# If a,b ‎are ‎unitary ‎equivalent,‎Dose ‎ ‎‎$‎\sigma(a)=‎\sigma(b)‎$‎ is true?

‎‎Let A‎ ‎is ‎an ‎unital‎‎ ‎algebra ‎and ‎‎$‎Ad‎~u:‎‎‎A\rightarrow ‎A~,~a‎\mapsto~‎uau‎^{*}‎‎$ ‎and u‎ ‎is ‎unitary ‎element ‎of A‎(‎$‎uu‎^{‎*‎}=‎u‎‎^{*}‎u=1‎$‎), ‎if ‎‎$‎b=‎uau‎^{‎*‎}‎‎$ ‎(a,b ‎are ‎unitary ‎equivalent),‎‎ Dose ‎ ‎‎$‎\sigma(a)=‎\sigma(b)‎$‎ is true?

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The same comment from another question of yours applies here as well. –  Ilya Jan 9 '13 at 9:27
It seems to be true since: $\sigma(uau^\ast)=\{\lambda\in C: \lambda-uau^\ast=u(\lambda-a)u^\ast\in \operatorname{sing}(A)\}=\{\lambda\in C: \lambda-a\in \operatorname{sing}(A)\}=\sigma(a)$
@Ali $\operatorname{sing}{A}$ was correct. A singular element of an algebra is an element which is not invertible. The spectrum consists of those $\lambda \in \mathbb{C}$ such $\lambda -a$ is not invertible, so singular. –  Martin Feb 7 '13 at 15:18