Imagine there's a set of ordered coefficients $\lambda_1>\lambda_2>\ldots>\lambda_n>0$ which I don't know. However, I know the set of relations $$ \sum_{i=1}^n\lambda_i^k(-1)^{i+1}=a_k $$ for $k=1$ to $n$, with known values $a_k$. Is there anything smart that I can do in order to determine the $\lambda_i$?

I realise that if it weren't for the $(-1)^{i+1}$ term in my sum, we could use Newton's identities/Viete's formula to determine the polynomial which has roots $\{\lambda_i\}$, but have so far failed to spot a way of making use of this in my problem.

An alternative way to view this problem is that there's an unknown Vandermonde matrix $V$ for which we have to solve $Vx=a$ given known $x,a$. (Indeed, $x=(1,-1,1,-1,1,\ldots,1)^T$.)

Original context: I have a real symmetric tridiagonal matrix H with 0 on the diagonal, and it is also centrosymmetric. So, it looks something like $$ H=\left(\begin{array}{ccccccc} 0 & J_1 & 0 & \ldots & 0 & 0 \\ J_1 & 0 & J_2 &&0 & 0 \\ 0 & J_2 & 0 && 0 & 0 \\ \vdots &&& \ddots &\vdots & 0 \\ 0 & 0 & 0 & \ldots & 0 & J_1 \\ 0 & 0 & 0 & \ldots & J_1 & 0 \end{array}\right). $$ For the $N\times N$ matrix, I want to fix $2k+1$ of the eigenvalues. If $2k+1=N$, this is a standard symmetric inverse eigenvalue problem, but I want to consider smaller values of $k$. You might formulate this generally, although I'm actually interested in the case where the eigenvalues are $0,\pm 1,\pm 2,\ldots \pm k$. The remaining eigenvalues are unspecified. They have to occur in $\pm\lambda$ pairs, and I want the smallest (positive) value to be larger than $k$. These will be determined by the additional constraint - that I want to fix the $N-1-2k$ central coupling strengths to be a specific set of values (again, in the specific case I'm looking at, I'm assuming they're all the same value $J$, and that $J$ happens to be in a range that allows a solution. I'm deferring the problem of determining what that range might be).

The way that I'm trying to approach this is to use some symmetry properties. If $S$ is the swap operator, we can evaluate $\text{Tr}(SH^k)$ both in terms of the eigenvalues of $H$ and in terms of the known coupling strengths. I then get a bunch of equations for my unknown eigenvalues which are those of the original problem statement. With those eigenvalues, I can run the standard inverse eigenvalue routine to determine the matrix $H$.

I've got some good numerical techniques for finding solutions based on this paper. I was just wondering if I could find solutions more directly.

  • 2
    $\begingroup$ For which values of $k$ does your relation hold? Do you know the values of the $a_i$ and you're looking for a solution in the form $\lambda_i(a_j)$? Your relation doesn't quite make sense because you have $i$ as a bound variable in the sum on the left-hand side and as a free variable on the $a_i$ on the right hand side. Did you mean to put $a_k$? $\endgroup$
    – Jojo
    Dec 16, 2020 at 19:05
  • $\begingroup$ yes, I did. Sorry! $\endgroup$
    – DaftWullie
    Dec 17, 2020 at 7:56
  • $\begingroup$ Thanks for clarifying, this is an interesting and complex problem. I don't think your intuition in writing it as a linear system $Vx=a$ will help; you need to solve coupled polynomial equations in the $\lambda_i$ so it's more complex than solving a linear system. I will spend some time to think about it and see what I can say. Do you have any context for the problem? It might help in finding a solution $\endgroup$
    – Jojo
    Dec 17, 2020 at 16:09
  • $\begingroup$ Sorry, I may be missing something, but if you can solve the form without the $(-1)^{i+1}$ factor, why not just define $\gamma_i = (-1)^{i+1} \lambda_i$ and solve? $\endgroup$ Jul 11, 2023 at 14:55
  • $\begingroup$ @ChrisLewis This substitution doesn't work when $k$ is even. $\endgroup$
    – DaftWullie
    Jul 12, 2023 at 6:47


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