unitary operator $U$ and $U_{\lambda}$

Question

I want prove that, $\lambda$ is an eigenvalue of a unitary operator $U: H \longrightarrow H$ if and only if $\overline{U_{\lambda}(H)} \neq H.$

For direct implication, since $\lambda$ is eigenvalue $U_{\lambda}(H)^{\perp}\neq 0$ and so $\overline{U_{\lambda}(H)}\neq H.$ But for the other implication i don't know how to conclude. could you give me a help please?

Answer 1

Write $\sigma_p(T)$ for the set of eigenvalues of $T$. Since $U$ is a unitary, $$\tag1 \lambda\in\sigma_p(U)\iff\overline\lambda\in\sigma_p(U^*). $$ This is easy: if $Ux=\lambda x$, then $x=U^*Ux=\lambda U^*x$. As $|\lambda|=1$, $U^*x=\overline\lambda x$.

You also need $(1)$ for the first implication (you don't explain how you did it).

If $\lambda\in\sigma_p(U)$, then $\overline\lambda\in\sigma_p(U^*)$. So $\ker U_\lambda^*\ne\{0\}$. Then $\overline{U_\lambda H}=(\ker U^*)^\perp\ne H$.

Conversely, if $\overline{U_\lambda H}\ne H$, then $\ker U_\lambda^*=(U_\lambda H)^\perp\ne\{0\}$. So $\overline\lambda\in\sigma_p(U^*)$, and then by $(1)$ we have $\lambda\in\sigma_p(U)$.