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2 votes
1 answer
541 views

Does spectral theory assume separability

On an infinite dimensional space, the spectral theorem for compact normal operators says that the eigenvectors form an orthonormal basis which, from wikipedia, is equivalent to the space being ...
8 votes
1 answer
844 views

A doubt about the parts of the spectrum of tensor products

Let $\mathcal{H}$ be any complex Hilbert space of infinite dimensional. By an operator $T$ I mean a linear bounded transformation from $\mathcal{H}$ into $\mathcal{H}$, i.e, $T:\mathcal{H}\rightarrow\...
1 vote
1 answer
213 views

Weyl asymptotics vs. form perturbations

Consider Hilbert spaces $V$,$H$; a closed quadratic form $a$ with domain $V$; and its associated operator $A$ on $H$. (If necessary, the form can be assumed to be coercive.) For the sake of simplicity,...
3 votes
1 answer
588 views

orthonormal basis of eigenvectors for laplacian on a concave polygon

I am interested in the Laplace operator $\Delta$ on a concave polygon. When the polygon is convex, it is known that $\Delta: H^2(\Omega) \rightarrow L^2(\Omega)$ is boundedly invertible. In addition, ...
3 votes
3 answers
3k views

Infinite hermitian matrix

Suppose we have a finite square n x n matrix of complex numbers H that is Hermitian and skew-symmetric: $H^\dagger = H$ and $H^T = -H$. (T denotes transpose, $\dagger$ denote conjugate transpose. I ...

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