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I have the following question. Let $T\colon \mathcal{H} \to \mathcal{H}$ be a bounded, self-adjoint operator on a Hilbert-space $\mathcal{H}$. By spectral theorem we know that there exists a measure space $(\Omega, \Sigma, \lambda)$ and a bounded measurable function $f : \Omega \rightarrow \mathbb{R}$ and a unitary operator $U \colon \mathcal{H} \to L^{2}(\Omega)$ such that $(UTU^{-1}) (\varphi) = f\cdot \varphi$. My question is: Is the spectrum of $T$ and the spectrum of $UTU^{-1}$ then the same, i.e. $\sigma(T) = \sigma(UTU^{-1})$ ? Hope this question is not too trivial? Thanks in advance.

mika

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    $\begingroup$ By definition the spectrum of $T$ is $$ \sigma(T) = \{\lambda \in \mathbb{C}\,:\,\lambda \operatorname{id}_\mathcal{H} - T \text{ is not invertible}\}. $$ Use this to verify that $\sigma(T) = \sigma(UTU^{-1})$. $\endgroup$
    – t.b.
    Jan 11, 2012 at 9:04

1 Answer 1

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By definition the spectrum of $T$ is $\sigma(T)=\{\lambda\in\mathbb{C}:\lambda 1_{\mathcal{H}}-T \mbox{ is not invertible }\}$.

Use this to verify that $\sigma(T)=\sigma(UTU^{-1})$.

Credit goes to t.b.

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