# The maximal eigenvalue of a symmetric Toeplitz matrix

Let $0\le x\le 1$ be a real number. Denote by $A_n(x)=(a_{ij})$ the $n$ by $n$ matrix such that $a_{ij}=x^{|i-j|}$ and let $\lambda_n(x)$ be the maximal eigenvalue of $A_n(x)$.

Is there any formula for $\lambda_n(x)$?

I am particularly interested in the following question.

Let $y_n$ be such that $\lambda_n(y_n)=\frac{n}4$. What is the value of $$\lim_{n\to \infty} (y_n)^n$$ if the limit exists?

• mathoverflow.net/questions/68099 Mar 25 '15 at 20:10
• @Steve Huntsman: thank you for the link. Mar 25 '15 at 20:17

Using the results in pages 59-63 of Rosenblum and Rovnyak (p. 62 in particular, and noting that your Toeplitz matrices are symmetric, hence normal, so the operator norm is what you want), it follows that $\lim_n \lambda_n(x) = \frac{1+x}{1-x}$. The non-asymptotic behavior is probably hard, but numerics clearly indicate (and it is probably not hard to show) that $\lambda_n(1) = n$. It might be possible to combine these observations, or more detailed ones along similar lines, to get good estimates for $\lambda_n(x)$.

• Thank you for your answer. $\lambda_n(x)\le \frac{1+x}{1-x}$ is a natural bound, because it is greater than the sum of coefficients in any row. Also you are right that $\lambda_n(1)=n$. Mar 25 '15 at 19:59

Here's another example of techniques used in determining the asymptotics for extrema in the spectra of a symmetric toeplitz matrix. The cited works may be of use as well.

I don't think that there is a closed-form expression, but there are some research results on the asymptotic behavior of eigenvalues of symmetric Toeplitz matrices: see the comments to Spectra of a Symmetric Toeplitz Operator for instance.

You might relate it to asymptotic evaluations of certain (skew) Schur polynomials, by interpreting your determinant via one of the Jacobi-Trudi identities. You will be interested in the second identity.

I suspect that a formula for $\lambda_n$ might exist and be reasonably nice. You might try a computer algebra system to see how far you can get. I tried for up to $n=8$ and still was getting exact answers (for each eigenvalue) for unknown $x$, cf. https://cloud.sagemath.com/projects/42c646bc-c7fc-41d8-b50d-efc995e3cd08/files/toeplitz.sagews

• I take this back: from mathoverflow.net/questions/68099/… it is seen that apparently for $n\geq 9$ there cannot be an explicit formula. Mar 26 '15 at 9:19