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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Positive semidefinite linear operator T over a unitary space V that satisfies $T^k=I$ where $k \geq 1$ must be identity?

I got the following question in an exam I got yesterday that I didn't managed to answer: Let $V$ be a finite dimensional unitary vector space and let $T:V \to V$ be a positive semidefinite linear ...
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66 views

boundary of a spectrum proof

Let $A$ be a closed unital subalgebra of banach algebra $B$. Prove that ${\delta}{\sigma}_{B}(x)$ is contained in ${\delta}{\sigma}_{A}(x)$ for every $x$ in $B$.
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23 views

Can we find an invertible projection in an arbitrary von Neumann algebra?

I am looking for an answer for this question: Let $\mathcal{A}$ be an arbitrary von Neumann algebra, can we say there is an invertible projection ($P\neq I$) in $\mathcal{A}$? I think, if there is ...
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109 views

renorm a Banach space to make an operator have spectral radius equal to norm

Let $X$ be an infinite-dimensional complex Banach space equipped with the norm $\lVert\cdot\rVert$, and let $T\in\mathcal{L}(X)$ a bounded linear operator on $X$. Let $r(T)$ denote the spectral ...
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57 views

homomorphism or not

Let $T$ be a bounded operator on $H$ and fix a vector $x\in H$. Define $f$ on the space of polynomials in $T$ by $f(p(T))=p(x)$. Is $f$ a homomorphism? Initally I thought it obvious but the subtelty ...
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Examples of operator theory on Hilbert space

$(1)$ If $T \in B(H)$ is self-adjoint and $T \neq 0$ then $T^n \neq 0$ $(a)$for $n=2,4,8,16,... (b)$ for every $n$ $(2)$ Show that any $T \in B(H)$ can be uniquely expressed as $T=T_1+iT_2$ ...
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71 views

Can we say $TT^{*}=T^{2}$ implies $T=T^{*}$?

Let $A$ be a $C^{*}$-algebra, Can we say $TT^{*}=T^{2}$ implies $T^{*}=T$? for $T\in A$ I am looking for a counterexample! Thanks
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28 views

homomorphism question

Let B(H) be the set of bounded linear operators on a hilbert space H. Let F be a unital commutative subspace of B(H). Give an example of a homomorphism h from F to the complex numbers such that h is ...
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56 views

Surjective homomorphism example

What is an example of a surjective homomorphism $B(H)\to\mathbb C$, where $B(H)$ is the set of bounded linear operators on a Hilbert space $H$, and $\mathbb C$ is the complex numbers.
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99 views

Behavior of the resolvent near the boundary of the spectrum

My question is, in some sense, a continuation of the question below. Isolated singularities of the resolvent Suppose $T\in B(H)$ has no eigenvalues, pick $x\in H$, $x\neq 0$, and consider the ...
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29 views

A couple of proofs on a spectrum

Let $T$ be a normal bounded operator. Let ${\lambda}$ be in $({\sigma}(T))$. Without invoking general algebra theories, show that: a) $p({\lambda},{\lambda}^*)$ is in $({\sigma}(T))$ for all ...
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37 views

How to find all the eigenvalues of a positive operator whose eigenvectors are positive semi-defintie?

A linear operator $T:\mathcal{H}_n\rightarrow \mathcal{H}_n$ is said to be positive if $T(\mathcal{P}_n)\subset\mathcal{P}_n$ where $P_n$ is the set of positive semi-definite matrices. For a positive ...
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1answer
44 views

Gelfand transform explicity

Let $T$ be a bounded normal operator. Let $A$ be the algebra generated by $T$ and $T^*$. What is the explicit Gelfand transform $G:A\to C(\sigma(T))$? My book says the image of $T$ is the ...
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1answer
35 views

How to prove that $A B A^* \leq \|B\| A A^*$ for operators A,B?

Let $A$, $B$ bounded operators on a Hilbert space $H$. Further let $B$ be self-adjoint. Then we have that $A B A^* \leq \|B\| A A^*$. I wanted to ask how to prove this inequality or where I can find ...
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44 views

Closed graph theorem question?

