TITLE: Eigenvalues of a tridiagonal trigonometric matrix
QUESTION [8 upvotes]: Let $D$ be the diagonal matrix w/alternating in sign diagonal entries: 
$$D_{kk}=(-1)^{k+1}\tan(\frac{k\pi}{2n+1}),$$
where $k=1,2,\dots n\in N$, and let $B$ be the $n$ by $n$ square $(0,1)$-matrix
$$B=
\begin{pmatrix}
 0     & 1 & 0 & \ldots & 0 \\
 1     & 0 & 1 & \ldots & 0 \\
 0     & 1 & \ddots & \ddots & \vdots \\
 \vdots    & \vdots & \ddots & 0 & 1 \\
 0     & 0  & \ldots & 1 & 1 \\
\end{pmatrix}.
$$
(a) Prove, that the eigenvalues of the product matrix $(-1)^{n+1}DB$ are
$$
 2\sin\left(\frac{k\pi}{2n+1}\right), \,\, k=1,2,\dots,n.
$$
The result follows from a continued fraction identity w/a lengthy proof and an exercise from the open Wiki book "On 2D Inverse Problems" (http://en.wikibooks.org/wiki/On_2D_Inverse_Problems), but a direct shorter proof w/some geometric intuition would be very useful.  
(b) The matrix $D$ is a discrete version of the operator $\frac{d^2}{dx^2}+2+\delta$. Is there differential/continuous/limiting equation version of the result in (a)?

REPLY [5 votes]: Let me use $n=5$ to show the result. It is easy to generalize the result for general $n$. For $n=5$, let $a_k=\tan\frac{k\pi}{11}$ and 
$$ A=\left(\begin{matrix}a_1&0&0&0&0\\
0&-a_2&0&0&0\\
0&0&a_3&0&0\\
0&0&0&-a_4&0\\
0&0&0&0&a_5\\
\end{matrix}\right),D=\left(\begin{matrix}0&1&0&0&0\\
1&0&1&0&0\\
0&1&0&1&0\\
0&0&1&0&1\\
0&0&0&1&1\\
\end{matrix}\right). $$
Then the corresponding characteristic polynomial is 
\begin{eqnarray*} p(\lambda)&=&\det(\lambda I-AD)=\left(\begin{matrix}\lambda&-a_1&0&0&0\\
a_2&\lambda&a_2&0&0\\
0&a_3&\lambda&-a_3&0\\
0&0&a_4&\lambda&a_4\\
0&0&0&-a_5&\lambda-a_5\\
\end{matrix}\right)\\
&=&\lambda^5-a_5x^4+(a_1a_2+a_2a_3+a_3a_4+a_4a_5)x^3-(a_1a_2a_5+a_2a_3a_5+a_3a_4a_5)x^2\\
&=&+(a_1a_2a_3a_4+a_1a_2a_4a_5+a_2a_3a_4a_5)x-a_1a_2a_3a_4a_5.
\end{eqnarray*}
Let
$$f(\lambda)=(\lambda-b_1)(\lambda-b_2)(\lambda-b_3)(\lambda-b_4)(\lambda-b_5)$$ 
where $b_k=2\sin\frac{k\pi}{11},k=1,2,3,4,5$. Now we show that $p(\lambda)$ and $f(\lambda)$ are have the same coefficient for each $x^k$, $k=0,1,2,3,4$ and hence $b_k,k=1,2,3,4,5$ are the eigenvalues of $AD$. For simplicity, we just show that the constant terms of these two polynomials and the coefficients of $x^4$ are the same, respectively, namely.
$$ b_1b_2b_3b_4b_4b_5=a_1a_2a_3a_4a_5, b_1+b_2+b_3+b_4+b_5=a_5 $$
and the rest will be tedious computations.
In fact, since
\begin{eqnarray*}
&&32\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}\\
&=&\frac{32\sin\frac{\pi}{11}\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}}{\sin\frac{\pi}{11}}\\
&=&\frac{16\sin\frac{2\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}}{\sin\frac{\pi}{11}}\\
&=&\frac{8\sin\frac{4\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}}{\sin\frac{\pi}{11}}=\frac{4\sin\frac{8\pi}{11}\cos\frac{3\pi}{11}\cos\frac{5\pi}{11}}{\sin\frac{\pi}{11}}\\
&=&\frac{-4\sin\frac{3\pi}{11}\cos\frac{3\pi}{11}\cos\frac{6\pi}{11}}{\sin\frac{\pi}{11}}=-\frac{2\sin\frac{6\pi}{11}\cos\frac{6\pi}{11}}{\sin\frac{\pi}{11}}=-\frac{\sin\frac{12\pi}{11}}{\sin\frac{\pi}{11}}=1,
\end{eqnarray*}
we have
\begin{eqnarray*}
b_1b_2b_3b_4b_4b_5&=&32\sin\frac{\pi}{11}\sin\frac{2\pi}{11}\sin\frac{3\pi}{11}\sin\frac{4\pi}{11}\sin\frac{5\pi}{11}\\
&=&(\tan\frac{\pi}{11}\tan\frac{2\pi}{11}\tan\frac{3\pi}{11}\tan\frac{4\pi}{11}\tan\frac{5\pi}{11})(32\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11})\\
&=&a_1a_2a_3a_4a_5.
\end{eqnarray*}
Using that
$$ \sum_{k=1}^N\sin k\theta=\frac{1}{2}\cot\frac{\theta}{2}-\frac{\cos(N+\frac{1}{2}\theta)}{2\sin\frac{\theta}{2}}  $$
we have
\begin{eqnarray*}b_1+b_2+b_3+b_4+b_5
&=&2(\sin\frac{\pi}{11}+\sin\frac{2\pi}{11}+\sin\frac{3\pi}{11}+\sin\frac{4\pi}{11}+\sin\frac{5\pi}{11})\\
&=&2\cdot\frac{1}{2}\cot\frac{\pi}{22}=\tan(\frac{\pi}{2}-\frac{\pi}{22})=\tan\frac{5\pi}{11}=a_5
\end{eqnarray*}