8
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Are you completely sure you need the factor $2$ in the denominator of tangent? $\endgroup$ – Caran-d'Ache May 28 '13 at 18:04
5
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ Do you have intuition on why the rest of the coefficients of the polynomials are equal? $\endgroup$ – DVD Jun 1 '13 at 6:16
  • $\begingroup$ The answer to this question may be helpful: math.stackexchange.com/questions/392908/… $\endgroup$ – DVD Jun 1 '13 at 6:25
  • $\begingroup$ @Daved Thank you for telling me this link. $\endgroup$ – xpaul Jun 1 '13 at 11:41
  • $\begingroup$ Actually, to show that the rest of the coefficients are the same is to use some identities like the following $$ \sum_{k=1}^n\sin\frac{k\pi}{2n+1}=\frac{1}{2}\cot\frac{\pi}{2(2n+1)} $$ and $$ \prod_{k=1}^n\cos\frac{k\pi}{2n+1}=\frac{1}{2^n} $$ and others. $\endgroup$ – xpaul Jun 1 '13 at 11:47
  • $\begingroup$ How has nobody upvoted this??? $\endgroup$ – Eleven-Eleven Feb 14 '16 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.