Eigenvalues of two symmetric tridiagonal Toeplitz matrices

Question

I am trying to find the eigenvalues of the following two $n \times n$ symmetric tri-diagonal Toeplitz matrices (let us call them $A$ and $B$ respectively):

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Note that the standard way of computing the eigenvalues is by computing the characteristic polynomial and then finding the roots of the same. This seems to be done in this paper, in Lemma $43$ --- moreover the characteristic polynomial of one of these matrices seems to be a function of the characteristic equation of another --- but the steps are not clear.

Answer 1

Consider the first matrix $A$ of dimension $n\times n.$ It suffices to determine the eigenevalues of $C=A-2I.$ Let $v_t=\left \{{\sin kt\over \sin t }\right \}_{k=1}^n $ for $0<t<\pi.$ Then $v_t\neq 0$ as the first coordinate is equal $1.$ We have $$(Cv_t)_k= \begin{cases} \displaystyle {\sin (2t)\over \sin t} & k=1\\ \displaystyle{\sin (k-1)t\over \sin t }+{\sin (k+1)t\over \sin t } & 2\le k\le n-1\\ \displaystyle{\sin (n-1)t\over \sin t } & k=n \end{cases}\\ = \begin{cases} \displaystyle 2\cos t & k=1\\ \displaystyle 2\cos t {\sin(kt)\over \sin t} & 2\le k\le n-1\\ \displaystyle 2\cos t {\sin(nt)\over \sin t}-{\sin(n+1)t\over \sin t} & k=n \end{cases} $$ Let $t_m=\displaystyle {m\pi\over n+1},$ $1\le m\le n.$ Then $\sin(n+1)t_m=0$ and $$Cv_{t_m}=(2\cos t_m) v_{t_m},\quad 1\le m\le n$$ In this way we have obtained $n$ different eigenvalues $2\cos t_m$ of the matrix $C.$ Hence the numbers $2+2\cos t_m,$ $1\le m\le n,$ are the eigenvalues of $A.$

Answer 2

For matrix $B$ it seems to be $2-2\cos\frac{k\pi}{n}$ , $ 0 \leq k \leq n-1$, at least for small $n$ values.

Answer 3

I am a bit late to the party but your B matrix is a perturbed 2-Toeplitz matrix, for which you can find analytical expressions for the characteristic polynomial (not sure you will be able to solve those analytically for your matrix). Check "The Characteristic Polynomial of Some Perturbed Tridiagonal k-Toeplitz Matrices" from Fonseca