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Let $T:\ell^2\to \ell^2$ such that $T(x_1,x_2,x_3,x_4,...)=(0,4x_1,x_2,4x_3,x_4....)$

then

$T^ *:\ell^2 \to \ell^2$ such that $T^*(x_1,x_2,x_3,x_4...)=(4x_2,x_3,4x_4,....)$,

hence $(T^*)^2(x_1,x_2,x_3,x_4,...)=(4x_3,4x_4,4x_5,...)$.

I have shown that if $|\lambda|<4$, then $\lambda$ is eigenvalue of $(T^*)^2$.

Now I have to show $\sigma(T)=\{\lambda \in\mathbb{C:|\lambda|\leq2}\}$

from the fact that $|\lambda|<4$ is eigenvalue of $(T^*)^2$. We have

$$\{\lambda\in\mathbb{C:|\lambda|\leq4}\} \subseteq \sigma(T^2).$$

Now in the hint given it say if $|\lambda|\leq2$, then $\lambda\in\sigma(T)$, otherwise $\lambda^2\notin \sigma(T^2)$.

But I am not getting this hint, how does $\lambda\notin \sigma(T)$ imply $\sigma^2\notin(T^2)$?

Could anyone help me please?

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  • $\begingroup$ Without further assumptions it doesn't. We have $-1 \notin \sigma(I)$, but of course $(-1)^2 \in \sigma(I^2)$. The implication in the other direction, $\lambda^2 \notin \sigma(T^2) \implies \lambda \notin \sigma(T)$ holds generally. And by the spectral mapping theorem, $\sigma(T^2) = \{ \lambda^2 : \lambda \in \sigma(T)\}$. So for $\lvert\lambda\rvert \leqslant 2$, at least one of $\lambda$ and $-\lambda$ belongs to $\sigma(T)$. Showing that both do (if the asserted result about $\sigma(T)$ is correct, I haven't checked) would need additional considerations. $\endgroup$ – Daniel Fischer Mar 10 '18 at 12:25
  • $\begingroup$ @DanielFischer please provide some help in this problem.math.stackexchange.com/questions/2633776/… $\endgroup$ – paarth Mar 10 '18 at 13:00
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What is always true is that $$\tag1\sigma(T^2)=\{\lambda^2:\ \lambda\in\sigma(T)\}=\sigma(T)^2,$$which follows easily from $T^2-\lambda^2=(T-\lambda)(T+\lambda)$. As Daniel mentioned, I don't see that you can immediately obtain $\sigma(T)$ from this.

What you get from $(1)$, since $\|T^2\|=4$, is that $$\tag2\sigma(T)\subset\{|\lambda|:\ |\lambda|\leq2\}.$$

For $T^*$, you can do a similar trick than you did for $T^{*2}$. Namely, if $|\lambda|<2$, then $$ T^*(\lambda,\frac{\lambda^2}4, \frac{\lambda^3}4,\frac{\lambda^4}{16},\frac{\lambda^5}{16},\frac{\lambda^6}{2^6},\frac{\lambda^7}{2^6}\ldots) =\lambda\, (\lambda,\frac{\lambda^2}4, \frac{\lambda^3}4,\frac{\lambda^4}{16},\frac{\lambda^5}{16},\frac{\lambda^6}{2^6},\frac{\lambda^7}{2^6}\ldots), $$ which shows that $$\tag3\{|\lambda|:\ \lambda<2\}\subset\sigma(T^*).$$ Now combining $(2)$ and $(3)$, $$ \sigma(T)=\overline{\sigma(T^*)}=\{|\lambda|:\ |\lambda|\leq2\}. $$

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