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How do i diagonalize the discrete laplace-operator $\triangle$ on $l^2(\mathbb{Z})$ (defined by e.g. $\triangle e_k = e_{k-1} + e_{k+1}$, with $e_{k}$ being the canoical basisvectors of $l^2(\mathbb{Z})$? I showed that $\triangle$ is linear, bounded and selfadjoint but I can't figure out how exactly to calculate the diagonalization. Do I have to calculate $\mathcal{F} \triangle \mathcal{F}^{-1}$ with $\mathcal{F} : l^2(\mathbb{Z}) \rightarrow L^2([0,2\pi])$, $(\mathcal{F} f)(x) = \sum_{k \in \mathbb{Z}} e^{ikx} f(k)$?

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Shouldn't $\Delta e_k$ be $e_{k-1}-2e_k+e_{k+1}$ instead? –  Rahul Jan 6 '13 at 16:33
    
Welcome to Math.SE! Although it's fine to ask questions as an unregistered user, it's best to register so that you won't lose access to your account when a cookie gets deleted. On the mathematical side, it's a good idea to look at the $L^2([0,2\pi])$ side, because translation becomes multiplication under FT. The discrete Laplacian ($\Delta$, not $\triangle$) corresponds to multiplication by $2\cos x-2$, which shows what the spectrum is, and also that there are no eigenvalues. –  user53153 Jan 6 '13 at 17:54
    
I was going to say $\vec{f} = (\dots, 1,1,1, \dots)$ and $\Delta \vec{f} = 2 \vec{f}$ but $\vec{f} \notin l^2(\mathbb{Z})$. –  john mangual Jan 7 '13 at 3:58
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