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Questions tagged [normal-operator]

For question concerning normal operators in Hilbert spaces.

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non-isomorphic unitary irreps are orthogonal, and related concepts

Let $ N $ be a normal matrix. Let $ v_i,v_j $ be eigenvectors of $ N $ with eigenvalues $ \lambda_i, \lambda_j $. If $ \lambda_i \neq \lambda_j $ are distinct then $ v_i \cdot v_j =0$ are orthogonal. ...
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Prove that N is a normal operator. [duplicate]

Prove that $N$ is a normal operator if and only if $\Vert Nx \Vert= \Vert N^{*}x \Vert$, for any x. For sufficiency, i can't solve. Here are the answers given by others: $N N^{*}=N^{*}N$ iff $\langle ...
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Spectral measure and eigenvalues/eigenvectors

Suppose $N$ is a normal operator on some $\mathbb{C}$-Hilbert space $\mathcal{H}$ and let $\mathrm{E}:\mathcal{B}(\sigma(N))\to B(\mathcal{H})$ be the associated spectral measure. Fix $\lambda_0\in\...
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Spectral measure of a complement

Suppose $\mathrm{E}$ is a spectral measure for some normal operator $N$ and $A\in\mathcal{B}(\sigma(N))$. I am unable to find an expression of the spectral measure of $A^c:=\sigma(N)\setminus A$ in ...
Oskar Vavtar's user avatar
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An operator that satisfies some condition is a normal operator

QUESTION: Let $H$ be Hilbert space, $T$ is a bounded operator on $H$, $TT^{\ast }\geqslant T^{\ast }T$, proof $T$ is normal operator($TT^{\ast } =T^{\ast }T$). I guess this question is missing ...
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Decomposition of normal operators into selfadjoint ones: $T=A+iB$

Assume we are given a densely defined and closed operator $T$ on a Hilbert space such that $D(T)\cap D(T^*)$ is dense. I am looking for sufficient conditions for $T+T^*$ to be selfadjoint as well as ...
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Range of Normal operator N is closed and its spectrum

When reading A course in functional analysis, Conway, I meet a problem about normal operator on Hilbert space. It says that If $N$ is a normal operator on Hilbert space $H$, then range of $N$ ($\...
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Question about proof that normal operators have invariant subspace

In Normal $T\in B(H)$ has a nontrivial invariant subspace, Haskell Curry said (for case (ii)) to pick two open, disjoint subsets of $\sigma(T)$ where $|\sigma(T)|\geq 2$. I don't think that this is ...
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A question about the criterion of diagonalizable normal operator on a Hilbert space (perhaps non-separable).

This is the excercise 2.11 in Chapter IX of A Course in Functional Analysis by John B. Conway: Suppose $H$ is a Hilbert space,$\,N$ is a bounded normal operator with associated spectral measure $E.$ ...
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Does unitary operator on a Hilbert space maps a star cyclic vector to a star cyclic vector?

Let $H$ be a infinite dimensional Hilbert space and $T: H \to H $ be a bounded normal star cyclic operator with star cyclic vector $e$ ( $T$ will be called star cyclic with star cyclic vector $e$ if $\...
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For a von Neumann algebra, if the elements are all normal, then it is commutative. [closed]

Let $\mathcal{A}$ be a von Neumann algebra, if the elements of $\mathcal{A}$ are all normal, then $\mathcal{A}$ is commutative. I want to know how to prove this, or where I can find the proof. Thank ...
Shaoze Pan's user avatar
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normal operator where the sum of eigenspaces is not equal to the entire Hilbert space.

Hi I'm reading about Hilbert spaces and normal operators. In this i found the following result: Let $H$ a Hilbert space of finite dimension and $T\in\mathcal{L}(H)$ a normal operator, then $$ H = \...
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Prove the operator $f$ is normal with the expression of the operator

When is $f$ a normal operator? Given that $f$ is defined as $f(\mathbf{x})=\mathbf{x}+c \mathbf{u}(\mathbf{v},\mathbf{x})$ being the notation $(,)$ the inner hermitian product between two vectors, $c\...
Acedium 20's user avatar
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Equivalent criterion for invertiblity of a normal operator in an Hilbert space.

