the spectrum of a bounded linear operator on $X\times X$

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.

• Have you studied Banach space theory/operator theory? The question has nothing to do with matrices. Feb 10 '19 at 11:50
• Yes I have, but I thought that the spectrum is calculated through the matrix of $T$ with respect to some basis? @KaviRamaMurthy Feb 10 '19 at 11:58
• 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. Feb 10 '19 at 12:03
• 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 Feb 10 '19 at 12:09
• @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? Feb 10 '19 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 $$T-\lambda I$$.

• 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? Feb 11 '19 at 9:05
• @mandella Yes, the matrix is merely a tool to guess what points could be in the spectrum. Yes, and $\pm\sqrt{2}$ are eigenvalues. Feb 11 '19 at 9:07
• 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? Feb 11 '19 at 9:11
• @mandella I think the matrix is ok, notice the minus in front. I added the explanation why it's bounded. Feb 11 '19 at 9:22
• @mandella I believe $1\pm \sqrt{2}$ is correct because $$\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix}1\pm\sqrt{2} \\ 1\end{bmatrix} = \begin{bmatrix}2\pm\sqrt{2} \\ \pm\sqrt{2}\end{bmatrix} = \pm\sqrt{2} \begin{bmatrix}1\pm\sqrt{2}\\ 1\end{bmatrix}$$ Feb 11 '19 at 9:35