Let $H$ be a Hilbert space. Let $A:\operatorname{dom}A\to H$ has a closed graph, where $\operatorname{dom}A$ is dense in $H$. Let $S\subseteq \operatorname{dom}A$ be dense. Is it true $A_{|S}$ has a ...
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30 views

Differential operator is self-adjoint

This is an exercise (13.8) in Rudin's Functional Analysis. Let an operator $T$ in $L^{2}(\mathbb R)$ be defined as follows: $\mathcal{D}(T)=\{f \textrm{ absolutely continuous}\in L^{2}(\mathbb{R}): ...
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33 views

Gelfand Transform in a specific case

What is the gelfand transform of an operator in the algebra generated by a bounded normal operator and it's adjoint? Thanks
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37 views

Unbounded extension of bounded operator

Is it possible to construct an unbounded extension of bounded densely defined operator? To be more concrete, let $\mathcal{H}$ be Hilbert space, $\mathcal{D}\subset\mathcal{H}$ - a dense subset, ...
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172 views

Integral representation for $\log$ of operator

How can one prove that $$ (\log\det\cal A=) \operatorname{Tr} \log \cal{A} = \int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr} e^{-s \mathcal{A}},$$ for a sufficiently well-behaved ...
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1answer
46 views

$ \|T_n\| \not \rightarrow \|T\|$ even if $ T_nx \rightarrow Tx, \forall x $

There is a theorem which states that given $X$ normed space, $Y$ Banach space on $\mathbb R, D \subseteq X$ dense and $T_n \in \mathcal L(X,Y)$ a bounded sequence s.t. $T_nz$ converges $\forall z \in ...
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42 views

Operators on a Hilbert space question

For a Borel measure $\mu$ define $\langle S_\mu x,y\rangle=\int_H\langle x,z\rangle \langle y,z\rangle \mu(z)$. An exercise in my book that I am reading says that I could find a $\mu$ s.t. $S_\mu$ ...
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2answers
89 views

Isolated singularities of the resolvent

Let $T$ be a bounded operator on $l_2$ such that there exists $\mu$ in the spectrum of $T$ which is an isolated point of the spectrum. We know that for any $x\in l_2$ the resolvent function ...
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$T:V\rightarrow V $ is over $\mathbb{R}$ , it's matrix is $A$, $A=PDP^*$. Is it true that $A$, $D$, and $P$ are in $M_{n \times n}(\mathbb{R})$

$T:V\rightarrow V$ is over $\mathbb{R}$ and $V$ of finite dimension $n$, and I know that it is orthogonally diagonalizable. The Matrix that represents it - call it $A$ ,in orthonormal basis is ...
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183 views

$T^*T=TT^*$ and $T^2=T$. Prove $T$ is self adjoint: $T=T^*$ [duplicate]

$V$ is an inner product space of finite dimension over $\mathbb{R}$, and $T:V\to V$ a linear transformation which is normal, that is, $T^*T=TT^*$. In addition $T^2=T$. Prove $T$ is self adjoint, that ...
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38 views

Adjoint of an operator question.

Let T be a normal operator. Prove that $\|T\|^{2n}=\|TT^*\|^n$ Has it got something to do with $\|T\|=\|T^*\|$?
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31 views

induction on powers of a norm

Let T be a self adjoint operator on a hilbert space. I wish to prove by induction that $||T^n||$=$||T||^n$. I have proved it for n=1 and n=2. So assume it is true for some n Then, ...
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77 views

Isometry between $l^p$ and $L^p$.

Consider $p\in[1,\infty)$ and the operator $T:l^p\rightarrow L^p([0,\infty))$: $$ Tx=\sum_{n=1}^\infty x_n\chi_{[n-1,n]} \qquad\forall\,x\in(x_1,x_2,\ldots,)\in l^p $$ Prove that $T$ is an isometry. ...
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45 views

operator on separable banach space whose spectrum and point spectrum is prescribed compact set

I am interested in obtaining the following paper: G. K. Kalisch, "On operators with large point spectrum," Scripta Math. 29 No. 3-4, (1973), 371-378. According to Ben Mathes, "Strictly Cyclic ...
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can we prove that a certain supremum of affine functions is frechet differentiable or at least continuous?

Let $X$ be a Hilbert space. Let $A\colon \operatorname{dom} A\to X$ be linear operator with closed graph (not necessarily bounded). Define $$ g\colon \operatorname{dom}A\to \mathbb{R}:x\mapsto ...
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55 views

Spectral decomposition

For a compact normal operator, the space can be written as the sum of generalized eigenspaces. So every element can be written as a linear combination of the eigenvectors, one from each eigenspace. ...
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52 views

Spectral decomposition for normal compact operator

My book says $Tx=(\alpha x_{\alpha})$ where the $\alpha$ are eigenvalues the of T. But the image of an operator is not in general a sequence. Do they mean these are the scalars in the linear ...
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If $T^2=TT^*$ then can i conclude that $T=T^*$?

let $B(H)$ be all bounded operator on Hilbert space H. If $T^2=TT^*$ then can i conclude that $T=T^*$? I think this is true if T is one to one. Can i construct an example that shows it is not true for ...
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0answers
30 views

spectral projections

Let $A$ be a von Neumann algebra, and $T$ be an hermitian element of $A$. Show that the spectral projections of $ T $ belong to $A$. Proof: the spectral projections of $ T $ commute with every ...
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1answer
33 views

