0
$\begingroup$

Following this question:Prove that $T$ is not compact but $T^2$ is compact operator

Consider the linear operator $T: \ell_2\to \ell_2$ defined by for $x=(\xi_1,\xi_2,\dots)\in \ell_2$ $$ Tx=(0,\xi_1, 0, \xi_3, 0, \xi_5, 0,\dots, ) $$

I want to obtain the eigenvalues and spectrum of $T$.

Clearly, let $\lambda$ be eigenvalue of $T$. Then consider equation $Tx=\lambda x$: $$ (0,\xi_1, 0, \xi_3, 0, \xi_5, 0,\dots, )=(\lambda \xi_1, \lambda \xi_2, \lambda \xi_3,\dots) $$ we get $\lambda =0$ and its eigenvector is $(1, 0, 1, 0,\dots)$. So $\sigma_p(T)=\{0\}$.

To get spectrum $\sigma(T)$. Since $\|T\|=1$, then $\sigma(T)\subset \{\lambda: |\lambda|\le 1\}$.

Note that $$(\lambda I-T)x=(\lambda \xi_1, \lambda \xi_2-\xi_1, \lambda \xi_3, \lambda \xi_4-\xi_3, \dots)$$

When $0<|\lambda|<1$, I want to show that $(\lambda I-T)$ is not surjective.

I try to consider $$(\lambda I-T)x=e_1$$. But this one does not work...


As $\lambda=0$, consider $(0I-T)x=0$. That is $(0, -\xi_1, 0, -\xi_3, \dots)=0$. If we choose $x_0=(0, 1, 0, 1, 0,\dots)\neq 0$, then $x_0\in ker(0I-T)$. So $0\in \sigma(T)$.

As $0<|\lambda|<1$. Is it still not injective?

$\endgroup$
4
  • $\begingroup$ @DominikS Thanks! So we can choose its eigenvector as $(1,0,1,0,\dots)$? $\endgroup$
    – Hermi
    Commented Dec 11, 2023 at 8:31
  • $\begingroup$ "But this one does not work..." Why does it not work? Don't you get a recurrence for all $\xi_i$? I presume it can always be solved, and the remaining question is whether the result is in $\ell^2$. $\endgroup$
    – DominikS
    Commented Dec 11, 2023 at 8:32
  • $\begingroup$ Eigenvectors for $\lambda=0$: I would say any vector with $\xi_1 = \xi_3 = \ldots = 0$ (i.e. all odd indices). This gives you a big eigenspace. $\endgroup$
    – DominikS
    Commented Dec 11, 2023 at 8:34
  • $\begingroup$ @DominikS Because I got $\lambda \xi_1=1, \lambda \xi_2-\xi_1=0, \lambda \xi_3=0,\dots $. So $\xi_1=1/\lambda, \xi_2=1/\lambda_2, \xi_3=\xi_4=\dots =0$... It will be in $\ell_2$. $\endgroup$
    – Hermi
    Commented Dec 11, 2023 at 8:34

1 Answer 1

4
$\begingroup$

By the spectral Mapping Theorem $\lambda \in \sigma(T)$ implies $\lambda^{2} \in \sigma (T^{2})=\{0\}%$ because $T^{2}=0$. Hence, there are no non-zero points in the spectrum of $T$.

Alternatively, note that $T^{2}=0$ implies that (for $\lambda \neq 0$) $(\lambda I-T)(\frac I {\lambda} +\frac T {\lambda^{2}})=(\frac I {\lambda}+\frac T {\lambda^{2}})(\lambda I-T)=I$ so $\lambda I-T$ is invertible.

$\endgroup$
3
  • $\begingroup$ Thanks! But how to get its spectrum without using spectral mapping theorem? $\endgroup$
    – Hermi
    Commented Dec 11, 2023 at 8:32
  • $\begingroup$ Just look at the map. It annihilates its image. The only possible eigenvalue is zero. $\endgroup$
    – Louis
    Commented Dec 11, 2023 at 9:11
  • $\begingroup$ @Hermi The trick that geetha290krm uses above is called the Neumann series (it's essentially a more general version of the familiar geometric series). You can find some more intuition here math.stackexchange.com/questions/4815027/… (sorry for the self-promotion). $\endgroup$ Commented Dec 11, 2023 at 10:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .