# Tridiagonal matrix w/trigonometric eigenvalues

Let $n$ be a natural number and $B$ be the $n\times n$ square matrix of $0$'s and $1$'s $$B=\begin{pmatrix} 0 & 1 & 0 & \ldots & 0 \\ 1 & 0 & 1 & \ldots & 0 \\ 0 & 1 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & 0 & 1 \\ 0 & 0 & \ldots & 1 & 1 \\ \end{pmatrix}.$$ Is there a way to find the $n\times n$ diagonal matrix D, such that the eigenvalues of the product matrix $DB$ are $$2\sin\left(\frac{k\pi}{2n+1}\right), \,\, k=1,2,\dots,n.$$ The pertaining old question: Eigenvalues of a tridiagonal trigonometric matrix

• Crossposted to physics.stackexchange.com/q/237740/2451 – Qmechanic Feb 16 '16 at 22:53
• pertains Chebyshev polynomials of the first and second kind... – DVD Feb 21 '16 at 22:11
• The proof of the existence of the diagonal matrix $D$ w/alternating in sign entries involves special matrices and continued fractions... – DVD Feb 24 '16 at 4:10
• Would you accept an answer where the angle is $\frac{2k\pi}{2N+1}$ or $\frac{(2k-1)\pi}{2N+1}$ for any$N$? Currently, I am getting cosine function but I might be able to fix that. – Vini Mar 19 '17 at 0:57
• @Vini Yes, seems like a good progress... – DVD Mar 28 '17 at 0:45