I recently learned something about spectrum in functional analysis and saw some examples. However I struggling with this when trying to understand how it can be used for the following example regarding Hilbert spaces. I will write the example below.
Let $H_1,H_2$ be Hilbert spaces, then the direct sum $H_1\oplus H_2$ is given by the following $$\langle (x_1,y_1)|(x_2,y_2)\rangle:=\langle x_1|x_2 \rangle_{H_1}+\langle y_1|y_2 \rangle_{H_2}.$$ Let $T_1\oplus T_2$ be given where $T_1\in B(H_1)$ and $T_2\in B(H_2)$. How can one find the spectrum of $T_1\oplus T_2$?
I have the definition of the spectrum as follows:
Definition: Let $A\in \mathscr{A}$, where $\mathscr{A}$ is a unital Banach algebra. The spectrum of $A$ denoted $\sigma(A)$ is $\{\lambda \in \mathbb{C}:A-\lambda I \text{ is not invertible in } \mathscr{A}\}$. Of course we have $I$ as our unit.
The identity operator on $H_1 \oplus H_2$ is the direct sum of the two identities, $I_1 \oplus I_2$. This the spectrum we're looking for is the set of $\lambda$ which make $$\lambda (I_1 \oplus I_2) - (T_1 \oplus T_2) = \lambda I_1 - T_1 \oplus \lambda I_2 - T_2$$ not have an inverse. The key observation is that when the inverse of a direct sum exists, it is the direct sum of the inverses. To see this, consider the block matrix representation of the operator sum: $$ T_1 \oplus T_2 = \begin{pmatrix} T_1 & 0 \\ 0 & T_2 \end{pmatrix} $$ Hence any $\lambda$ in either $\sigma(T_1)$ or $\sigma(T_2)$ falls in the spectrum of $T_1 \oplus T_2$: the spectrum of the sum is the union of the spectra!