Let $T$ be a self-adjoint invertible operator on $\mathcal{H}$ with a continuous spectrum, means the spectral measure is nonatomic. For which class of invertible operators $V$( with continuous spectrum) other than unitary $VTV^{*}$ also have continuous spectrum?
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2$\begingroup$ Obviously it works for nonzero scalar multiples of unitaries. $\endgroup$– Robert IsraelCommented Apr 29, 2019 at 12:31
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$\begingroup$ I dont want that @Israel!! Something nontrivial I am expecting $\endgroup$– user136400Commented Apr 29, 2019 at 12:32
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4$\begingroup$ "Something nontrivial I am expecting" -- you should not expect other people to do the work of formulating a well-defined question for you. $\endgroup$– Yemon ChoiCommented Apr 29, 2019 at 15:16
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$\begingroup$ @Yemon I formulated the correct question, somebody answered it in the wrong way, I made him understand the mistake!! $\endgroup$– user136400Commented Apr 30, 2019 at 7:25
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2 Answers
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In any infinite-dimensional Hilbert space, the only such operators $V$ that work for all $T$ are scalar multiples of the identity.
Suppose $V$ is not a scalar multiple of a unitary. Then there are linearly independent vectors $v$, $w$ such that $V^* V v = w$. Let $T$ be a self-adjoint invertible operator such that $Tw = v$. Then $VTV^* (Vv) = V T w = V v$, so $V T V^*$ does not have continuous spectrum. All that remains is to show that such $T$ can be chosen with continuous spectrum, which is not difficult.
answered Apr 29, 2019 at 13:50
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The spectrum of an operator $T$ is invariant under conjugation by invertible operators $V$, i.e. $$\sigma(T)=\sigma(VTV^{-1}).$$ So, if your operator has continuous spectrum also its conjugate $VTV^{-1}$ does, because the spectral measures are the same.
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$\begingroup$ yes I know $\sigma(T)=\sigma(VTV^{-1})$, but my question about $VTV^{*}$ not $VTV^{-1}$ $\endgroup$ Commented Apr 29, 2019 at 13:11
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$\begingroup$ Ok, so perhaps you may want to slightly rephrase your question so that it is clearer. $\endgroup$– JohnCommented Apr 29, 2019 at 13:16