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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.

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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!

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    $\begingroup$ 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. $\endgroup$ Commented Nov 11, 2022 at 18:15
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    $\begingroup$ "spectra" is a plural for spectrum $\endgroup$ Commented Nov 11, 2022 at 18:16
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    $\begingroup$ 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 $A$. $\endgroup$ Commented Nov 11, 2022 at 18:20
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    $\begingroup$ Characteristic polynomials don't generalize to infinite dimensions, because determinants don't either. Computing infinite spectra is hard in general. $\endgroup$ Commented Nov 11, 2022 at 19:08
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    $\begingroup$ 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. $\endgroup$ Commented Nov 11, 2022 at 20:59
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$\sigma(T_{1}\oplus T_{2})=\sigma(T_{1})\cup \sigma(T_{2})$ because $S\oplus T$ is invertible iff $S$ is invertible and $T$ is invertible.

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  • $\begingroup$ Yes thanks! I will now try to compute each $\sigma(T_1)$ and $\sigma(T_2)$. $\endgroup$ Commented Nov 11, 2022 at 18:35
  • $\begingroup$ 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. $\endgroup$ Commented Nov 12, 2022 at 10:38

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