Spectrum of direct sum of bounded operators

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!
• You have to find the spectra of $T_1$ and $T_2$ somehow or another. If these are finite-dimensional operators, you can use characteristic polynomials. Commented Nov 11, 2022 at 18:15
• Most linear algebra resources focus on finite-dimensional operators; you should check out an intro textbook or wikipedia for a more detailed explanation than I can provide. But in short, the characteristic polynomial of a matrix $A$ is $\det(\lambda I - A)$ as a function of $\lambda$. The roots are the spectrum of $A$. Commented Nov 11, 2022 at 18:20
• There isn't a way to compute the spectrum of a generic infinite-dimensional operator as far as I know. So $\sigma(T_1 \oplus T_2) = \sigma(T_1) \cup \sigma(T_2)$ is the best characterization of the spectrum you're going to get. Commented Nov 11, 2022 at 20:59
$$\sigma(T_{1}\oplus T_{2})=\sigma(T_{1})\cup \sigma(T_{2})$$ because $$S\oplus T$$ is invertible iff $$S$$ is invertible and $$T$$ is invertible.
• Yes thanks! I will now try to compute each $\sigma(T_1)$ and $\sigma(T_2)$. Commented Nov 11, 2022 at 18:35
• Ok so it didn’t work for me. How would one pick a matrix and compute the spectra of each $T_1$ and $T_2$? Assuming we are working with finite dimensional spaces. One can use the equation $\det(\lambda I-A)=0$ for some matrix $A$ but then what. I cannot plug a random matrix and solve this. Commented Nov 12, 2022 at 10:38