Solve the following nonlinear algebraic equation

Question

I am trying to find the explicit equation for the eigenvalues of a very specific toeplitz finite-dimensional operator (a matrix of rank N). I have arrived at the following eigenvalue equation:

$$f_1(\lambda)^N - f_2(\lambda)^N = -\frac{\lambda^2}{1-\lambda}(f_1(\lambda)^{N-1} - f_2(\lambda)^{N-1}),$$

Where $f_{1,2}(\lambda) = 1 \pm \sqrt{1 + \lambda^2}$. In the above it is assumed that $N > 1$.

Is it possible to solve this analytically?

I must admit that this is not a kind of a textbook problem, so no guarantee that it is solvable explicitly, but I will be happy to hear any advice on solving this.

UPDATE: As Jean Marie suggested, I provide the form of a toeplitz operator $A$ that has the eigenvalue equation equivalent to the above: $A_{k,l} = i^{|k-l|}$, where $i$ is an imaginary unit, $k,l$ - indices of a matrix representing an operator $A$.

So, for $N=4$ it is:

$$ A^{(4)} = \begin{pmatrix} 1 & i & -1 & -i \\ i & 1 & i & -1 \\ -1 & i & 1 & i \\ -i & -1 & i & 1 \end{pmatrix}$$

There is something I don't understand... because, by an immediate recurrence, one has $f_1(\lambda)^N - f_2(\lambda)^N=(-1)^{N-1}\left(\frac{\lambda^2}{\lambda-1}\right)^{N-1}(f_1(\lambda)^1- f_2(\lambda)^1)=(-1)^{N-1} 2\left(\frac{\lambda^2}{\lambda-1}\right)^{N-1} \sqrt{1-\lambda^2}$ which is not a polynomial. — Jean Marie, Jun 12 '21 at 13:11
Dear Jean Marie! Not, actually the eigenvalue equation is the following: $\dfrac{(f_1^N - f_2^N)(1-\lambda)+(f_2^Nf_1 - f_1^Nf_2)}{f_1-f_2}=0$, but you can easily show that this is equivalent to what I wrote. — Sl0wp0k3, Jun 12 '21 at 13:27

Jean Marie · Accepted Answer · 2021-06-13T00:26:15.210

2

In fact, this matrix has an inverse; let us call it $B^{(N)}$, which is simpler to work with because it is tridiagonal and symmetric with the following form (where $h:=\frac12$ and $k:=-\frac12 i$):

$$B^{(N)}=\begin{pmatrix} h&k&0&0&\ddots&&\\ k&0&k&0&\ddots&&\\ 0&k&0&k&\ddots&&\\ &\ddots&\ddots&\ddots&\ddots&\ddots&\\ &&\ddots&k&0&k&0\\ &&\ddots&0&k&0&k\\ &&\ddots&0&0&k&h\\ \end{pmatrix}$$

(an upper and a lower diagonal filled with $k$ ; the main diagonal with $B_{k,k}=0$ but $B_{1,1}=B_{N,N}=h$ ; all other entries equal to $0$).

What is the interest of having a symmetric tridiagonal matrix ? Its eigenvalues can be obtained in particular by an algorithm recursively computing the characteristic polynomial (just as you have attempted to do for the direct matrix $A^{(N)}$).

Here is a graphical representation of the eigenvalues of $B^{(N)}$ for values of $N=2,3,4,\cdots 12.$

Fig. 1: An example: the $4$ eigenvalues of $B^{(4)}$ are represented by $4$ blue stars.

One sees in particular that the real part of these eigenvalues is positive and tends to $0$ when $N \to \infty$ whereas their imaginary part belongs to interval $[-1,1]$.

Of course, don't forget to take the inverses of these eigenvalues !

Remark: for a generalization, see my answer to this question.

edited Jun 13 '21 at 00:26

answered Jun 12 '21 at 14:28

Jean Marie

88,997

I had made an erroneous first redaction because... I hadn't the right matrix. Now it is fixed. – Jean Marie Jun 13 '21 at 00:30
1

Dear @Jean Marie, nice to see you here again, and thank you for such a detailed answer! A nice reference, I will definitely look closer to this. I have a feeling that I am really close to diagonalize the matrix, even though this can not be done completely analytically, it looks like it relies on very few constants that can be obtained numerically. I will provide a more detailed answer once I am done, but yours is of big interest as well! – Sl0wp0k3 Jun 13 '21 at 11:23
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One minor comment: if you consider $h$'s on both edges of a main diagonal as "perturbations", you can find the eigenvalues perturbatively! this is because if $h=0$, then this matrix can be diagonalized exactly, in physics this is known as tight-binding model (or a linear chain of $N$ identical masses connected with nearest-neighbors with springs). So the tight-binding problem (eigenvalues and eigenvectors) will be the $0^{\text{th}}$-order approximations, and a power series in terms of $h$ can be computed. – Sl0wp0k3 Jun 13 '21 at 11:27
I thought also to these exceptional values as "perturbations" in particular with the physics interpretation you mention. – Jean Marie Jun 13 '21 at 12:46

Solve the following nonlinear algebraic equation

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