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Questions tagged [operator-theory]

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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Equivalent Definitions of the Operator Norm

How do you prove that these four definitions of the operator norm are equivalent? $$\begin{align*} \lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for ...
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2answers
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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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1answer
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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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2answers
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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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2answers
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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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3answers
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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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3answers
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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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1answer
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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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2answers
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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 $g(t)=e^{ikt}$. ...
3
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1answer
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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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3answers
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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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2answers
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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 $\mathcal I$ ...
3
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1answer
448 views

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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1answer
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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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2answers
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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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2answers
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When is the image of a linear operator (between Banach spaces) closed?

Let $T: X \longrightarrow Y$ be a continuous linear map between two Banach spaces. When is $\operatorname{Ran}(T)$ a closed subspace? What theorems are there? Thanks :)
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3answers
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Norm of an inverse operator: $\|T^{-1}\|=\|T\|^{-1}$?

I am a beginner of funcional analysis. I have a simple question when I study this subject. Let $L(X)$ denote the Banach algebra of all bounded linear operators on Banach space X, $T\in X$ is ...
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3answers
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If a linear operator has an adjoint operator, it is bounded

This is a question I'm struggling with for a while: Let $H$ be a Hilber space. Let $T,S: H\rightarrow H$ be linear operators (not neccessarily bounded) such that for every $x,y\in H$: $\langle Tx,y\...
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2answers
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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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3answers
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Compact operators on an infinite dimensional Banach space cannot be surjective

I am reading a book about functional analysis and have a question: Let $X$ be a infinite-dimensional Banach-space and $A:X \rightarrow X$ a compact operator. How can one show that $A$ can not be ...
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1answer
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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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1answer
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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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2answers
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Translating Tarski's Axiomatization/Logic of $\mathbb R$ to the Theory of Magnitudes

Update: This has become a project, but I need help. All answers will now be definitions, propositions, theorems, etc. that build on the theory. I will marks some of my own answers as community wiki so ...
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1answer
109 views

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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2answers
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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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2answers
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Spectrum of shift-operator

Hoi, consider the Hilbertspace $l^2$ and the Left and Right-shift operator \begin{align*} L(x_1,x_2,\cdots) &= (x_2,x_3,\cdots)\\ R(x_1,x_2,\cdots) &= (0,x_1,x_2,\cdots ) \end{align*} I know ...
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2answers
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Spectral radius of the Volterra operator

The Volterra operator acting on $L^2[0,1]$ is defined by $$A(f)(x)=\int_0^x f(t) dt$$ How can I calculate the spectral radius of $A$ using the spectral radius formula for bounded linear operators: $$\...
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3answers
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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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How to prove that two non-zero linear functionals defined on the same vector space and having the same null-space are proportional?

Let $f$ and $g$ be two non-zero linear functionals defined on a vector space $X$ such that the null-space of $f$ is equal to that of $g$. How to prove that $f$ and $g$ are proportional (i.e. one is a ...
8
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1answer
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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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2answers
541 views

A characterization of trace class operators

Let $H$ be a separable Hilbert space and let $T\in B(H)$, such that $\displaystyle \sum_{j=1}^\infty\langle T\xi_j,\eta_j\rangle$ converges for any choice of orthonormal bases $\{\xi_j\}$, $\{\eta_j\}$...
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1answer
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Positive operator has a positive spectrum?

Let $T : \operatorname{dom}(T) \rightarrow H $ be a positive self-adjoint operator, is it then true that $\sigma(T) \subset [0,\infty)$? This is something that sounds natural and I guess that it is ...
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1answer
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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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1answer
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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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2answers
627 views

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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1answer
442 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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1answer
710 views

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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2answers
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Square root of positive operator on a complex Hilbert space

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $...
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1answer
460 views

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:=...
172
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6answers
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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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4answers
10k views

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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11answers
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What is an operator in mathematics?

Could someone please explain the mathematical difference between an operator (not in the programming sense) and a function? Is an operator a function?
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1answer
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Difference between an isometric operator and a unitary operator on a Hilbert space

It seems that both isometric and unitary operators on a Hilbert space have the following property: $U^*U = I$ ($U$ is an operator and $I$ is an identity operator, $^*$ is a binary operation.) What ...
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2answers
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Why no trace operator in $L^2(\Omega)$?

We have trace operator which allows us to define boundary values of an $H^1$ function. This is because of the fact that $C^\infty$ is dense in $H^1$ under the $H^1$ norm, I believe. I'm sure either $...
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4answers
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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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1answer
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Operator norm and tensor norms

I have a linear operator $A\in\mathcal{L}(X,Y)$ where $X$ and $Y$ are some Banach spaces (or Hilbert spaces would also do, if that simplifies the answer.). The operator norm of $A$ is given by $$ \|A\...
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1answer
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Norm of self-adjoint operator

I am trying to prove that $\|A\|=\sup_{||x||=1}|\langle x,Ax\rangle|$ for some selfadjoint bounded operator A on a Hilbertspace. Can anyone give me a hint how to prove it.
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1answer
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A linear operator between Banach spaces is weakly continuous iff norm continuous?

Claim : a linear function $T$ between Banach spaces is weakly continuous iff norm continuous? Okay, So I think I have realised weakly continuous implies norm continuous. As weakly continuous implies ...
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6answers
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Bounded operator that does not attain its norm

What is a bounded operator on a Hilbert space that does not attain its norm? An example in $L^2$ or $l^2$ would be preferred. All of the simple examples I have looked at (the identity operator, the ...
6
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1answer
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Show $T$ is invertible if $T'$ is invertible where $T\in B(X)$, $T'\in B(X')$

Seems simple enough but I can't quite get it. $X$ is a complex Banach space, and $T\in B(X)$, $T'\in B(X')$ is its adjoint. Suppose $T'$ is invertible. How can we show that $T$ is invertible? I have ...