While studying functional analysis I started with spectral theory and while doing some exercises of spectrum in my source book, I came up with this one:
Find the spectrum and resolvent of the following operator in $A:l^{2}\rightarrow l^{2}$ such that $Ax=(0, x_{1},x_{2},0,x_{4},x_{5},\dots, x_{3n+1},x_{3n+2},\dots)$
First I tried doing the following:
Take $x,y\in l^{2}$ and $(Ax-\lambda)x=y$, so $Ax=\lambda x$.
Now, for all $(x_{1},x_{2},x_{3},\dots)$ we have that:
$\lambda x_1=0,$
$\lambda x_2=x_1,$
$\lambda x_3=x_2,$
$\lambda x_4=0,$
$\lambda x_5=x_4,$
$\lambda x_6=x_5,$
$\vdots$
$\lambda x_{3n+1}=0 $
$\lambda x_{3n+2}=x_{3n+1}$
$\lambda x_{3n+3}=x_{3n+2}$
From these we have we obtain that for $\lambda=0$ then $\{0\}$ is part of the spectrum, but now, how should I proceed to find the $\lambda$ which are not zero and still part of the spectrum?
Regarding this problem, I asked my professor for a way to find this $\lambda\neq 0$ and he told me that I should play with the powers of $A$, but I don't understand what to do with this information.
Any kind of help would be highly appreciated.
Best Answer
Recall that $\lambda\notin \sigma(A)$ iff $A-\lambda I:\ell^2\rightarrow \ell^2$ is a bijection. In this case we call $(A-\lambda I)^{-1}$ the resolvent. If $\lambda \neq 0$, then we can formally write $$(A-\lambda I)^{-1}=\lambda^{-1}(\lambda^{-1}A-I)^{-1}=-\lambda^{-1}(I-\lambda^{-1}A)^{-1}.$$
In fact, if one of the expressions exist, then so do the other. However, the RHS looks like the result of a geometric series. Hence, we might expect that we can write
$$ (I-\lambda^{-1}A)^{-1} = \sum_{n=0}^\infty (\lambda^{-1}A)^n $$ for $\vert\lambda\vert $ sufficiently large. Indeed, this is the so-called Neumann series and one can prove that it converges if $\Vert \lambda^{-1}A \Vert<1$ (just like for the geometric series). Indeed, we have $\Vert (\lambda^{-1}A)^n \Vert \leq \Vert \lambda^{-1}A \Vert^n$ and hence we get
$$ \left\Vert \sum_{n=0}^\infty (\lambda^{-1}A)^n \right\Vert \leq \sum_{n=0}^\infty \Vert(\lambda^{-1}A)^n \Vert \leq \sum_{n=0}^\infty \Vert \lambda^{-1} A \Vert^n. $$ Thus, for $\Vert \lambda^{-1}A\Vert<1$ the series converges. In particular, for $\vert \lambda \vert > \Vert A \Vert$ the series converges. However, if the series converges absolutely, then we check that this series really gives the inverse $(I-\lambda^{-1}A)^{-1}$ as we know it for the geometric series. Indeed, we have \begin{align*} (I-\lambda^{-1}A) \left(\sum_{n=0}^\infty (\lambda^{-1}A)^n\right) &= \sum_{n=0}^\infty (\lambda^{-1}A)^n - (\lambda^{-1}A) \sum_{n=0}^\infty(\lambda^{-1}A)^n \\ &= \sum_{n=0}^\infty (\lambda^{-1}A)^n - \sum_{n=1}^\infty (\lambda^{-1}A)^n = I. \end{align*} Similarly one shows that $$ \left(\sum_{n=0}^\infty (\lambda^{-1}A)^n\right)(I-\lambda^{-1}A) =I. $$ Therefore, we have $$ (I-\lambda^{-1}A)^{-1} = \sum_{n=0}^\infty (\lambda^{-1}A)^n $$ for $\vert \lambda \vert > \Vert A \Vert$. This means that $\lambda$ cannot be in the spectrum if $\vert \lambda \vert > \Vert A \Vert$.
Now what happens for small $\lambda$? The norm $\Vert \lambda^{-1} A \Vert$ is too large to apply the argument from before. Here we need to use another property of $A$. We note that $A^3 =0$. Thus, we get $$ \sum_{n=0}^\infty (\lambda^{-1}A)^n = I + (\lambda^{-1}A) + (\lambda^{-1}A)^2. $$ We can hope that this series still gives us the inverse. Indeed, $$ (I-\lambda^{-1}A)\left(I + (\lambda^{-1}A) + (\lambda^{-1}A)^2I + (\lambda^{-1}A) + (\lambda^{-1}A)^2\right) = I - (\lambda^{-1}A)^3 =I $$ and similarly if we commute the two operators. This means that $(A-\lambda I)$ is invertible for all $\lambda \neq 0$ and hence only $\lambda=0$ can be in the spectrum. Note that we could have done this argument from the start, but it would have been unclear why we would try to do that.
We finally check that $0$ is in the spectrum. In fact it is even an eigenvalue as $A(0,0,1,0,0, \dots )=(0, \dots)$.