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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 …
Jochen Glueck's user avatar
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 …
Jochen Glueck's user avatar
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 …
Jochen Glueck's user avatar
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 …
Jochen Glueck's user avatar