Let $T \in B(H)$ be a bounded operator. Is $\sigma(T)=\sigma{(\overline{T^{*}})}$ true for $T$?
$\textbf{TRY-}$ I have proved it is true for normal operator but could not do it for bounded operator. Any help please! Thanks
2026-03-27 00:02:55.1774569775
Show $\sigma(T)=\sigma{(\overline{T^{*}})}$
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Hint: $T-\lambda I$ is invertible with inverse $A \in \mathcal{B}(H)$ iff $$ I = (T-\lambda I)A = A(T-\lambda I) $$ The above holds iff $$ I = A^{\star}(T^{\star}-\overline{\lambda}I) = (T^{\star}-\overline{\lambda}I)A^{\star}. $$