1##.
     
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