$U$ is unitary mapping on a Hilbert Space $\mathcal H$, where $U = \lambda I - T$ and $T$ is compact operator. Show that $U$ can be diagonalized.
To start with, is the condition '$T$ is normal' or '$T$ is symmetric' not needed? It seems strange that those conditions do not exist in the diagonalization question.
Anyhow, I tried to show that $U$ is also compact. The reason is that under the condition that '$T$ is normal' or 'symmetric', $U$ is also normal and thus $U$ can be diagonalized. (the previous exercise was that if $T$ is normal and compact, then $T$ can be diagonalized). However, defining a bounded sequence $\{f_n\}_n$ s.t. $\{Tf_{n_k}\}_k$ converges doesn't seem to be enough to conclude that $Uf_{n_k}$ also converges to some function in $\mathcal H$...
Any rough ideas to start with this exercise would be grateful. Thank you in advance.
-this exercise is from Stein's Real Analysis Chapter4 Exercise 35-
Additionally, I've got a question about the term 'Unitary'
According to the Stein's textbook, 'Unitary mapping $U$' on $\mathcal H$ is defined as follows:
$U$ is linear
$U$ is bijective
$\Vert Uf \Vert_{\mathcal H} = \Vert f \Vert_{\mathcal H}$
However, I found another definition of 'Unitary Mapping' in wiki: $U^{*}U = UU^{*} = I$. Using this definition, I figured out how to show the normality of $T$. But, I found it difficult to verify the same thing with the definition of Stein's book. I'd like to understand the relationship between the Stein's definition and the other one ($U^{*}U = UU^{*} = I$).