The eigenvalues and spectrum of $T$

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?

• @DominikS Thanks! So we can choose its eigenvector as $(1,0,1,0,\dots)$? Commented Dec 11, 2023 at 8:31
• "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 $\ell^2$. Commented Dec 11, 2023 at 8:32
• 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. Commented Dec 11, 2023 at 8:34
• @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 $\ell_2$. Commented Dec 11, 2023 at 8:34

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.