The spectrum of the operators

Question

Let $X, Y$ be the Banach space, and $T_{1}: X\rightarrow X$ and $T_{2}: Y\rightarrow Y$ be the bounded linear operators. Then what is the relationship between $\sigma(T_{1})$, $\sigma(T_{2})$ and $\sigma(T_{1} \oplus T_{2})$?

My answer is :$\sigma(T_{1} \oplus T_{2})=\sigma(T_{1})\cap\sigma(T_{2})$. Is it correct?

Answer 1

It should be $\sigma(T_1 \oplus T_2)=\sigma(T_1) \cup \sigma(T_2)$.

$T_1 \oplus T_2 - \lambda I$ is invertible iff $T_1 - \lambda I$ and $T_2 - \lambda I$ is invertible. Hence $\lambda \notin \sigma(T_1 \oplus T_2) \iff \lambda \notin \sigma(T_1) \cap \sigma(T_2)$

So $\sigma(T_1 \oplus T_2) = \sigma(T_1) \cup \sigma(T_2)$

(I assume $\sigma(T)$ is the spectrum of $T$. If you meant the resolvent, then your answer is right).