Suppose I have an operator $ H $ that is self-adjoint and densely defined on $ L^2 (\mathbb{R} , \mathbb{C}) \times L^2 (\mathbb{R} , \mathbb{C}) $.
Suppose further that $ L^2 (\mathbb{R} , \mathbb{C}) $ is an invariant subspace of $ H $.
Now, suppose that the spectrum of $ H $ consists purely of the point spectrum. I am thinking about the Quantum Harmonic Oscillator, for example.
Suppose further that the restriction of $ H $ onto $ L^2 (\mathbb{R} , \mathbb{C}) \times 0 $ and $ 0 \times L^2 (\mathbb{R} , \mathbb{C}) $ has purely simple eigenvalues.
Is it true that the spectrum of $ H $ will be the union of the point spectrum of the restriction? If so, how could one prove it?
Proof Attempt
Let $ \lambda \in \sigma(H)= \sigma_p (H) $. Then,there is a $ u=(u_1,u_2) \in L^2 (\mathbb{R} , \mathbb{C})\times L^2 (\mathbb{R} , \mathbb{C}) $ such that \begin{equation}\label{key} Hu= \lambda u. \end{equation} Now, \begin{equation} H = \begin{pmatrix} H_{11} & 0 \\ 0 & H_{22}. \end{pmatrix} \end{equation} for $ H_{11} $, $ H_{22} $ self adjoint operators on $ L^2 (\mathbb{R} , \mathbb{C}) $.
Then, \begin{align} H_{11} u_1 &= \lambda u_1, \\ H_{22} u_2 &= \lambda u_2. \end{align}
and so certainly $ \sigma(H) \subset \sigma(H_{11}) \cup \sigma(H_{22}) $.
For the other direction, is it correct to simply take $(u,0)$ and $(0,u)$?
I feel like some form of what I am trying to prove- a statement about reconstructing the spectrum of an operator from its restrictions on invariant subspaces which are orthogonally decomposed- should be a standard proposition or theorem in a Functional analysis textbook, but I can't seem to find it.
Your argument is fine. You don't need to restrict the type of the elements of the spectrum, though.
From $$ H-\lambda I=\begin{bmatrix} H_{11}-\lambda I&0\\0& H_{22}-\lambda I\end{bmatrix} $$ it is easy to see (proof below) that $H-\lambda I$ is invertible if and only if both $H_{11}-\lambda I$ and $H_{22}-\lambda I$ are. So $$ \rho(H)=\rho(H_{11})\cap \rho(H_{22}),$$ and taking complements you get $$\sigma(H)=\sigma(H_{11})\cup\sigma(H_{22}).$$
We want to show that $\begin{bmatrix} X&0\\0& Y\end{bmatrix}$ is invertible if and only if $X$ and $Y$ are.
If $AX=XA=I$ and $BY=YB=I$, then $\begin{bmatrix} A&0\\0& B\end{bmatrix}$ is an inverse. Conversely, if $$ \begin{bmatrix} X&0\\0& Y\end{bmatrix}\begin{bmatrix} A&B\\ C& D\end{bmatrix}=\begin{bmatrix} A&B\\ C& D\end{bmatrix}\begin{bmatrix} X&0\\0& Y\end{bmatrix}=\begin{bmatrix} I&0\\0& I\end{bmatrix}, $$ looking at the diagonal coordinates we get the equalities $$ XA=AX=I, \qquad YD=DY=I. $$ So $X,Y$ are invertible.