Spectrum in terms of two spectra in bounded Hilbert spaces.

Question

If we have $H_1,H_2$ as hilbert spaces, then $$H_1\oplus H_2:=\{(h,k):h\in H_1, k\in H_2\}$$ then it is also a Hilbert space with the following, $$(h_1,k_1)+(h_2,k_2):=(h_1+h_2,k_1+k_2)$$ and $$\langle (h_1,k_1),(h_2,k_2) \rangle:=\langle h_1,k_1\rangle+\langle h_2,k_2\rangle$$ and so on... Now let $T_1\in H_1$ and $T_2\in H_2$ then we have $T_1\oplus T_2\in B(H_1\oplus H_2)$ defined by the following, $$(T_1\oplus T_2)(h,k):=(T_1h,T_2k)$$ One can think $T_1\oplus T_2$ as a diagonal matrix acting on column vectors, i.e. $$\begin{pmatrix} T_1 & 0 \\ 0 & T_2 \\ \end{pmatrix}\begin{pmatrix} h \\k \end{pmatrix}=\begin{pmatrix} T_1h \\ T_2k \end{pmatrix}$$ So I believe we still can compute the spectrum of $\sigma (T_1\oplus T_2)$ in terms of $\sigma(T_1)$ and $\sigma(T_2)$ but still I'm stuck. I've discussed with few others that we can use linear algebra methods but what kind of matrix should I use? Should we use the norm of $T_1\oplus T_2$?

Answer 1

You have $$ \sigma(T_1\oplus T_2)=\sigma(T_1)\cup\sigma(T_2). $$ All you need to show is that $T_1\oplus T_2$ is invertible if and only if $T_1$ and $T_2$ and both invertible.

If you write $$ \begin{bmatrix} A&B\\ C&D\end{bmatrix} \begin{bmatrix} T_1&0\\0&T_2\end{bmatrix} =\begin{bmatrix} I&0\\0&I\end{bmatrix} ,\qquad \begin{bmatrix} T_1&0\\0&T_2\end{bmatrix} \begin{bmatrix} A&B\\ C&D\end{bmatrix} =\begin{bmatrix} I&0\\0&I\end{bmatrix} $$ you get $$ AT_1=T_1A=I,\qquad BT_2=T_1B=0,\qquad CT_1=T_2B=0,\qquad DT_2=T_2D=I. $$ The first and last equations give you $A=T_1^{-1}$, $D=T_2^{-1}$, and then $B=C=0$. So, when $T_1\oplus T_2$ is invertible then $T_1,T_2$ are invertible and $(T_1\oplus T_2)^{-1}=T_1^{-1}\oplus T_2^{-1}$.

Conversely, if $T_1,T_2$ are both invertible, then $T_1\oplus T_2$ is invertible and $(T_1\oplus T_2)^{-1}=T_1^{-1}\oplus T_2^{-1}$.