Correction: Non-Self-Adjoint Toeplitz Matrices Whose Principal Submatrices Have Real Spectrum

Correction to: Constr Approx (2019) 49:191–226 https://doi.org/10.1007/s00365-017-9408-0

In [3, Eq. (7)], the false equality

$$\begin{aligned} \langle u,T_{n}(\overline{a})u\rangle =\langle f_{u},\overline{a}f_{u}\rangle _{\gamma } \end{aligned}$$

has been used in the course of the proof of Theorem 8. If the terms are interpreted using the notation from [3], the right-hand side equals the integral

$$\begin{aligned} \frac{1}{2\pi \textrm{i}}\int _{-\pi }^{\pi }\overline{a(\gamma (t))}f_{u}(\gamma (t))\overline{f_{u}(\gamma ^{*}(t))} \frac{\dot{\gamma }(t)}{\gamma (t)}\textrm{d}t, \end{aligned}$$

while the left-hand side coincides with $\langle u,T_{n}^{*}(a)u\rangle$ and can be expressed as the integral

$$\begin{aligned} \frac{1}{2\pi \textrm{i}}\int _{-\pi }^{\pi }\overline{a(\gamma ^{*}(t))}f_{u}(\gamma (t))\overline{f_{u}(\gamma ^{*}(t))} \frac{\dot{\gamma }(t)}{\gamma (t)}\textrm{d}t. \end{aligned}$$

The two integrals do not coincide in general. Unfortunately, this error is an essential problem for the idea of the proof of Theorem 8, that was based on a contour integral representation of the quadratic form of a Toeplitz matrix $T_{n}(a)$, where the integration path is chosen as the Jordan curve $\gamma$ on which the symbol a is real-valued.

Recall that [3, Thm. 1], which is the main result of the paper, claims that the following 3 statements are equivalent:

• (i) $\Lambda (b)\subset {\mathbb {R}}$,

• (ii) $b^{-1}({\mathbb {R}})$ contains a Jordan curve,

• (iii) $\textrm{spec}(T_{n}(b))\subset {\mathbb {R}}$ for all $n\in \mathbb {N}$,

where b is a Laurent polynomial, $T_{n}(b)$ the $n\times n$ Toeplitz matrix given by the symbol b, and $\Lambda (b)$ is the set of limit points of eigenvalues of $T_{n}(b)$, as $n\rightarrow \infty$; see [3] for details. The logic of the proof of Theorem 1 was to establish implications (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(i). The currently unproven Theorem 8 was used to prove (ii)$\Rightarrow$(iii). As we are not able to find an alternative proof at the moment, the implication (ii)$\Rightarrow$(iii) remains unproven.

However, we strongly believe that the currently unproven implication, or at least its weaker form (ii)$\Rightarrow$(i), is true. The implication (ii)$\Rightarrow$(iii) is supported by numerous numerical experiments that we have performed. Professors Yi Zhang and Hao-Yan Chen, to whom we are sincerely grateful for noticing the latter error in the proof, expressed a similar opinion.

Recall, by [3, Def. 2], that the Laurent polynomial b is said to belong to the class $\mathscr {R}$, i.e. $b\in \mathscr {R}$, if and only if $\Lambda (b)\subset {\mathbb {R}}$. We list several places in [3], where arguments based on the claim of [3, Thm. 8] have been used:

• (a) Remark 10.

• (b) Sec. 4.1: Example 1. The symbol $b(z)=z^{-1}+az$, where $a\in {\mathbb {C}}{\setminus }\{0\}$, belongs to the class $\mathscr {R}$, if and only if $a>0$.

• (c) Sec. 4.2: Example 2. The symbol $b(z)=z^{-1}+\alpha z+\beta z^{2}$, where $\alpha \in {\mathbb {C}}$ and $\beta \in {\mathbb {C}}{\setminus }\{0\}$, belongs to the class $\mathscr {R}$, if and only if $\beta \in {\mathbb {R}}{\setminus }\{0\}$ and $\alpha ^{3}\ge 27\beta ^{2}$.

