What is the trace of the square of the matrix $T_{ij}=t_{i-j}$, with $t_k=t_{-k}=c^k/k$? I want to know the sum of the squares of the eigenvalues of the traceless, symmetric, complex, $NxN$ Toeplitz matrix $T_{ij}=t_{i-j}$, with $t_k=t_{-k}=c^k/k$, $t_0=0$. The Szegoe Limit Theorem may provide the answer, but for large $N$, since the generating function is a high-order trigonometric polynomial, it may be difficult to obtain a closed form expression in terms of $N$ and a complex number $c, |c|>1$.
What is the best way to proceed? Thanks!
 A: In any Toeplitz matrix $T$ with $T_{ij}=t_{i-j}$, one has
\begin{align*}
\operatorname{tr} T^2
&=\sum_{i=1}^n\sum_{j=1}^n T_{ij}T_{ji}\\
&=\sum_{i=1}^n\sum_{j=1}^n t_{i-j}t_{j-i}\\
&=\sum_{k=-(n-1)}^{n-1}t_kt_{-k}(n-|k|).
\end{align*}
So, since $t_kt_{-k}=\frac{c^k}{k^2}$ for $k\neq 0$ and $t_0=0$, one has
$$\operatorname{tr}T^2=2\sum_{k=1}^{n-1}(n-k)t_kt_{-k}=2\sum_{k=1}^{n-1}\frac{c^{2k}(n-k)}{k^2}.$$
We claim that, for $|c|>1$,
$$\operatorname{tr}T^2=\frac{2c^{2n}}{n(1-c^{-2})}(1+o(1)).$$
Indeed,
$$\frac{\operatorname{tr}T^2}{2c^{2n}}=\sum_{j=1}^{n-1}\frac{jc^{-2j}}{(n-j)^2};$$
we see
\begin{align*}
\frac{c^2}{(c^2-1)^2}-n^2\sum_{j=1}^{n-1}\frac{jc^{-2j}}{(n-j)^2}
&=\sum_{j=n}^\infty jc^{-2j}-\sum_{j=1}^{n-1} jc^{-2j}\left(\frac{n^2}{(n-j)^2}-1\right)\\
&=O(nc^{-2n})+O\left(\sum_{j=1}^{n-1}\frac{j^2nc^{-2j}}{(n-j)^2}\right)\\
&=O(nc^{-2n})+O\left(\sum_{j<n^{1/4}}\frac{j^2nc^{-2j}}{n^2}+\sum_{j\geq n^{1/4}}^{n-1}n^3c^{-2j}\right)\\
&=O(nc^{-2n})+O\left(n^{-1/4}\right)+O\left(n^4c^{-2n^{1/4}}\right)\\
&=o(1).
\end{align*}
So, as $n$ grows, $\operatorname{tr}T^2$ has magnitude growing like some constant times $c^{2n}/n$, and variable argument (roughly $2n\phi$ plus some constant, if $c=e^{i\phi}$).
