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Let $x_i=(\sin i\pi/n,\cdots,\sin (n-1)i\pi/n)$ for $i=1,\cdots,n-1$. I want to show that $x_i \cdot x_j=\delta_{ij} n/2$. Why is it true? I tried $\sin a \sin b=-[\cos(a+b)-\cos(a-b)]/2$ but don't know how to clean the terms.

Edit This is from 1d poisson equation on interval: $\Delta u = f, u(0)=u(1)=0$. If we partition the interval and use approximation $\Delta u(t) \simeq (u(t+h)+u(t-h)-2u(t))/h^2$, we have a matrix equation $Ax=b$ where $A$ is the tridiagonal matrix with $a_{ii}=2, a_{i,i+1}=a_{i+1,i}=1$. $x_i$ are eigenvectors of $A$.

Gobi
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  • Eigenvalues of what laplacian? – Mariano Suárez-Álvarez Jun 14 '13 at 05:22
  • In any case, if $V$ is a vector space with an inner product and $f:V\to V$ is a self-adjoint linear map, then eigenvectors of $f$ corresponding to distinct eigenvalues are automatically orthogonal. Maybe this is the explanation you want? – Mariano Suárez-Álvarez Jun 14 '13 at 05:24
  • @MarianoSuárez-Alvarez This is from the 1d poisson equation on interval: $\Delta u = f, u(0)=u(1)=0$. Using finite difference method gives them as eigenvectors of corresponding matrix. – Gobi Jun 14 '13 at 05:54
  • Add that information to the body of the question itself (ideally, giving details abput what particular discretization you used and/or the actual matrix). Context almost always helps! – Mariano Suárez-Álvarez Jun 14 '13 at 05:55
  • Just note that, the Laplace operator is a symmetric operator which implies the eigenvectors are orthogonal. – Mhenni Benghorbal Jun 14 '13 at 06:21

2 Answers2

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Your matrix is symmetric, so it is self-adjoint with respect to the usual inner product on $\mathbb R^n$. It follows automagically from that that eigenvectors of that matrix corresponding to distinct eigenvalues are orthogonal —this should be proved in any textbook which deals with inner-product spaces.

This leaves you with computing the norms of the specific eigenvectors you have...

Mariano Suárez-Álvarez
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Did you mean $\sin a \sin b = [\cos (a-b) - \cos(a+b)]/2$?

First, show that $\cos {i\pi /n} + \cos {2i\pi/n} + \ldots \cos {(n-1)i\pi /n} = 0$ for all $i \neq 0$. Then using the above identity gives the required result.

polkjh
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  • Useful links to compute it: http://math.stackexchange.com/questions/195398/prove-that-cos2-pi-n-cos4-pi-n-cdots-cos2k-1-pi-n-1 and http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange.27s_trigonometric_identities – Gobi Jun 14 '13 at 08:38