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Consider a real and symmetric tridiagonal matrix with zero diagonals and where subdiagonals and superdiagonals are equal to 1 except the (1,2)-th component being equal to $a$, i.e.,

$$\begin{bmatrix}0 & a & 0 & 0 & \cdots & 0\\1 & 0 & 1 & \cdots & 0\\ 0 & 1 & 0 & 1 & \cdots & 0\\ & & & \ddots \\0 & \cdots & 0 & 0 & 1 & 0\end{bmatrix}.$$

For $a=1$ the spectrum is 2 $\cos (\pi j / (n+1)), j=1, \ldots, n$. I want to determine its spectrum for $a \neq 1$.

Thanks in advance!

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    $\begingroup$ After symmetrizing this, it is a scalar times the matrix in this question: mathoverflow.net/questions/141322/… $\endgroup$
    – Douglas Zare
    Commented May 23, 2014 at 12:31
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    $\begingroup$ should be on mse $\endgroup$
    – username
    Commented May 23, 2014 at 15:14
  • $\begingroup$ Asymptotically, the behaviour of the roots of the characteristic polynomial is the same as for when $a=1$. $\endgroup$
    – Per Alexandersson
    Commented Oct 10, 2014 at 23:26

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Your matrix is not symmetric.

Let $P_n(x)=\Pi_{j=1}^n(x-2\cos(\dfrac{\pi j}{n+1}))$ and let $Q_n(x)=\det(xI_n-U_n)$ where $U_n$ is the considered matrix that depends on $a$. Then $Q_n(x)=(1-a)xP_{n-1}(x)+aP_n(x)$.

EDIT: moreover $P_n(x)=\dfrac{T(n+2,x/2)-T(n,x/2)}{x^2/2-2}$ where $T_n$ is the $n^{th}$ Chebyshev polynomial. That implies $Q_n(x)=\dfrac{(x^2-2a)T(n,x/2)+(a-2)xT(n-1,x/2)}{x^2/2-2}$. To find a closed from for the roots of the numerator is hopeless. Indeed, if $n=16$ and $a=3$, then $Q_{16}$ is irreducible and has a non-solvable Galois group. In particular, according to the Kronecker-Weber theorem, the roots are not in an extension $\mathbb{Q}(\omega)$, where $\omega$ is a root of unity.

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