Let the (finite dimensional) Hilbert space $\mathcal{H}$ be the direct sum of $\mathcal{H}_A$ and $\mathcal{H}_B$.
Let $A$ be a linear operator on $\mathcal{H}_A$ and $B$ be a linear operator on $\mathcal{H}_B$. Let $A = \sum_j \lambda_j^A |\psi_j^A\rangle \langle \psi_j^A|$ and $B = \sum_j \lambda_j^B |\psi_j^B \rangle \langle\psi_j^B|$ be the eigendecomposition of the two operators.
What is the eigendecomposition of $A \oplus B$?
From the definition it follows: $$(A \oplus B) = \sum_j \lambda_j^A |\psi_j^A\rangle \langle \psi_j^A| \oplus \sum_k \lambda_k^B |\psi_k^B \rangle \langle\psi_k^B|$$ and $$(A \oplus B) = \sum_j \sum_k (\lambda_j^A |\psi_j^A\rangle \oplus \lambda_k^B |\psi_k^B\rangle)(\langle\psi_j^A| \oplus \langle\psi_k^B|)$$ Not sure, though, if the next step is correct: $$(A \oplus B) = \sum_j \sum_k \lambda_j^A \lambda_k^B (|\psi_j^A\rangle \oplus |\psi_k^B\rangle)(\langle\psi_j^A| \oplus \langle\psi_k^B|)$$
The eigendecomposition of $A \oplus B$ is just the sum of the two eigendecompositions, that is,
$$A \oplus B = \sum_j \lambda_j^A | \psi_j^A \rangle \langle \psi_j^A | + \sum_k \lambda_k^A | \psi_k^A \rangle \langle \psi_k^A |$$
Where I identified $H_A$ as a subspace of $H = H_A \oplus H_B$ by $H_A \ni h \mapsto h \oplus 0 \in H_A \oplus H_B$ and similarly identified $H_B$ as a subspace of $H = H_A \oplus H_B$ by $H_B \ni h \mapsto 0 \oplus h \in H_A \oplus H_B$.
To prove the above decomposition is the eigendecomposition, hint: Show that if $\psi_j^A$ is an eigenvector of $A$, then $\psi_j^A \oplus 0$ is an eigenvector of $A \oplus B$ with the same eigenvalue. Similarly, if $\psi_k^B$ is an eigenvector of $B$, then $0 \oplus \psi_k^B$ is an eigenvector of $A \oplus B$ with the same eigenvalue. Moreover, if $\{e_j\}$ and $\{f_k\}$ are orthonormal bases of $H_A$ and $H_B$, respectively, then $\{e_j \oplus 0\}_j \cup \{0 \oplus f_k\}_k$ is an orthonormal basis of $H_A \oplus H_B$.