Let $A_i \in \mathbb{R}^{d_i\times d_i}$ be a finite number of real matrices and $A = \bigoplus_{i=1}^n A_i$ their direct sum. If $u_i \in \mathbb{R}^{d_i}$ is an eigenvector of $A_i$ with eigenvalue $\lambda_i$ then trivially $u = 0 \oplus \cdots \oplus u_i \oplus \cdots \oplus 0$ is an eigenvector of $A$ with the same eigenvalue.
I'm not sure how to prove the converse: if $A$ has an eigenvalue $\lambda$ then there exists a vector $u$ of the form $u = 0 \oplus \cdots \oplus u_i \oplus \cdots \oplus 0$ such that $Au = \lambda u$. It seems intuitive but I'm not sure how to justify it formally. Any ideas?
If $Au = \lambda u$ for some $u = u_1 \oplus \cdots \oplus u_n$ then $u_i \neq 0$ for some $i$, which implies $A_i u_i = \lambda u_i$ and so $u' = 0 \oplus \cdots \oplus u_i \oplus \cdots 0$ is also an eigenvector of $A$.