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Limit of a matrix sequence

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the limit as ##n \to \infty ## of ##U_n(a) =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & a/n \\ 0 & -a/n & 1 \end{pmatrix}^n##, for any real ##a##.

    2. Relevant equations


    3. The attempt at a solution

    I find ##U =\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos a & \sin a \\ 0 & -\sin a & \cos a \end{pmatrix}## but I'm not too sure. Do you think it is correct ?

    I wrote ##U_n(a) = \begin{pmatrix} 1 & 0 \\ 0 & (C_n(a))^n \end{pmatrix} ##, where ## C_n(a) = \begin{pmatrix} 1 & a/n \\ -a/n & 1 \end{pmatrix}##

    Then I diagonalized ##C_n(a)## in ##M_2(\mathbb{C})## so that

    ##U_n(a) = \begin{pmatrix} 1 & 0 \\ 0 & P \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & (D_n(a))^n \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & P^{-1} \end{pmatrix} ##

    with ##D_n(a) = \begin{pmatrix} 1+ia/n & 0 \\ 0 & 1 - ia/n \end{pmatrix} ##, ##P = \begin{pmatrix} i & i \\ -1 & 1 \end{pmatrix}##, and ##P^{-1} = \frac{1}{2i}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix} ##

    and then ##(D_n(a))^n \to \begin{pmatrix} e^{ia} & 0 \\ 0 & e^{-ia} \end{pmatrix}##
     
  2. jcsd
  3. Feb 10, 2016 #2

    fresh_42

    Staff: Mentor

    How did you get the limits? Beside the ##1## all terms include increasing powers of ##\frac{1}{n}##.
     
  4. Feb 10, 2016 #3
    Honestly I have extrapolated to the complex numbers the fact that ## (1+x/n)^n \to e^x ## for any real ##x##. I didn't think too much about it. But now you ask I would say something like

    ## | (1+ ia/n)^n - e^{ia}| = | \sum_{k=0}^n (\frac{n(n-1)...(n-k+1)}{n^k} - 1) i^k \frac{a^k}{k!} - \sum_{k> n}i^k \frac{a^k}{k!}| ##

    and by the triangle inequality
    ##| (1+ ia/n)^n - e^{ia}| \le \sum_{k=0}^n |\frac{n(n-1)...(n-k+1)}{n^k} - 1| \frac{|a|^k}{k!} + \sum_{k> n} \frac{|a|^k}{k!}##

    The second term converges to 0 as ##n\to \infty## as it is the rest of a convergent sum
    The first term also converges to 0 as for any ##\epsilon >0## there exist ##N\in\mathbb{N}## such that ## n\ge N \Rightarrow |\frac{n(n-1)...(n-k+1)}{n^k} - 1| \le \epsilon ## and the term ## \sum_{k=0}^n \frac{|a|^k}{k!} ## is bounded by ## e^{|a|} ##.

    Hope it is correct
     
  5. Feb 10, 2016 #4

    fresh_42

    Staff: Mentor

    You cannot take the real sequence and pretend the complex would behave in the same way.
    Btw ##\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n(n-1) \cdot ... \cdot (n-k+1)}{1 \cdot ... \cdot k} ≠ \frac{n(n-1) \cdot ... \cdot (n-k+1)}{n^k} ##.
    But you can take your formulas and calculate ##C_n(a)^n = P \cdot D_n(a)^n \cdot P^{-1}##.
    This results in sums of ##(1+iα)^n## and ##(1-iα)^n## (with some ##±i## coefficients) and ##α=\frac{a}{n}##.
    Now expand both with the binomial formula and look what cancels out due to ##i^n##.
    What's left should converge and since ##U## is real the limit should as well be real.

    My first approach was simply calculating some powers of ##U, C##, resp.
    If I made no mistakes there are many terms of increasing powers of ##α##.
     
    Last edited: Feb 10, 2016
  6. Feb 10, 2016 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Well done, clearly presented and a pleasure to read. To satisfy #2 you might mention that ##(1 + ia/n)^n \to e^{ia}##, etc.
     
  7. Feb 10, 2016 #6

    fresh_42

    Staff: Mentor

    Edit: I've seen it now. Sorry. (Where has my delete option gone to?)
     
  8. Feb 11, 2016 #7
    Thank you !
     
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