# How to show this matrix has eigenvalues $\lambda_{j}=4\sin^2{\frac{j\pi}{2(n+1)}}$?

Show that the $n\times n$ tridiagonal matrix $$A=\begin{bmatrix} 2&-1&0&0&0\\ -1&2&-1&0&0\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0&0&-1&2&-1\\ 0&0&0&-1&2 \end{bmatrix}$$

has the eigenvalues $$\lambda_{j}=4\sin^2{\dfrac{j\pi}{2(n+1)}},j=1,2,\cdots,n$$

• joriki has written down the eigenvectors. If you don't understand his answer, you may ping him first to see if he has time to give you more details. – user1551 May 8 '13 at 9:36
• yes, I have some questions, and don't have solve it,Thank you – math110 May 8 '13 at 10:20

So we need to solve the following problem: $Ax =\lambda x$.
Or $$cx_{j−1} + ax_j + bx_{j+1} = \lambda x_j , \ j = 1,\ldots ,m$$ $$x_0 = x_{m+1} = 0$$ where $a=2, \ b=c=-1$ which is equivalent to $$cx_{j−1} +(a-\lambda)x_j + bx_{j+1} = 0 , j = 1,\ldots ,m$$ $$x_0 = x_{m+1} = 0$$ Its solution is represented in the form: $$x_j = \alpha r_1^j+\beta r_2^j$$ where $r_1^j, \ r_2^j$ are solutions of the characteristic polymonial $f(r)=b r^2+(a-\lambda)r+c$: $$r_1=\frac{-(a-\lambda)+\sqrt{(a-\lambda)^2-4b c}}{2b}$$ $$r_2=\frac{-(a-\lambda)-\sqrt{(a-\lambda)^2-4b c}}{2b}$$ We can find the unknown coefficients by using the initial condition: $$x_0 = \alpha+\beta = 0 \Rightarrow \alpha=-\beta$$ $$x_j = \alpha (r_1^j-r_2^j)$$ $$x_{m+1} = \alpha(r_1^{m+1}-r_2^{m+1})=0 \Rightarrow \left(\frac{r_1}{r_2}\right)^{m+1}=1$$ $$r_1 r_2=\frac{c}{b}$$ $$\left(\frac{r_1}{r_2}\right)^{m+1}=\left(\frac{b r_1^2}{c}\right)^{m+1}=1$$ Plugging in $b=c=-1$ leads us to $r^{m+1}=1$, so $$r_{1,j}=e^{\pi i \left(\frac{j}{n+1}\right)}$$ $$r_{2,j}=e^{-\pi i \left(\frac{j}{n+1}\right)}$$ But one can see that $r_{1,j}+r_{2,j}=\frac{\lambda_{j}-a}{b}=2-\lambda_{j}$. Using the Euler's formula one can obtain: $$\lambda_{j}=2\left(1-\cos\left(\frac{j\pi}{n+1} \right)\right)=4\sin^2\left(\frac{j\pi}{2(n+1)} \right)$$
Hint: Find $\det (A-\lambda I)$ by induction on size of $A$ (you will get a recurrent equation of order 2).