Let $G$ be an operator on Hilbert space $H$ such that $\ker G$ is different from $\{0\}$.
Let also $P$ be the orthogonal projection onto $\ker G$, $G_1$ the restriction of $G$ on $\ker P$ and
$G_2$ the restriction of $G$ on $\operatorname{Im} P$.
My question is: can I say that $\sigma(G) = \sigma(G_{1})\cup\sigma(G_{2})$?
Best Answer
This is a general result : if $E$ is a Banach space, $F,G$ closed subspaces of $E$ such that $E = F \oplus G$ and if $T : E \to E$ is a bounded operator which leaves invariant $F$ and $G$, then $\sigma(T) = \sigma(T_{|F}) \cup \sigma(T_{|G})$.
The proof is easy if you consider resolvant sets.
If $\lambda \in \mathbb{C}$ is such that $T - \lambda I_E$ is invertible, then it follows immediately that $T_{|F} - \lambda I_F$ is one-to-one. Then, fix $y \in F$. By assumption, there is $x \in E$ with $(T-\lambda)x = y$. Writing $x = x_F + x_G$, with $x_F \in F$ and $x_G \in G$, we have $(T - \lambda)x_F = y$. (because $y$ belongs to $F$, and the sum $F \oplus G$ is direct)
This then shows that $T_{|F} - \lambda I_F$ is invertible. The same goes for $T_{|G} - \lambda I_G$.
Reciprocally, if $T_{|F} - \lambda I_F$ and $T_{|G} - \lambda I_G$ are invertible, we can easily build an inverse for $T - \lambda I_E$.