I'm considering $(N+1\times N+1)$-matrices of the form

\begin{equation} \newcommand{\iu}{\mathrm{i}} \newcommand{\euler}{\mathrm{e}} A = \begin{pmatrix} 1 & \euler^{\iu\phi} & \euler^{\iu2\phi} &\dots& \euler^{\iu N\phi} \\ \euler^{\iu\phi} & 1 & \euler^{\iu\phi} & \dots & \euler^{\iu (N-1)\phi} \\ \euler^{\iu2\phi} & \euler^{\iu\phi} & 1 & & \vdots \\ \vdots & \vdots& &\ddots& \euler^{\iu\phi}\\ \euler^{\iu N\phi} & \euler^{\iu(N-1)\phi} & \dots &\euler^{\iu\phi}& 1 \end{pmatrix}, \end{equation} i.e., matrices $(A)_{nm} = \euler^{\iu\phi|n-m|}$.

I want to have a (if possible closed form) expression for the eigenvalues. Mathematica readily finds them, even for $N = 30$. Also the structure of these matrices is relatively simple. Thus I assume there are probably known results for something like the eigenvalues and eigenvectors of $A$. However, I wasn't able to find my matrix in the list of matrices.

My questions are:

  • For $\phi \in \mathbb{R}$ are these matrices called and what are, for a general $N$, the eigenvalues of $A$? This question is likely related to this question (thanks to omnomnomnom)
  • If there is no closed form, is there are an approximate expression for the eigenvalues and eigenvectors for $N\to\infty$?
  • $\begingroup$ Is $\phi$ a real number? $\endgroup$ Jan 17, 2020 at 8:16
  • $\begingroup$ I assume this doesn't matter (to mathematica it doesn't), but if this helps you may take $\phi \in \mathbb{R}$. $\endgroup$
    – manthano
    Jan 17, 2020 at 8:18
  • 2
    $\begingroup$ As far as the pattern of entries go, your matrix is symmetric and Toeplitz. There are algorithms to speed up the computation of eigenvalues/eigenvectors of such matrices, but there is apparently no closed form for their eigenvalues in general. If there's something special about your matrix, it would have to use the fact that the entries have the form $e^{in\theta}$. $\endgroup$ Jan 17, 2020 at 8:25
  • 2
    $\begingroup$ Actually, following a few more links, I see that this question is exactly the same as yours, where they have taken $\alpha = e^{i \phi}$. $\endgroup$ Jan 17, 2020 at 8:27
  • $\begingroup$ Thank you for your research and comments @Omnomnomnom, I adapted my question accordingly. $\endgroup$
    – manthano
    Jan 17, 2020 at 8:38

1 Answer 1


Following answer is probably as close to a solution as one can get within reasonable effort. Much of what I'm about to write is based on following two questions and answers therein:

Diagonalizing a [...] Symmetric Toeplitz matrix

[...] determinant of a tridiagonal matrix with constant diagonals

Part 1 - Inverse of $A$

$A$ possesses an inverse, given by

\begin{equation} (A^{-1})_{nm} = \frac{1}{1-e^{2i \phi}} \begin{cases} 1 & n = m = 1, N \\ 1+e^{i\phi} & n = m \\ - e^{i\phi} & |n-m| = 1 \\ 0 & \text{otherwise} \end{cases}, \end{equation} i.e., the inverse of $A$ is a tridigonal matrix with near constant diagonals, except for $(A^{-1})_{1,1}$ and $(A^{-1})_{N,N}$.

Part 2 - Characteristic Polynomial of $A^{-1}$ For the calculation of the characteristic Polynomial

$$\chi(\lambda) = \det\big(A^{-1} - \lambda\big)$$

it's useful to consider the substitution

$$\lambda = \frac{1}{1-e^{2 i \phi}} \mu + \frac{1+e^{2i\phi}}{1-e^{2i\phi}}.$$

With that, we have $\det(A^{-1} - \lambda) = 0 \Leftrightarrow \det(\tilde{A}^{-1} - \mu) = 0$ with

\begin{equation} \tilde{A}^{-1} = \begin{cases} -e^{2i\phi} & n = m = 1,N \\ -e^{ i\phi} & |n-m| = 1 \\ 0 & \text{otherwise} \end{cases} \end{equation}

Now, by defining the auxilliary $N\times N$ matrix $B_N$ which contains $-e^{i\phi}$ on the first of diagonals and $0$ on all other elements (i.e. $(B_N)_{n,m} = (\tilde{A}^{-1})_{n,m}$ expect for $(B_N)_{1,1} = (B_N)_{N,N} = 0$), one derives

\begin{equation} \det(\tilde{A}^{-1} - \mu) = (e^{2i\phi} - \mu)^2 \det(B_{N-2}) + 2 e^{2i\phi} (e^{2i\phi} - \mu) \det(B_{N-3}) + e^{4i\phi} \det(B_{N-4}). \end{equation}

For $\det(B_N)$ one can derive a recursion relation, like in the second linked question, which gives $\det(B_N)$ in terms of Chebyshev polynomials. For $\mu^2 \ne 4 e^{2i\phi}$ and after one further substitution $\tilde{\mu} = \mu e^{-i\phi}$ one finds the characteristic Polynomial

\begin{align} &\det(\tilde{A}^{-1} - \mu) = 0 \Leftrightarrow \\ &\color{blue}{(e^{i\phi} - \tilde{\mu})^2 U_{N-2}(\tilde{\mu}/2) - 2(e^{i\phi} - \tilde{\mu}) U_{N-3}(\tilde{\mu}/2) + U_{N-4}(\tilde{\mu}/2) = 0}, \end{align} where $U_N$ is the Chebyshev Polynomial of the second kind.

Part 3 - short discussion

With the knowledge of the solutions of the characteristic equation for $\tilde{\mu}$ on directly finds the the eigenvalues $\lambda$ of $A^{-1}$ and, with that, for $A$. $U_N(x)$ is a polynomial of degree $N$ with $N$ distinct roots $x_k = \cos(\pi k/(N+1))$ with $k=1,\dots,N$. One may hope, that these are sufficiently close to the actual roots and do pertubation theory on those roots (e.g. by using Newton's method to calculate corrections to $x_k$). However, we are calculating the eigenvalues of $A^{-1}$ and some of the (approximative) eigenvalues $x_k$ lie close to $0$ (namely for $k$ close to $N/2$). Thus, small corrections to those, mean enormous corrections to the resulting approximative eigenvalue of $A$. Therefore finding approximative eigenvalues with the above stated "algorithm" is not well suited.

Finally, there is one more option, to get to a "simpler" eigenvalue equation for $A^{-1}$. By substituting once more $\tilde{\mu} = \nu + \nu^{-1}$ the Chebyshev Polynomials reduce to $U_N(\tilde{\mu}/2) = \nu^{N-1}-\nu^{-N+1}$. After rearranging some powers one ends up with

\begin{align} &\det(\tilde{A}^{-1} - \mu) = 0 \Leftrightarrow \\ &\nu^{2N}(e^{i\phi} + \nu)^2 - (1 + e^{i\phi}\nu)^2 = 0. \end{align}

Notice, this is now a polynomial of degree $2N+2$, however $\nu = \pm 1$ is a solution, which we already excluded, since we assumed $\mu^2 \ne 4e^{2i\phi}$ and all other roots come in pairs: if $\nu_k$ is a solution, so is $\nu_k^{-1}$. However, they provide the same $\tilde{\mu}$ after resubstitution. All in all, while this polynomial equation is much more compact, it didn't helped me find a closed form solution or an good approximation of the original problem (this is, the eigenvalues of $A$).


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