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Consider the $N\times N$ matrix $$ M = \left(\begin{array} \\ 0 & 1 & & 0 \\ 1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & 1 & 0 \\ \end{array}\right) $$

which comes from the adjacency matrix of a graph corresponding to a one-dimensional chain of $N$ nodes with dangling ends. A cartoon of this graph is $$\circ -\circ -\circ -\circ -\cdots-\circ -\circ$$

It turns out that if you plot a histogram of its eigenvalues, it appears to fit exactly with an arcsine distribution $$f(x) = \frac{1} {\pi \sqrt{4-x^2}}, \vert x \vert < 2 $$ which is exactly what one would expect from the free convolution of the binomial distribution $$ p(x) = \frac 1 2 \left( \delta\left(x-1\right) + \delta \left(x+1\right)\right)$$ with itself.

Is this mere coincidence, or evidence of something deeper? I feel like this must be some example of a known result out there.

I've gotten as far as figuring out how $\pm 1$ shows up; you can write $M$ as the sum of two pieces $$ M = A + B $$ $$ A = \left(\begin{array}{cccccc} 0 & 1\\ 1 & 0\\ & & 0 & 1\\ & & 1 & 0\\ & & & & \ddots\\ & & & & & \ddots \end{array}\right) = \sigma_x \oplus \sigma_x \oplus \cdots $$ $$ B = \left(\begin{array}{cccccc} 0\\ & 0 & 1\\ & 1 & 0\\ & & & 0 & 1\\ & & & 1 & 0\\ & & & & & \ddots \end{array}\right) = [0] \oplus \sigma_x \oplus \sigma_x \oplus \cdots $$

where $\sigma_x$ is the Pauli sigma matrix which of course has eigenvalues $\pm 1$. It must be that these two matrices are freely independent in the $N\rightarrow \infty$ limit, and possibly even for finite $N$ also, so that this reduces to the free convolution described above.

I may be reading too much into this, but it's interesting to me that this is a completely deterministic matrix problem with free probabilistic characteristics. I'm not at all familiar with the algebraic aspects of free probability theory, let alone what the graph theoretic relationships would be.

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The discrete Laplacian on a chain approximates the Laplacian on $[0, 1]$, so its eigenvalues approximate frequencies of standing waves with vanishing (?) boundary conditions. In fact using this idea one can explicitly write down eigenfunctions and eigenvalues. –  Qiaochu Yuan Sep 25 '11 at 18:21
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Why do you call o-o-...-o a cartoon of a graph? Isn't it a legitimate presentation of a graph? –  Hans Stricker Sep 25 '11 at 22:46
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3 Answers 3

up vote 4 down vote accepted

I believe the relation between deterministic graphs and free probability you mentioned is not something generic. In fact, the main property of your matrix $M$ which makes connection with free probability (at the best of my knowledge) is not to be the adjacency matrix of some graph, but a Jacobi matrix related to some orthogonal polynomials, which themselves come from random matrix models.

Let me try to develop : We first need some computations.

We have to assume $N$ even to make things properly. The Chebychev (monic) polynomials $(T_k)_{k\geq0}$ of the first kind satisfy $T_k(x)=x^k+\ldots$ and are orthogonal for the weight $$ w(x)=\frac{1}{\sqrt{1-x^2}}, $$ defined on $[-1,1]$, namey for any $k\neq l$ $$ \int_{-1}^1T_k(x)T_{l}(x)w(x)=0. $$ Their Jacobi matrix (associated with its recurrent coefficients) is actually $M/2$. For our purpose its enough to know that the zeros of $T_N$ are actually the eigenvalues of $M/2$. Moreover, a formula due to Heine yields $$ T_N(x)=\int_{-1}^1\ldots\int_{-1}^1\prod_{i=1}^N(x-x_i)\prod_{1\leq i < j \leq N}|x_i-x_j|^2\prod_{i=1}^Nw(x_i)dx_i. $$ The change of variables $x_i=\cos\theta_i$ gives $$ T_N(x)=\int_{0}^{2\pi}\ldots \int_0^{2\pi}\prod_{i=1}^N(x-\cos\theta_i)\prod_{1\leq i < j \leq N}|\cos\theta_i - \cos\theta_j|^2\prod_{i=1}^Nd\theta_i $$ and by Weyl formula $$ T_N(x)=\int_{\mathcal{U}_N}\det(xI_N-\frac{U+U^*}{2}) dU = \mathbb{E}_{Haar}\Big(\det(xI_N-\frac{U+U^*}{2})\Big) $$ where $dU$ stands for the Haar measure of the unitary group $\mathcal{U}_N$.

Conclusion : The random matrix $U+U^*$, with $U$ distributed according to Haar, has for mean eigenvalues the zeros of $T_N(x/2)$, and equivalently the eigenvalues of $M$. Thus they should have the same limiting distribution as $N\rightarrow\infty$ as soon as that the limiting distribution of $U+U^*$ is deterministic.

One one hand, the limiting distribution of $M$ is indeed known to be the arcsine distribution (note it is also the limiting distribution of the zeros of $T_N$ as $N\rightarrow\infty$, which is known to minimize the logarithmic energy $$ \iint \log\frac{1}{|x-y|}d\mu(x)d\mu(y) $$ over all probability measure $\mu$ on $[-1,1]$, a classical statement in potential theory).

On an other hand, by the invariance property of the Haar measure, the distribution of $U+U^* $ is the same than $A+VAV^*$, with $V$ also distributed according to Haar, which is known to converge by Voiculescu Theorem towards $\mu_A\boxplus\mu_A$, where $\mu_A$ is the limiting distribution of your matrix $A$, namely $\mu_A=\frac{1}{2}(\delta_1+\delta_{-1})$.

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I would agree that there is something special about the matrix structure that allows this correspondence. I was not aware of Heine's formula, thanks. This deserves to be accepted as an answer, although I wonder if there is something more interesting that points toward possible generalizations. –  Jiahao Chen May 11 '12 at 18:01
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I'd like to add that in general, your matrix is a special case of symmetric tridiagonal matrices of the form

$$ \begin{bmatrix} a & b & \cdots & \cdots & 0\\\\ b & a & b & \cdots & 0\\\\ & \ddots & \ddots & \ddots\\\\ 0 & \cdots & &a & b\\\\ 0 & \cdots & &b & a \end{bmatrix} $$

The eigenvalues of this matrix are $$ \lambda_k = a + 2b\cos(k\pi/(n+1)),\qquad 1 \le k \le n. $$

The eigenvectors have a similar nice closed form.

$$ v_{ik} = \sin\left(\frac{ik\pi}{n+1}\right),\qquad 1\le i,k \le n. $$

These facts can probably be quickly derived by looking at the characteristic polynomial.

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Thanks for crossing one item off my to-do list. –  Jiahao Chen Sep 26 '11 at 17:52
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In short: tridiagonal Toeplitz matrices and Chebyshev polynomials are intimately related. –  J. M. Nov 21 '11 at 16:22
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If $\phi_n$ denotes the characteristic polynomial of the path on $n$ vertices then $$ \phi_{n+1}(t) = t\phi_n(t) - \phi_{n-1}(t), $$ from which you can show that $$ \phi_n(2\cos(\zeta)) = \frac{\sin(n+1)\zeta}{\sin(\zeta)}. $$ So your observation is not a surprise from a graph theoretical viewpoint. I have nothing useful to say about free probability.

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