If we consider $X\neq\{0\}$ to be a complex Banach space then the product $X\times X$ is a Banach space with the norm $\|(x,y)\|=\|x\|+\|y\|$. $T(x,y)=(x + y,x - y)$ is then a bounded linear operator on $X\times X$. Find $\sigma(T)$ and the subsets $\sigma_p(T),\sigma_c(T), \sigma_r(T)$, where $\sigma(T)$ is the spectrum of operator $T$, and the subsets are the point spectrum, the continuous spectrum, and the residual spectrum, respectively.

I am stuck at this problem. I know that the spectrum is supposed to be the set of eigenvalues, which I am supposed to find with matrices I believe. So I'd have to compute $T-\lambda I$ but I am not sure how the matrix of $T$ looks in this space? Then of course I'd have to compute the subsets which seems even harder.

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    $\begingroup$ Have you studied Banach space theory/operator theory? The question has nothing to do with matrices. $\endgroup$ – Kavi Rama Murthy Feb 10 at 11:50
  • $\begingroup$ Yes I have, but I thought that the spectrum is calculated through the matrix of $T$ with respect to some basis? @KaviRamaMurthy $\endgroup$ – mandella Feb 10 at 11:58
  • $\begingroup$ Your line of thinking completely fails in the case of infinite dimensional spaces. I am sorry to say this, but if I post an answer to this question you may not be able to understand it. $\endgroup$ – Kavi Rama Murthy Feb 10 at 12:03
  • $\begingroup$ Ah, I see. Of course I have the literature but I made a mistake. Maybe I will be able to follow the answer, so any help is appreciated. @KaviRamaMurthy $\endgroup$ – mandella Feb 10 at 12:09
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    $\begingroup$ @KaviRamaMurthy Why are you saying that there are no eigenvalues? $\lambda=\pm \sqrt{2}$ and $y=(\pm\sqrt{2}-1)x$ are the solution of $T(x, y)=\lambda(x, y)$, aren't they? $\endgroup$ – Andrey Gorbunov Feb 10 at 13:42

Consider the matrix $$\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}$$ This isn't the matrix of $T$ w.r.t. to any basis but it can still be useful because formally$$T\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}, \quad \forall\begin{bmatrix}x \\ y \end{bmatrix} \in X \times X $$

Its characteristic polynomial is $\lambda^2-2$ so the eigenvalues are $\pm \sqrt{2}$.

Check that for $\lambda\ne \pm\sqrt{2}$ we have $$(T-\lambda I)^{-1}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}1-\lambda & 1 \\ 1 & -1-\lambda \end{bmatrix}^{-1}\begin{bmatrix}x \\ y \end{bmatrix} = -\frac1{\lambda^2-2}\begin{bmatrix}\lambda+1 & 1 \\ 1 & \lambda-1 \end{bmatrix}^{-1}\begin{bmatrix}x \\ y \end{bmatrix} $$ and this is a bounded linear map so $\lambda \notin \sigma(T)$.

On the other hand, for $\lambda = \pm\sqrt{2}$, we have $$T\begin{bmatrix}(1\pm\sqrt{2})y \\ y\end{bmatrix} = \pm\sqrt{2}\begin{bmatrix}(1\pm\sqrt{2})y \\ y\end{bmatrix}$$ for any $y \in X, y \ne 0$ so $\pm \sqrt{2} \in \sigma_p(T)$.

Therefore $\sigma(T) = \sigma_p(T) = \{\pm\sqrt{2}\}$ and $\sigma_r(T) = \sigma_c(T) = \emptyset$.

We have \begin{align} \left\|(T-\lambda I)^{-1}\begin{bmatrix}x \\ y \end{bmatrix}\right\| &= \frac1{|\lambda^2-2|}\left\|\begin{bmatrix}(\lambda+1)x+y \\ x+(\lambda-1)y \end{bmatrix}\right\| \\ &= \frac1{|\lambda^2-2|}(\|(\lambda+1)x+y \| + \|x+(\lambda-1)y\|)\\ &\le \frac1{|\lambda^2-2|}(|\lambda+1|\|x\|+\|y\| + \|x\|+|\lambda-1|\|y\|)\\ &\le \frac{\max\{1+|\lambda+1|, 1+|\lambda-1|\}}{|\lambda^2-2|}(\|x\|+\|y\|)\\ &= \frac{\max\{1+|\lambda+1|, 1+|\lambda-1|\}}{|\lambda^2-2|}\left\|\begin{bmatrix} x \\ y\end{bmatrix}\right\|\\ \end{align} so $(T-\lambda I)^{-1}$ is bounded. However, this also follows from the Bounded Inverse Theorem once you verify that $(T-\lambda I)^{-1}$ is indeed the algebraic inverse of the bounded map $T-\lambda I$.

  • $\begingroup$ So the matrix can be used to express $T$ although it isn't $T$ itself? And you conclude that the residual and continuous spectrum are empty because for $\lambda \neq \pm \sqrt{2}$ we get that $\lambda\notin\sigma(T)$ right? $\endgroup$ – mandella Feb 11 at 9:05
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    $\begingroup$ @mandella Yes, the matrix is merely a tool to guess what points could be in the spectrum. Yes, and $\pm\sqrt{2}$ are eigenvalues. $\endgroup$ – mechanodroid Feb 11 at 9:07
  • $\begingroup$ When you calculate $(T-\lambda I)^{-1}$, should the matrix at the end when you find the inverse have the ${-1}$? and also why is it bounded? $\endgroup$ – mandella Feb 11 at 9:11
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    $\begingroup$ @mandella I think the matrix is ok, notice the minus in front. I added the explanation why it's bounded. $\endgroup$ – mechanodroid Feb 11 at 9:22
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    $\begingroup$ @mandella I believe $1\pm \sqrt{2}$ is correct because $$\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix}1\pm\sqrt{2} \\ 1\end{bmatrix} = \begin{bmatrix}2\pm\sqrt{2} \\ \pm\sqrt{2}\end{bmatrix} = \pm\sqrt{2} \begin{bmatrix}1\pm\sqrt{2}\\ 1\end{bmatrix}$$ $\endgroup$ – mechanodroid Feb 11 at 9:35

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