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Let $\epsilon_1,\ldots,\epsilon_n$ be a sequence of signs and $M(t)$ be the tridiagonal matrix whose diagonal entries are $\epsilon_1 t,\ldots, \epsilon_n t$ and off-diagonal entries equal to $1$. Is it true that the number of real zeroes of $P(t)=\det M(t)$ is equal to the absolute value of $\epsilon_1+\dots+\epsilon_n$?

I checked this numerically on various cases, tried to apply the Sturm algorithm but couldn't prove it. Thanks for any help!

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For $\epsilon_1=\epsilon_2=-1$ and $\epsilon_3=\epsilon_4=\epsilon_5=1$ you get the counterexample $\operatorname{det}M(t)=t(t - 1)^2(t + 1)^2$.

Another example, with simple real roots, is $\epsilon_1=\epsilon_5=-1$ and $\epsilon_2=\epsilon_3=\epsilon_4=1$ with $\operatorname{det}M(t)=(t - 1)t(t + 1)(t^2 + 1)$.

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  • $\begingroup$ Thanks for this perfect answer, do you have an example with $n$ even? I am specially interested in the case when the sequence is palindromic, that is $\epsilon_k=\epsilon_{n+1-k}$. $\endgroup$
    – Julien Marché
    Commented Jun 2, 2023 at 21:17
  • $\begingroup$ @JulienMarché Yes, for $\epsilon_1,\ldots,\epsilon_8=-1, -1, 1, 1, 1, 1, -1, -1$ we have $\operatorname{det}M(t)= (t - 1)(t + 1)(t^3 - t - 1)(t^3 - t + 1)$. $\endgroup$
    – Peter Mueller
    Commented Jun 2, 2023 at 21:33
  • $\begingroup$ Thanks again, this is very helpful. althought this is not a good news for my problem... $\endgroup$
    – Julien Marché
    Commented Jun 3, 2023 at 7:47

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