• (d) Sec. 4.3: Example 3. The symbol $b(z)=z^{-r}(1+az)^{r+s}$, where $r,s\in \mathbb {N}$ and $a\in {\mathbb {R}}{\setminus }\{0\}$, belongs to the class $\mathscr {R}$.

• (e) The second paragraph of Sec. 4.4.

The claim of [3, Rem. 10] was drawn as a direct consequence of Theorem 8 and remains open, too. The other points (b)–(e) comprise concrete examples of symbols, which belong to $\mathscr {R}$ for specific restrictions of the parameters. As the main argument for these claims, we found in [3] an explicit parametrization of the Jordan curve $\gamma$, for which $b\circ \gamma$ is real-valued in each of the cases. We will prove at least partly the claims without using Theorem 8.

The equivalence in (b) can be verified directly since the eigenvalues of $T_{n}(b)$ can be computed fully explicitly

$$\begin{aligned} \lambda _{k}=2(-1)^{n}\sqrt{a}\cos \frac{\pi k}{n+1},\quad k=1,2,\dots , n, \end{aligned}$$

with the standard branch of the square root. This example also exhibits (iii), if $a>0$.

The point (c) is left conjectural as inessential for the paper in its generality. Its relevant particular case, $b\in \mathscr {R}$ for $\alpha =3a^{2}$ and $\beta =a^{3}$, with $a\in {\mathbb {R}}{\setminus }\{0\}$, will be checked as a a special case of the point (d) in the end of this corrigendum. Lastly, a description of a possible construction of further examples of not necessarily banded Toeplitz matrices with real spectra from the point (e) is heavily based on the falsely proven Theorem 8 and also remains open at this point.

As a final correction, we briefly indicate the proof of the implication

$$\begin{aligned} b(z)=\frac{1}{z^{r}}(1+az)^{r+s} \text{ with } r,s\in \mathbb {N}\; \text{ and } \; a\in {\mathbb {R}}{\setminus }\{0\} \quad \Rightarrow \quad b\in \mathscr {R}. \end{aligned}$$

without using the fact that b is real on a Jordan curve. This is the claim (d). In fact, one can prove the stronger claim that the eigenvalues of $T_{n}(b)$ are all real, for all $n\in \mathbb {N}$, i.e. (iii). The argument relies on the theory of osciallatory matrices [2] and has been used for the special case $r=s=1$ in the proof of [1, Thm. 2.8]. Without loss of generality, we may assume $a=1$ since two Toeplitz matrices $T_{n}(b)$ and $T_{n}(b_{a})$, where $b_{a}(z):=b(az)$, are similar, if $a\ne 0$. Thus, suppose $b(z)=z^{-r}(1+z)^{r+s}$. Notice the bidiagonal matrix $T_{n}(1+z)$ is totally non-negative, see [2, Def. 4, p. 74]. Since $T_{n}(b)$ is a submatrix of $T_{n+r+s}^{r+s}(1+z)$, it is also totally non-negative; see [2, p. 74]. Further, without going into details, let us mention the determinant of $T_{n}(b)$ ban be explicitly computed in terms of binomial coefficients as follows:

$$\begin{aligned} \det T_{n}(b)=\prod _{j=0}^{r-1}\left( {\begin{array}{c}n+s+j\\ s\end{array}}\right) \bigg /\left( {\begin{array}{c}s+j\\ j\end{array}}\right) =\prod _{j=0}^{r-1}\frac{(n+s+j)!j!}{(n+j)!(s+j)!}. \end{aligned}$$

The determinant is obviously non-vanishing and therefore $T_{n}(b)$ is non-singular. By [2, Thm. 10, p. 100], $T_{n}(b)$ is oscillatory and hence eigenvalues of $T_{n}(b)$ are all positive (and simple), see [2, Thm. 6, p. 87].