Normal operators are an important class of operators in operator theory and spectral theory.We encounter it a lot in functional analysis.The following is my definition for normal operators: Defn. Let ...
Kishalay Sarkar's user avatar
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Spectral decomposition of a normal operator: non-distinct eigenvalues

A normal operator $T$ in some inner product space usually has spectral decomposition $T=\sum_j\lambda_j P_j$ where $P_j$ are orthogonal projections and $\lambda_j$ are usually distinct values often ...
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is this (unbounded) operator normal?

I have some question regarding normal operators. I have been given the following definition for the case of bounded linear operators $A\in\mathcal{A}(H)$: $A^*A=AA^*$. My question is: what happens if ...
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An exercise about normal operators

Given a normal operator $T\in\mathcal{L}(H)$ with $||T||\leq 1$, let $M\subseteq H$ the closed subspace defined by $M=\{x\in H|T(x)=x\}$ and $P$ the orthogonal projection on $M$. Prove that for all $x\...
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Ambiguous step in theorem 7.20 of Sheldon Axler’s Linear Algebra Done Right 3rd ed.

Hello, I have attached theorem 7.20 in the link above. Context: In the picture, V denotes a finite-dimensional inner product space over F, and F denotes R (field of real numbers) or C (field of ...
Khaled A's user avatar
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prove that the generator of normal operator semigroup is normal

This problem is from "A Course on Functional Analysis II" by Zhang Gongqing and Guo Maozheng, which is the textbook of a functional analysis course I'm taking. Precisely speaking, let T(t) ...
country road's user avatar
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How to prove that $\text{Ker}(A)=\text{Ker}(A^2)$ if $A$ is a normal operator?

Prove that $\text{Ker}(A)=\text{Ker}(A^2)$ if $A$ is a normal operator. I know that that $\text{Ker}(A)\subseteq \text{Ker}(A^2)$, which I proved as follows: $x\in \text{Ker}(A) \implies Ax=0\implies(...
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Normal operator equal to its norm and that of its adjoint

I am coming up with this equality over and over again in Linear ALgebra Done Right by Axler and I cant seem to derive it. I've tried looking up something similar here, the closest I've got is this:...
Paul Stone's user avatar
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Real normal operators whose characteristic polynomial split

Let $V$ be a finite dimensional inner product vector space over $\mathbb{R}$ and let $T : V \to V$ be a normal operator (that is, $T T^\ast = T^\ast T$) whose characteristic polynomial splits over $\...
Eduardo Longa's user avatar
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Normal operator $T$, $\sigma(T) \subset \{0,1\} \Rightarrow$ $T$ is an orthogonal projection

Let $T$ be a bounded normal operator acting from a Hilbert space $H$ to itself. I must check that if the spectrum of $T$ lies in the set defined in the title, then $T$ is an othogonal projection. I ...
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Spectral resolution of the identity of $S = U^*TU$

Suppose $T,U$ are bounded operators on a Hilbert space $\mathcal{H}$ with $T$ normal and $U$ unitary. Now we define $S = U^*TU$. Then I want to find the spectral resolution of the identity $E_S$ for $...
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if $T$ is normal and a projection in a finite dimensional vector space $V$, then $T$ is an orthogonal projection

Let $T$ be a normal operator ($TT^* = T^*T$) defined in a finite dimensional vector space $V$ with inner product ($<v, w>$). Then, if $T$ is a projection ($T^2 = T$) then $T$ is an orthogonal ...
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If $T$ is normal, is ($T^2$ is compact $\Rightarrow T$ compact) true?

