2
$\begingroup$

I have a sequence of $N$-by-$N$ matrices with specific entries $J_{ij}=J(i-j,N)$ that depend on $i-j$ (Töplitz matrix), but also on $N$ (not just extending the Töplitz matrix. Numerically, I find that the eigenvalue (EV) distribution approaches the following distribution for large $N$: The eigenvalues 1, -1, -1/3 and 1/3 each appear with multiplicity N/4. I would like to prove this, but don't know how I could do this. Are there some standard techniques that I could try to apply?

The explicit expression pf $J(d,l)$is given by \begin{align*} \frac{\sin \left(\pi d \left(\frac{1}{l}+\frac{3}{4}\right)\right)+i \cos \left(\frac{1}{4} \pi d \left(3-\frac{4}{l}\right)\right)+\cos \left(\frac{\pi d}{4}\right) \left(2 \sin \left(\frac{\pi d}{l}\right)-i\right)+\sin \left(\frac{3 \pi d}{4}\right)}{l \left(-1+e^{\frac{3 i \pi d}{l}}\right)\left(2 e^{\frac{i \pi d (l+2)}{l}} \sin \left(\frac{\pi d}{4}\right)\right)^{-1}}\ \end{align*} unless the denominator is zero. In this case, $J(d,l)$ is zero.

Where does this matrix come from?

  1. I take the $2N$-by-$2N$ diagonal matrix with eigenvalues \begin{align} E=(\underbrace{1,\cdots,1}_{N/2},\underbrace{1,-1,-1,1,-1,-1,\cdots,1,-1,-1}_{3N/2}) \end{align}
  2. I then take the discrete Fourier transform of this matrix, namely with entries \begin{align} J_{ij}=\frac{1}{2N}\sum^{2N}_{k=1}e^{i\frac{2\pi k(i-j)}{2N}}E_k \end{align} where $E_k$ are the entries of the eigenvalue row shown above.
  3. Obviously, the $2N$-by-$2N$ matrix has still the same eigenvalues because the Fourier transform just expresses the same linear map in a different basis. However, now I restrict to the upper left $N$-by-$N$ matrix. This matrix has exactly the entries shown above and its spectrum in the limit $N\to\infty$ behaves as indicated, namely eigenvalues $1,-1,1/3,-1/3$ each appear with multiplicity $N/4$.
  • 0
    Without knowing the explicit expression of $J_{ij}$ as a function of $i-j$ and $N$, it looks hard... – 2017-02-09
  • 0
    I'm happy to provide the explicit expression. – 2017-02-09
  • 0
    Woooh... Reassure me: are all the "i" (row) indices ? I fear that some of them mean $\sqrt{-1}$, like in the exponential ? – 2017-02-09
  • 0
    All i refer to the imaginary unit. I wrote the function $J(d,l)$ where $d=i-j$ and $l=N$ is the size of the matrix. I used $l$ instead of $N$ because I copied the expression from Mathematica where $N$ can't be used as variable name... – 2017-02-09
  • 0
    1) This clarification about $i$ was a good thing 2) Moreover, I think you should "groove" a little your expression. I am almost sure that it can be presented under a simpler form. For example, it is not classical to leave an inverse in the denominator. – 2017-02-09
  • 0
    That's right, I had it in the numerator, but the expression didn't fit very well - apart from that I'm not sure how much one can simplify it. Anyway, I don't expect people to solve the problem for me, but I just want to know if there are some standard techniques to study the eigenvalue distribution in the limit. I used Szegö theorem successfully in another case, but here it doesn't work. – 2017-02-09
  • 0
    1) Do you know the following document (http://ee.stanford.edu/~gray/toeplitz.pdf)? 2) Is your matrix circulant or close to be circulant ? In this case you could make connexion with DFT (Discrete Fourier Transform)... – 2017-02-09
  • 0
    Hi JeanMarie, Yes, I used the reference when I proved the eigenvalue distribution in a different limit (namely, first taking $l$ to $\infty$ independent of $N$ and then taking the $N\to\infty$ limit). This worked extremely well. However, it seems to be more tricky for the case where all matrix elements change in the limit. My matrix would only be circulant if $N=2l$, but in this case the eigenvalues will all be $\pm1$. I'm interested in $N=l$ in the limit $N\to\infty$. – 2017-02-09
  • 0
    The simplicity of your limit spectrum $\{\pm 1, \pm \dfrac13\}$ is puzzling and should be a guide, but under the condition that you explain a little the context in which you have built this matrix. – 2017-02-09
  • 0
    I agree and I added an explanation of how the matrix is constructed. It is the result of a discrete Fourier transform of a specific spectrum, but then restricted to a submatrix... – 2017-02-10

0 Answers 0