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9
votes
Accepted
Removing the interior of spectrums
The answer is no, in general. Here is a counterexample:
Let $A$ be the algebra of bounded linear operators on $\ell^2(\mathbb{N})$, and let $a \in A$ be the left shift on $\ell^2(\mathbb{N})$. Then t …
6
votes
Accepted
Regarding an element being self adjoint
The golden rule for conjectures in operator theory:
Every ad-hoc conjecture is most likely false for $2 \times 2$-matrices. :-)
So here's a $2 \times 2$-counterexample for the question:
Let $A = \math …
4
votes
1
answer
468
views
Bicommutant theorem for commutative operator algebras
Let $\mathcal{B}(H)$ denote the space of bounded linear operators on a complex Hilbert space $H$. The von Neumann bicommutant theorem says:
Theorem. Suppose that $\mathcal{A}$ is a $C^*$-subalgebra o …
10
votes
Accepted
Trace norm of operators obtained by restricting the matrix of a trace class operator
Here's an algorithm for testing an ad-hoc conjecture $C$ about Hilbert space operators. :-)
Set up the runtime environment correctly by loading the information "Most conjectures are false" into short …
14
votes
1
answer
683
views
Criterion for a Banach algebra to be finite dimensional
Let $A$ be a Banach algebra (say, complex and unital) and suppose that every (closed) commutative subalgebra of $A$ is finite dimensional.
Question. Does it follow that $A$ is finite dimensional?
Re …