# Tagged Questions

Operator theory is the branch of functional analysis that focuses on bounded linear operators, but it includes closed operators and nonlinear operators. Operator theory is also concerned with the study of algebras of operators.

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### How could we define the factorial of a matrix?

Suppose I have a square matrix $\mathsf{A}$ with $\det \mathsf{A}\neq 0$. How could we define the following operation? $$\mathsf{A}!$$ Maybe we could make some simple example, admitted it makes any ...
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### Operator vs function

Could someone please explain the MATHEMATICAL difference between an operator and a function? I am not talking about these in terms of coding but rather the mathematical difference. Is operator also a ...
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### Why do zeta regularization and path integrals agree on functional determinants?

When looking up the functional determinant on Wikipedia, a reader is treated to two possible definitions of the functional determinant, and their agreement is trivial in finite dimensions. The first ...
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### Operators with finite spectrum

Suppose that $T$ is a bounded operator with finite spectrum. What happens with the spectrum of $T+F$, where $F$ has finite rank? Is it possible that $\sigma(T+F)$ has non-empty interior? Is it always ...
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### Differential equations and Fourier and Laplace transforms

Why do both the Fourier transform and the Laplace transform appear in the study of differential equations? I've never understood why there are some situations where the Fourier transform is used and ...
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### Is it possible to characterize completeness of a normed vector space by convergence of Neumann series?

If $X$ is a normed vector space and if for each bounded operator $T \in B(X)$ with $\| T\| < 1$, the operator ${\rm id} - T$ is boundedly invertible, does it follow that $X$ is complete? Context: ...
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### Factorial of a matrix: what could be the use of it?

Recently on this site, the question was raised how we might define the factorial operation $\mathsf{A}!$ on a square matrix $\mathsf{A}$. The answer, perhaps unsurprisingly, involves the Gamma ...
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### Does an unbounded operator $T$ with non-empty spectrum have an unbounded spectrum?

It's well known that the spectrum of a bounded operator on a Banach space is a closed bounded set (and non-empty)on the complex plane. And it's also not hard to find unbounded operators which their ...
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### Compact sets as point spectrum of a bounded operator

It is well known that if $K$ is any compact set in $\mathbb{C}$, then there exist a bounded linear operator $T:l_2\to l_2$ such that $\sigma(T)=K$. My questions are: Q1) Does there exist $T$, a ...
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### A property of exponential of operators

Let $X$ be a Banach space. $A\in B(X)$ is a bounded operator. we can define $e^{tA}$ by $$e^{tA}=\sum_{k=0}^{+\infty}\frac{t^kA^k}{k!}$$ I am interested in this property: If $x\in X$, such that the ...
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### Gelfand-Naimark Theorem

The Gelfand–Naimark Theorem states that an arbitrary C*-algebra $A$ is isometrically *-isomorphic to a C*-algebra of bounded operators on a Hilbert space. There is another version, which states that ...
Consider $T\colon\ell^2\to\ell^2$ an operator such that $Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?