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Consider the discrete second derivative with Dirichlet boundary conditions on $\mathbb C^n$.

Its eigendecomposition is fully known: see wikipedia

It seems like the lowestlargest eigenvalue $\lambda_1$ is one with a fast decaying eigenfunction, by this I mean that at the first coordinate $\vert v_{1,1} \vert \le Cn^{-3/2}.$ The first $1$ indicates the eigenfunction, the second one the coordinate.

A priori there is no reason to have this type of decay, at the first coordinate, I guess.

Is there a way to prove this without(!) using that the eigenfunctions are explicitly known?-Thus, can one show this directly from the matrix?

Consider the discrete second derivative with Dirichlet boundary conditions on $\mathbb C^n$.

Its eigendecomposition is fully known: see wikipedia

It seems like the lowest eigenvalue $\lambda_1$ is one with a fast decaying eigenfunction, by this I mean that at the first coordinate $\vert v_{1,1} \vert \le Cn^{-3/2}.$

A priori there is no reason to have this type of decay.

Is there a way to prove this without(!) using that the eigenfunctions are explicitly known?-Thus, can one show this directly from the matrix?

Consider the discrete second derivative with Dirichlet boundary conditions on $\mathbb C^n$.

Its eigendecomposition is fully known: see wikipedia

It seems like the largest eigenvalue $\lambda_1$ is one with a fast decaying eigenfunction, by this I mean that at the first coordinate $\vert v_{1,1} \vert \le Cn^{-3/2}.$ The first $1$ indicates the eigenfunction, the second one the coordinate.

A priori there is no reason to have this type of decay, at the first coordinate, I guess.

Is there a way to prove this without(!) using that the eigenfunctions are explicitly known?-Thus, can one show this directly from the matrix?

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Consider the discrete second derivative with Dirichlet boundary conditions on $\mathbb C^n$.

Its eigendecomposition is fully known: see wikipedia

It seems like the lowest eigenvalue $\lambda_N$$\lambda_1$ is the one with the fastesta fast decaying eigenfunction, by this I mean that at the first coordinate $\vert v_{1,1} \vert \le Cn^{-3/2}.$

A priori there is no reason to have this type of decay. 

Is there a way to prove this without(!) using that the eigenfunctions are explicitly known?-Thus, can one show this directly from the matrix?

Consider the discrete second derivative with Dirichlet boundary conditions on $\mathbb C^n$.

Its eigendecomposition is fully known: see wikipedia

It seems like the lowest eigenvalue $\lambda_N$ is the one with the fastest decaying eigenfunction. Is there a way to prove this without(!) using that the eigenfunctions are explicitly known?-Thus, can one show this directly from the matrix?

Consider the discrete second derivative with Dirichlet boundary conditions on $\mathbb C^n$.

Its eigendecomposition is fully known: see wikipedia

It seems like the lowest eigenvalue $\lambda_1$ is one with a fast decaying eigenfunction, by this I mean that at the first coordinate $\vert v_{1,1} \vert \le Cn^{-3/2}.$

A priori there is no reason to have this type of decay. 

Is there a way to prove this without(!) using that the eigenfunctions are explicitly known?-Thus, can one show this directly from the matrix?

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Bound on decay Decay of eigenfunctions for Laplacian

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