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Let $\{a_n\}_{n=1}^{\infty} \subset \mathbb C$ be a sequence such that $|a_n|=r$ for all $n \geq 1$ and $r \geq 0$

Define $T: \ell^2 \to \ell^2$ by

$T(x_1,x_2,x_3,...) = (0,a_1x_1,a_2x_2,....) $ with $(x_1,x_2,....) \in l^2$

Determine the spectrum of T.

The spectrum of an operator is defined as $\{\lambda \in \mathbb{C}: T-\lambda I \ \text{is not invertible}\}$

I know how to calculate the eigenvalues of a matrix, but here I have really no idea to start.

Can anyone help?

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  • $\begingroup$ Definitions are your friends. How do we define the spectrum of linear operator $T$? $\endgroup$
    – hardmath
    Commented Dec 31, 2016 at 12:42
  • $\begingroup$ I edited my post with the definition $\endgroup$
    – Yuhe
    Commented Dec 31, 2016 at 12:47
  • $\begingroup$ Okay, so $(T-\lambda I)$ applied to $(x_1,x_2,\ldots)$ gives a sequence of complex numbers. Can we reconstruct $(x_1,x_2,\ldots)$ from that sequence? For simplicity, start with the case $\lambda = 0$. $\endgroup$
    – hardmath
    Commented Dec 31, 2016 at 12:52
  • $\begingroup$ What do you mean by reconstruct? We are getting $(-\lambda, a_1x_1-\lambda,....)$ aren't we? $\endgroup$
    – Yuhe
    Commented Dec 31, 2016 at 12:57
  • $\begingroup$ No, $\lambda I$ applied to $(x_1,x_2,\ldots)$ is $(\lambda x_1, \lambda x_2,\ldots)$, not $(\lambda,\lambda,\ldots)$. $\endgroup$
    – hardmath
    Commented Dec 31, 2016 at 13:00

1 Answer 1

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Note first that $\|T\|=r$, and so $\sigma(T)\subset\{\lambda:\ |\lambda|\leq r\}$.

It is very easy to check that $T$ has no eigenvalues. This makes it usually easier to deal with $T^*$. We have $$ T^*x=(\overline{a_1}x_2,\overline{a_2}x_3,\ldots). $$ Assume $|\lambda|<r$. Define a sequence $\{c_n\}$ by $c_1=1$ and $c_{n+1}=\dfrac{c_n}{\overline{a_n}}$ and put $x=(\lambda c_1,\lambda^2c_2,\lambda^3c_3,\ldots).$ Then $x\in\ell^2$, since $|c_n|=r^{n-1}$ and so $$ \sum_{n=1}^\infty |x_n|^2=r\sum_{n=1}^\infty\left|\frac\lambda r\right|^n<\infty. $$ Now $$ T^*x=(\overline{a_1}x_2,\overline{a_2}x_3,\ldots) =(\overline{a_1}c_2\lambda^2,\overline{a_2}c_3\lambda^3,\ldots) =_(c_1\lambda^2,c_2,\lambda^3,\ldots)=\lambda x. $$ So all such $\lambda$ are eigenvalues of $T^*$. We obtained $$ \{\lambda:\ |\lambda|<r\}\subset\sigma(T^*)\subset\{\lambda: |\lambda|\leq r\}. $$ As the spectrum is always closed, we get $\sigma(T^*)=\{\lambda:\ |\lambda|\leq r\}$. Finally, since the spectrum of $T^*$ is obtained from the spectrum of $T$ by conjugating the elements, we have $$ \sigma(T)=\{\lambda:\ |\lambda|\leq r\}, $$ that is, $\sigma(T)$ is the ball of radius $r$ centered at zero.

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  • $\begingroup$ I had to handle with a similar exercise and your answer was very helpful! Although I would appreciate if you could help me understand the last step of your proof. I know $\;σ(T^*)=\overline {σ(T)}=\{ \; \bar z \in \mathbb C: z \in σ(T)\}\;$. How do I proceed? Thanks in advance! $\endgroup$ Commented Jun 11, 2017 at 19:57
  • $\begingroup$ It's just that $T-\lambda I$ is invertible if and only if its adjoint is invertible. $\endgroup$ Commented Jun 11, 2017 at 21:25
  • $\begingroup$ You say $\;z \notin σ(T)\;$ iff $\;z \notin σ(T^*)\;$, don't you? But this means $\;σ(T)=σ(T^*)\;$. Is this true? What am I missing here? $\endgroup$ Commented Jun 11, 2017 at 21:42
  • $\begingroup$ Nope. The adjoint of $T-zI $ is $T^*-\bar z I $. $\endgroup$ Commented Jun 11, 2017 at 22:00
  • $\begingroup$ Oh yes! My mistake! Got it now! Thanks a lot for your time $\endgroup$ Commented Jun 11, 2017 at 22:03

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