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I have a linear map that is defined by $$T:\text{lin}(1,...,x^m) \rightarrow \text{lin}(1,...,x^m) \text{ with}$$ $$x^k \mapsto 2w(k-m)x^{k+1}+(k^2-k-w^2)x^k-2kwx^{k-1}+(k-k^2)x^{k-2}$$

Let me give a few comments on this: Apparently the signs of the coefficients of the terms of $x^{k+1}$ and $x^{k-1}$ have always the same sign. If we would neglect the term $(k-k^2)x^{k-2}$, then this would mean that we have a tridiagonal matrix satisfying $2w(k-m) \cdot (-2kw) \geq 0,$ which would mean that we would definitely know that this matrix would have a real spectrum. Since we also have this term with the either zero or positive coefficient $(k-k^2)$, I would like to know if we can still be sure to find for any $m \in \mathbb{N}$ real eigenvalues.

If anything is unclear about this question, please let me know.

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  • $\begingroup$ Have you checked any cases by hand (or computer)? A numerical check for $2\leq m\leq 10$ will give you a good hint. Also, do you mean $x^{-1}=x^{-2}=0$ in your formula? $\endgroup$
    – Joonas Ilmavirta
    Commented Sep 28, 2014 at 15:38
  • $\begingroup$ well, if I plug in $T(1)$, then $k=0$, so $x^{-1}$ and $x^{-2}$ vanish. And if I plug in $T(x)$, then $k=1$, so $x^{-1}$ vanishes. and yes, I have checked a few cases, although not all the ones that you suggested, cause the equations become so nasty for $m \ge 4$, that you cannot tell in general, if the eigenvalues are real for all $w$ or not. $\endgroup$
    – BaoLing
    Commented Sep 28, 2014 at 15:58

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