I am studying Spectral Theory and Functional Analysis. I know that in a Hilbert space H, if $T$ is self-adjoint then we have: $$T^2 \text{ compact }\Rightarrow T \text{ compact}$$ I want to know if ...
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Spectral Theorem Multiplicationoperatorversion applied to a simple explicit example

At the moment I am trying to understand the following formulation of the spectral theorem better: For a bounded, normal operator $T$ on a Hilbert space $H$, the pair $(H,T)$ is unitarily equivalent to ...
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Does $\mathbb{C}^n$ "contain" $\mathbb{R}^n$?

I have a question of the form "Let $T : V → V$ be a normal operator, where $V$ is a finite dimensional inner product space over $\mathbb{R}$. Show that if $T^3(v) = 0$ for some $v \in V$, then $T(...
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The isometry $S$ in polar decomposition of linear operator $L=S\sqrt{L^*L}=SP$ has the same complex eigenvectors as $P$ iff $L^*L=LL^*$.

Suppose $e_1,e_2,\cdots,e_n$ and $f_1,f_2,\cdots,f_n$ are orthonormal bases of vector space $V$ and that the singular value decomposition of a linear operator $L:V\to V$ is $$ Lv=\sum_{i=1}^{n}s_i\...
barbatos233's user avatar
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Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint

Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint I am not quite sure how to do this proof. My first idea was to use the ...
randommathuser's user avatar
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$\text{ran}\,N$ is closed if and only if $0$ is no limit point of $\sigma(N)$, for a normal bounded operator $N$ on a hilbert space $H$

This question has been asked once in this forum, but I don't understand some things in the proof of one direction. I marked them in bold: ${\Longleftarrow}:$ Let $\lambda\in\sigma(N) $ be an ...
test's user avatar
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How to show that $\lim\limits_{j \to \infty} \lambda_j = 0$ and $P_j$'s are all finite rank projections?

I am studying some properties of compact normal operators. I have found that if $T$ is a compact normal operator on an infinite dimensional Hilbert space $\mathcal H$ then $\sigma (T) = \sigma_p(T) \...
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Adjoint of the direct sum of operators vs. direct sum of their adjoints

Let $\mathcal{H}$ be an infinite-dimensional complex Hilbert space which decomposes as the direct sum of a countable family of Hilbert spaces $\{\mathcal{H}_n\}_{n\in\mathbb{N}}$, namely \begin{...
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Are normal operator with positive self-adjoint part boundedly invertible?

Consider an unbounded normal operator $A$ with domain $\mathcal{D}(A)$ on a Hilbert space $\mathcal{H}$, i.e. its adjoint $A^*$ satisfies $\mathcal{D}(A^*)=\mathcal{D}(A)$ and, for all $x\in\mathcal{D}...
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Proving that a Normal operator with real eigenvalues is self adjoint

I came up with this solution to the problem. Does this make sense and is there a better way to going about this problem? Let $\lambda$ be the real eigenvalue of a normal Matrix $N$. Then, $\overline{\...
not_danish's user avatar
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Proving that $0 \in \sigma (A)$ whenever $T$ is normal and $0 \in \sigma (T).$

Let $T$ be a bounded normal operator on a Hilbert space $\mathcal H$ with $0 \in \sigma (T).$ Then $0 \in \sigma (A),$ where $A = T^* T.$ $\textbf {My Attempt} :$ Since $T$ is normal and $0 \in \...
Anil Bagchi.'s user avatar
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An operator that is diagonalizable but not normal.

I'm looking for an example of operator $T \in L(\mathbb{C}^2)$ that is diagonalizable but not normal. But, I think any diagonalizable $T \in L(\mathbb{C}^2)$ is normal because $$M(T)= \begin{bmatrix} ...
john's user avatar
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If C intertwines two positive operators, does C intertwines their square root?

