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Hello,

Let $A_n = (a_{k-j};\;k,j = 0,1,\ldots,n-1)$ be a sequence of $n\times n$ Toeplitz matrices, with eigenvalues $(\lambda_{n,i};\;i = 0,1,\ldots,n-1)$.

If $A_n$ were a sequence of Hermitian Toeplitz matrices, and if $\sum_k|a_k|<\infty$, then Szego theorem states that for any continues function $F(\cdot)$ on $[\alpha,\beta]$ we have $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}F(\lambda_{n,k}) = \frac{1}{2\pi}\int_0^{2\pi}F(f(\xi))d\xi $$
where $$ f(\xi) = \sum_{k = -\infty}^{\infty}a_ke^{ik\xi} $$ and $\alpha = \text{ess}\inf f$ and $\beta = \text{ess} \sup f$.

If however $A_n$ are not-Hermitian, then the above hold only for polynomial functions, namely $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}p(\lambda_{n,k}) = \frac{1}{2\pi}\int_0^{2\pi}p(f(\xi))d\xi $$ where $p(\cdot)$ is some polynomial function.

My question: is there any result regard the asymptotic behavior of the modulus of the eigenvalues for the non-Hermitian case, namely $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}|\lambda_{n,k}| = ?. $$

EDITION: due to the thankful Alexandersson comment, I add the factor $1/n$...

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  • $\begingroup$ I believe that limit is infinite, why do you think that the absolute values of all eigenvalues converges to 0? $\endgroup$
    – Per Alexandersson
    Commented Jan 1, 2013 at 20:35
  • $\begingroup$ @Per Alexandersson I wrote $?$ and not $0$ :) $\endgroup$
    – Josh
    Commented Jan 1, 2013 at 20:39
  • $\begingroup$ Yes, but for the sum to converge at all, you need $|\lambda_{n,k}|\to 0$, which seems highly unlikely. $\endgroup$
    – Per Alexandersson
    Commented Jan 2, 2013 at 7:36
  • $\begingroup$ @Per Alexandersson For large $k$ yes. Hmm I'm not sure why its not make sense for you. I mean, you can say the same thing if we would chose $p(x)=x^2$, but in this case it is known that the sum will converge. $\endgroup$
    – Josh
    Commented Jan 2, 2013 at 7:56
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    $\begingroup$ Is there some normalization factor missing? Maybe it should be $1/n \sum_{k=0}^n |\lambda_{n,k}|$? If that is the case, then your sum will most likely converge, and it will be the center of mass for the limit of point measures that the eigenvalues create. This measure is, at least for the banded case, known, I think. $\endgroup$
    – Per Alexandersson
    Commented Jan 5, 2013 at 22:11

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To create a partial answer, eigenvalues of banded Toeplitz matrices accumulate on some real algebraic curves in $\mathbb{C}$, (this has been proved by Schmidt and Spitzer around 1970.). Now, if we also attach a point-mass at each eigenvalue, (for fixed matrix size) with equal mass at each point, and with total mass one, we get a probability measure. These measures converge in a certain sense to some limit measure, see for example the book by Bender and Böttcher .

Now, the sum $c_n = \frac{1}{n} \sum_{k=1}^n |\lambda_{n,k}|$ can then be interpreted as the center of mass mean distance to 0 of all the eigenvalues from matrix of size $n \times n.$ Say that the associated point measure is called $\mu_n$ (mass $1/n$ at each eigenvalue), then we know that $\mu_n \to \mu$ for some $\mu$ in some sense.

Note that $c_n = \int_{\mathbb{C}} |z| d\mu_n(z)$ and then what you are looking for is $c = \int_{\mathbb{C}} |z| d\mu(z)$.

As an explicit example, taking your matrix to be tridiagonal, will give rise to characteristic polynomials which are also a family of orthogonal polynomials, w.r.t some measure (the limit of point measures, actually), see e.g. this paper i just googled.

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  • $\begingroup$ Nice, I will look into the references. $\endgroup$
    – Josh
    Commented Jan 6, 2013 at 9:24

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