 Solution

If $u$ is invertible, applying the algebra involution to $u^{-1}u=u{u}^{-1}=1$ givesso $u^{\ast }$ is invertible and $(u^{\ast }{)}^{-1}=(u^{-1}{)}^{\ast }$. The converse follows by applying $\ast$ again. Sincewe have $\lambda \notin {\sigma }_A(x)$ if and only if $\overline{\lambda }\notin {\sigma }_A(x^{\ast })$. Taking complements and using that complex conjugation is a bijection givesThe conjugation bar in the printed PDF is essential; this is not an assertion that the two spectra always coincide.