Let $A$ and $B$ be positive invertible $n\times n$ matrices and $C$ be any $n\times n$ matrix such that $ A^2C= CB^2 $. Does this implies $AC=CB$ ? I know the answer in a particular case when $A=B$ (...
syam krishnan's user avatar
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2 answers
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$T: V \rightarrow V$ is a linear operator, and $TT^*=T^3$, $\text{Ker}(T) = U, W=U^⊥$, Prove that $T_{|_W}$ is normal

I proved that $W$ is $T^*-\text{invariant}$, and that $W$ is $T-\text{invariant}$. I couldn't find a good way to prove that the operator $T_{|_W}$ is normal. here's what I did so far: $U=\text{Ker}(T)...
Oac Roland's user avatar
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Invariant subspace of $T$ (normal) is also an invariant subspace of $T^\ast$.

I am struggling with the following question Let $V$ be a finite dimensional vector space and $T:V\rightarrow V$ be a linear normal operator ($T^\ast T = TT^\ast$), and $W$ an invariant subspace of $T$...
Theorem's user avatar
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Why if T is normal linear map, then image of T if the same as image of T*?

If $T$ is a normal linear map from vector space $V$ to itself, then image of $T$ is the same as image of $T^*$. Can anyone help me to prove this?
Ehsan Poursaeed's user avatar
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Exercise with normal operators.

Let $T \in Hom(ℂ^3)$, $T$ a normal operator and $T(1,1,1)=(2,2,2)$. If $(x,y,z) \in ker(T)$, then $x+y+z=0$. I was trying to solve this question, one of my attempts was using the fact that if $T$ is a ...
Aquila's user avatar
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How to understand "A possesses a complete orthonormal set of eigenvectors"?

A matrix A is said to be normal if $AA^H=A^HA$. Theorem of normal matrix is: A matrix A is normal if and only if A possesses a complete orthonormal set of eigenvectors. I don't know how to understand &...
Mariana's user avatar
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Let $S$ be symmetric operator on unitary $n$-dimensional space

(1) Prove that $S-iE$ and $S+iE$ are regular operators (2) prove that $U=i(S+iE)^{-1} (S-iE)$ is unitary operator and that none of its eigenvalues is $1$ Ok so first part is fairly easy. As for the 2. ...
Gregory's user avatar
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1 answer
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Let $L$ be a linear operator on vector space $R^3$ defined with $L(x,y,z)=(-x+2y+(3-c)z,(c+1)x-y+2z,2x+2y-z),$ with $c$ from $R$.

Prove that $L$ is symmetric operator of euclidean space $R^3$ with standard dot product if and only if $c=1$. So I think i proved that if $c=1$ then $L$ is a symmetric operator. Its matrix in standard ...
Gregory's user avatar
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Norm convergence of a series of operators.

Let $\mathcal H$ be a infinite dimensional Hilbert space and $T \in \mathcal L (\mathcal H)$ be a compact normal operator. Let $\sigma (T) = \{\lambda_1,\lambda_2, \cdots\} \cup \{0\}$ be the spectrum ...
math maniac.'s user avatar
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1 answer
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Spectral theory for compact normal operators.

Let $T$ be a compact normal operator on an infinite dimensional Hilbert space $\mathcal H.$ Then I came across the fact that the spectrum $\sigma (T)$ of $T$ is either finite or countably infinite ...
Anacardium's user avatar
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Similarity Transformation when the Linear Operator is not written in the standard basis doesn’t lead to a Diagonal Matrix. Why?

Consider a diagonalizable linear operator $T$, and let $\{\alpha\}$ be the basis of eigenvectors of $T$. Let $\{\beta\}$ be the standard basis for $\mathbb R^n$ and let $\{\gamma\}$ be another ...
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Doubt in Hoffman and Kunze Linear Algebra Section 8.5

Here is the proof from the book I was reading Now, I know about linear transformations and how in its matrix representation , the columns are coordinates of the image of basis vectors in the ...
Combat Miners's user avatar
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Is my proof correct? It can't be this simple

I am just starting to learn about linear algebra. My lecturer gave me the following exercise, and said 'it's quite an interesting exercise': $\text{Show that the identity operator $\hat{\mathbb{1}}$ ...
Christoffer Corfield Aakre's user avatar