Abelian von Neumann algebra

Let $M$ be a family of commutative normal operators which is closed under adjoint. clearly $M\subset M'$, but I do not know why $M'\subset M^{''}$? and how can conclude that $M^{''}$ is abelian?
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Convergence in the strong/weak operator topology: nets versus sequences

Let $H$ be a separable infinite dimensional Hilbert space, with orthonormal basis $(e_n)_{n=1}^\infty$. Consider the operators $U_n$ on $H$ such that $U_ne_k = e_1$ if $n=k$ and $U_ne_k=0$ if $n\neq ...
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42 views

Derivative on a function of tensor products

Assume I have defined an operator $A \otimes B$ on a $H \otimes L^2(\mathbb R^d)$ where $H$ is a Hilbert space as in Reed/Simon p. 299. $A$ is an operator on $H$ and $B$ is an operator on $L^2(\mathbb ...
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1answer
28 views

sequence of closed subspaces is not strictly increasing

Let H be an infinite dimensional hilbert space and $A_{n}$ a sequence of closed subspaces. Prove the sequence is not strictly increasing. I suppose it's sort of intuitive but what is a formal proof?
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69 views

If A unitary matrix and orthogonally diagonalizable why there is a basis in whichthe linear trans. matrix is diagonal?

If $A$ is a $n\times n$ unitary matrix (above the complex field) and is orthogonally diagonalizable, why does it mean that the is an orthonormal basis $\mathbb C$ in which the matrix that represent ...
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2answers
40 views

Operator Theory Textbook Question

I read the following excerpt in my course textbook: Now, I'm led to believe that $P:X\rightarrow X$ as above is bounded iff $M$ and $N$ are both closed. I understand the if direction, but I can't ...
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21 views

Question about operator algebra

I'm not certain about the rules of operator algebra, and I am wondering if these statements are equivalent $$\left(z^2\frac{d}{dz}-2z\right)\cdot\left(z^2\frac{d}{dz}-2z\right)=$$ ...
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22 views

Positive elements below projections

Let $a$ be a positive element in $A$, where $A$ is a $C^*$-algebra. Let $p\in A$ a projection and suppose $a\leq p$. Is it true that $ap=pa$? If yes, shouldn't we have $ap=pa=a$, since ...
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Adjoint of unbounded Operators: Product and Sum

When precisely does equality hold for sum and product: $$S^*+T^*\subseteq (S+T)^*$$ $$S^*T^*\subseteq (TS)^*$$ So far I checked that for the sum the seemingly weaker condition implies the stronger ...
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Exercise about compact operator.

In $X=\ell^p$, $p\in[1,\infty]$ we consider: $$ T(x_1,x_2,x_3,\ldots)=(0,x_1,0,x_3,\ldots) $$ Prove that $T$ isn't a compact operator and that $T^2$ is a compact operator. I think I solved the second ...
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33 views

orbits of a bounded linear operator

What interesting or strong results are there concerning orbits of an operator and invariant susbpaces (either in banach or hilbert space)? Obviously, I know that an operator T has an invariant ...
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2answers
86 views

Reference for a Proof of Weyl-Von-Neumann Theorem

I'm looking for a reference for the proof of the Weyl Von Neumann theorem, however there seems to be two (or the two might be the same). There's the one which is stated in Conways, A Course in ...
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16 views

partial differential operators

As we know, there exists a semigroup for partial differential operators $A = \sum_{i,j=1}^N D_i(a_{ij}(\cdot)D_j)$, see (Klaus-Jochen Engel, Rainer Nagel, one-parameter semigroups). My question is ...
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1answer
88 views

Bounded Inverse Theorem

$A$ is a bounded linear operator from $X$ to $Y$ (both Banach spaces). Show that if there exists $k > 0$ such that $\|Ax\| \geq k\|x\|$, for all $x$ then $\operatorname{range}(A)\,$ is closed. My ...
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2answers
165 views

Gateaux and Frechet derivatives and related notions

Let $X$ and $Y$ be normed real vector spaces, and $f : X \to Y$ a map. Let's say that: G) $f$ is Gateaux differentiable at $x_0 \in X$ if for all directions $v \in X$ the limit $f'(x_0)(v) := ...
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1answer
22 views

unbounded operator with no invariant subspace

Is there an example of an unbounded operator on a hilbert space with no invariant subspace? Or is there some other reason why we are only interested in bounded case.
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Condition for vector to be in the domain of unbounded operator.

Let $P$ be unbounded self-adjoint operator on some Hilbert space $\mathcal{H}$. We assume that the limit $$ \lim_{\epsilon \searrow 0} \|\exp(-\epsilon^2 P^2/2) P\psi\| $$ exists and is finite. Does ...