Eigenvalues of a tridiagonal trigonometric matrix

Question

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)?

Answer 1

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*}