# 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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### 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 ...
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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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### 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 ...
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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 ...
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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 ...
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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} ...
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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 $A=B$ (...