The following result is confusing me; I have not been able to find a clear proof that I understand.
Suppose $T$ is a bounded linear operator over a Hilbert space $\mathcal H$. Then $\sigma(T^*)=\{\overline\lambda:\lambda\in\sigma(T)\}$.
What I've done so far: I know that if $\lambda\in\sigma(T)$ then the map $T-\lambda I$ is not invertible. However, I don't know how to use this definition to conclude that $\sigma(T^*)=\{\overline\lambda:\lambda\in\sigma(T)\}$. If I know that $T-\lambda I$ is not invertible, then it could be the case that $T-\lambda I$ is not injective, or the case that $T-\lambda I$ is not surjective.
From linear algebra there is a theorem that states that, on a finite-dimensional vector space, a number $\lambda$ is not a eigenvalue for $T$ if and only if $\overline\lambda$ is not an eigenvalue for $T^*$, because if $T-\lambda I$ is invertible, then $$ (T-\lambda I)^{-1}(T-\lambda I)=(T-\lambda I)(T-\lambda I)^{-1}=I, $$ which gives $$ (T^*-\overline\lambda I)((T-\lambda I)^{-1})^*=((T-\lambda I)^{-1})^*(T^*-\overline\lambda I)=I. $$ This means $\lambda$ is an eigenvalue for $T$ if and only if $\overline\lambda$ is an eigenvalue for $T^*$.
But since we are dealing with $\sigma(T)$ over a not-necessarily finite dimensional Hilbert space, I don't believe this proof really applies. Any help on this result would be greatly appreciated!
Suppose $T-\lambda I$ has a bounded inverse. Call the inverse $S$. Then $S(T-\lambda I)=(T-\lambda I)S=I$. This implies that $(T-\lambda I)^{*}S^{*}=S^{*}(T-\lambda I)^{*}=I$ or $(T^{*}-\overline {\lambda }I)^{*}S^{*}=S^{*}(T^{*}-\overline {\lambda} I)^{*}=I$ and $S^{*}$ is a bounded operator. Thus $T^{*}-\overline {\lambda} I% R$ has a bounded inverse. Similarly you can prove then converse also. Hence $T-\lambda I$ and $T^{*}-\overline {\lambda }I$ have the same resolvent sets which implies that they have the same spectrum.
[I have used the fact that $(AB)^{*}=B^{*}A^{*}$].