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 to prove that an operator is compact?

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?
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Question about Fredholm operator

$X,Y$ are Banach spaces and $A\in B(X,Y)$ is a Fredholm operator (that is, the dimensions of ker($A$) and coker($A$) are both finite), then are closed linear subspaces ker($A$) and Im($A$) ...
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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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Question about Angle-Preserving Operators

This an exercise out of Spivak's "Calculus on Manifolds". Edit: There was a typo in the exercise as is noted below in the answers. The statement has been edited to reflect this. Given $x,y\in\...
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Compactness of a bounded operator $T\colon c_0 \to \ell^1$

Pitt Theorem says that any bounded linear operator $T\colon \ell^r \to \ell^p$, $1 \leq p < r < \infty$, or $T\colon c_0 \to \ell^p$ is compact. I know how to prove this in case $\ell^r \to \...
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Easy Proof Adjoint(Compact)=Compact

I am looking for an easy proof that the adjoint of a compact operator on a Hilbert space is again compact. This makes the big characterization theorem for compact operators (i.e. compact iff image of ...
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Sum of Closed Operators Closable?

Let $A$ and $B$ be closed operators on a (separable complex) Hilbert space with dense domains $D(A)$ and $D(B)$ respecitvely. Then, we may define the operator $A+B$ on $D(A)\cap D(B)$. In general, ...
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A compact operator is completely continuous.

I have a question. If $X$ and $Y$ are Banach spaces, we have to prove that a compact linear operator is completely continuous. A mapping $T \colon X \to Y$ is called completely continuous, if it maps ...
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Is there a formula similar to $f(x+a) = e^{a\frac{d}{dx}}f(x)$ to express $f(\alpha\cdot x)$?

Using the Taylor expansion $$f(x+a) = \sum_{k=0}^\infty \frac{a^k}{k!}\frac{d^k }{dx^k}f(x)$$ one can formally express the sum as the linear operator $e^{a\frac{d}{dx}}$ to obtain $$f(x+a) = e^{a\...
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Norm of a symmetric matrix equals spectral radius

How do I prove that the norm of a matrix equals the absolutely largest eigenvalue of the matrix? This is the precise question: Let $A$ be a symmetric $n \times n$ matrix. Consider $A$ as an ...
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Closure of the invertible operators on a Banach space

Let $E$ be a Banach space, $\mathcal B(E)$ the Banach space of linear bounded operators and $\mathcal I$ the set of all invertible linear bounded operators from $E$ to $E$. We know that $\mathcal I$ ...
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$\operatorname{Range}T$ is a closed subspace.

Let $X,Y$ two Banach spaces. If $T \in \mathcal{B}(X,Y)$ study if $\operatorname{Range}T$ is a closed subspaces. How can I prove this fact ? What theorems can I use ? thanks :)
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What is operator calculus?

I watched the excellent interview with Richard Feynman: http://www.youtube.com/watch?v=PsgBtOVzHKI In the interview Feynman mention that he at young age re-invented operator calculus. I have searched ...
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Spectral Measures: Pushforward

This thread is Q&A. Problem Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$ And its spectral measure: $$...
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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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Exponential of a function times derivative

Exponential of a derivative $e^{a\partial}$ is simply a shift operator, i.e. \begin{equation} e^{a\partial}f(x)=f(a+x) \end{equation} This can be easily verified from a Taylor series \begin{equation} ...
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Resolvent: Definition

Given a Banach space. Consider linear operators: $$T:\mathcal{D}(T)\to E:\quad T(\kappa x+\lambda y)=\kappa T(x)+\lambda T(y)$$ (No other assumptions on the operator!) Denote for shorthand: $$R_\...
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Every Hilbert space operator is a combination of projections

I am reading a paper on Hilbert space operators, in which the authors used a surprising result Every $X\in\mathcal{B}(\mathcal{H})$ is a finite linear combination of orthogonal projections. The ...
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Proving Stone's Formula for Constructively obtaining the Spectral Measure for $A=A^\star$

Let $A$ be a bounded or unbounded selfadjoint linear operator on a complex Hilbert space $H$ with spectral representation $A=\int_{\sigma}\lambda \, dE(\lambda)$ given by the Spectral Theorem for ...
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For positive invertible operators $C\leq T$ on a Hilbert space, does it follow that $T^{-1}\leq C^{-1}$?

I need the following result. I think it's quite obvious but I don't know how to prove that: Let $C, T : \mathcal{H} \rightarrow \mathcal{H}$ be two positive, bounded, self-adjoint, invertible ...
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Prove that $\sigma(AB) \backslash \{0\} = \sigma(BA)\backslash \{0\} $

Prove that $\sigma(AB) \backslash \{0\} = \sigma(BA)\backslash \{0\} $. Where $A,\ B$ are bounded operators in Banach space and $\sigma$ denotes spectrum.
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331 views

Complementability of von Neumann algebras

Is every von Neumann algebra complemented in its bidual? It is certainly true for commutative von Neumann algebras as their spectrum is hyperstonian. Is it 1-complemented?
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If space of bounded operators L(V,W) is Banach, V nonzero, then W is Banach (note direction of implication)

Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But Wikipedia writes that ...
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Normal Operators: Numerical Range

Disclaimer: As I realized in the comments that this works for normal operators I decided to modify this question. Besides, I got the proof now - thanks to T.A.E.! Prove that for normal operators the ...
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Summary: Spectrum vs. Numerical Range

This thread is only Q&A! Given a Hilbert space $\mathcal{H}$. Consider operators: $$T:\mathcal{D}(T)\to\mathcal{H}$$ Denote for shorthand: $$\Omega\subseteq\mathbb{C}:\quad\langle\Omega\rangle:=...
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Determining the action of the operator $D\left(z, \frac d{dz}\right)$

This question was motivated by a question by Tobias Kienzler and its wonderful answers. I begin as in the linked question... Using the Taylor expansion $$f(z+a) = \sum_{k=0}^\infty \frac{a^k}{k!}\...
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$\exp(A+B)$ and Baker-Campbell-Hausdorff

A few years ago, I did research in quantum mechanics, specifically dealing with generalized displacement operators. In such musings, BCH lights (or gets in, depending on your viewpoint) the way. A ...
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Proof Complex positive definite => self-adjoint

I am looking for a proof of the theorem that says: A is a complex positive definite endomorphism and therefore is A self-adjoint. Does anybody of you know how to do this?
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Positive operator is bounded

For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded. I don't know how to do that, maybe it's an ...
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If $(I-T)^{-1}$ exists, can it always be written in a series representation?

If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm < $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$. Thinking in terms of a ...
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Normal Operators: Meet

Given a Hilbert space $\mathcal{H}$. Normal Operators: $$\mathcal{N}(\mathcal{H}):=\{N:N^*N=NN^*\}$$ Borel Calculus: $$\mathcal{B}(N):=\{\eta({N}):\eta\in\mathcal{B}(\mathbb{C},\mathbb{C})\}$$ ...
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Hilbert's Inequality

Could you help me to show the following: The operator $$ T(f)(x) = \int _0^\infty \frac{f(y)}{x+y}dy $$ satisfies $$\Vert T(f)\Vert_p \leq C_p \Vert f\Vert_p $$ for $1 <p< \infty$ where $$...
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Operator: not closable!

Is there an operator between Banach spaces with the following properties: $$T:\mathcal{D}(T)\subseteq X\to Y:\text{ injective, dense range, continuously invertible, not closable!}$$ (Note that the ...
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generalized functions & operators

I am dealing with a function $f(r) $that behaves like ~ $\frac{1}{r}$ when approaching zero. When I take the Laplacian of this guy and then integrate the result ([0,$\infty$]) I get some additional ...
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Spectral Measures: Reducibility

Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$ And its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\...
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Proving an operator is compact exercise

Suppose $(a_{ij})_{i,j\in \mathbb N}$ satisfy $\sum_{i,j}|a_{ij}|^2<\infty$ and define $A:\ell ^2 \rightarrow \ell ^2$ by $(Ax)_i)=\sum _j a_{ij}x_j$. I need to prove $A$ is compact. Unfortunately,...
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Finite dimensional $C^*$-algebras [closed]

Show that if a $C^*$-algebra $A$ is reflexive as a Banach space, then $A$ must be finite dimentional. I tried to solve it; but, I could not. please help me for this exercise. Thanks a lot!
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Normal Operators: Transform

Given a Hilbert space $\mathcal{H}$. Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$ Construct the operator: $$Q:=(1+N^*N)^{-1}:\quad Z:=N\sqrt{Q}$$ Then it is ...
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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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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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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 ...
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Selfadjoint compact operator with finite trace

I have a compact selfadjoint operator $T$ on a separable Hilbert space. For some fixed orthonormal basis, the operator's diagonal is in $\ell^1(\mathbb{N})$. Can we conclude that $T$ is trace ...
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Graph of symmetric linear map is closed

A homework problem: Let $H$ be a Hilbert space. Let $T:H\rightarrow H$ be a symmetric linear map ($\langle Tx,y\rangle=\langle x,Ty\rangle$). Show that $S$ is bounded. My attempt: I'd ...
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Fourier transform as diagonalization of convolution

I've read this in a lot of places but never quite got how this is true or meant. Let's say we have a convolution Operator $$ A_f(g) = \int f(\tau)g(t-\tau)d\tau $$ and apply it to $g(t)=e^{ikt}$. ...
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Spectral radius inequality

Suppose $A,B \in M(n \times n, \mathbb{C})$ or $ A,B \in M(n \times n, \mathbb{R}) $. Under wich hypothesis can I state that: $\rho(AB) \leq \rho(A)\rho(B)$ ?
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Square root is operator monotone

This is a fact I've used a lot, but how would one actually prove this statement? Paraphrased: given two positive operators $X, Y \geq 0$, how can you show that $X^2 \leq Y^2 \Rightarrow X \leq Y$ (or ...
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Showing that the orthogonal projection in a Hilbert space is compact iff the subspace is finite dimensional

Suppose that we have a Hilbert Space $H$ and $M$ is a closed subspace of $H$. Let $T\colon H\rightarrow M$ be the orthogonal projection onto $M$. I have to show that $T$ is compact iff $M$ is finite ...
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Prove that $T$ is an orthogonal projection

Let $T$ be a linear operator on a finite-dimensional inner product space $V$. Suppose that $T$ is a projection such that $\|T(x)\| \le \|x\|$ for $x \in V$. Prove that $T$ is an orthogonal projection. ...
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There are compact operators that are not norm-limits of finite-rank operators

Given an example of a Banach space for which There are compact operators that are not norm-limits of finite-rank operators. Tanks for answer
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$T$ surjective iff $T^*$ injective in infinite-dimensional Hilbert space?

Let $T:H_1\rightarrow H_2$ be a bounded linear operator where $H_1$ and $H_2$ are Hilbert spaces. The Hilbert-adjoint is defined to the the operator $T^*:H_2\rightarrow H_1$ such that $\langle Tx,y\...