I'm supposed to find the spectrum of an operator $T$ on the Banach space $\ell^1$, where $\|x_n\|=\sum_{n=1}^\infty |x_n|$ and $T\{x_n\}=\{x_2,x_1, x_4, x_3, x_6, x_5, x_8,x_7, \ldots \}$. I have found that the point spectrum of the operator is $\sigma_p=\{1\}$ by analyzing $\{x_2-\lambda x_1, x_1-\lambda x_2, x_4-\lambda x_3, x_3-\lambda x_4, x_6-\lambda x_5,\ldots \} = \{0,0,0,0,0,\ldots \}$ (is this correct?). I've also shown that $\|T\|=1$, so $\sigma _T\subset \{\lambda \in \mathbb{C}, |\lambda |\le 1\}$. How do I proceed to show the whole spectrum?
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1$\begingroup$ From the Hilbert space case we should expect that $\sigma(T) \subseteq \{ \lambda \in \mathbb{C} \ : \ \vert \lambda \vert =1 \}$ (as $T$ is an isometry). I suggest you to try the following trick. Using $A^2=Id$, we get $$ (A - \lambda Id) (A-\mu Id) = - (\mu + \lambda) (A- \frac{1+\mu \lambda}{\mu + \lambda} Id). $$ To show that $A- \lambda Id$ is a bijection it is enough to find $\mu$ s.t. $\vert \mu \vert >1$ and $$\left\vert \frac{1+\lambda \mu}{\mu + \lambda} \right\vert>1$$ and then use the fact that $\sigma(T) \subseteq \{ \lambda \in \mathbb{C} \ : \ \vert \lambda \vert \leq 1\}.$ $\endgroup$– Severin SchravenApr 13, 2020 at 19:32
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$\begingroup$ A slightly more useful observation is $$ (A- \lambda Id)(A+ \lambda Id) = A^2 - \lambda^2 Id = (1-\lambda^2) Id. $$ Which allows us to directly compute the inverse in most cases. The only cases where it is not invertible you will get point spectrum. $\endgroup$– Severin SchravenApr 13, 2020 at 22:27
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2 Answers
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You could calculate $T^2$ and in the process observe that you haven't got the complete point spectrum. Minimal polynomials also work in infinite dimension.
answered Apr 13, 2020 at 19:58
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$\begingroup$ +1 Nice hint. Should have read it before writing a comment $\endgroup$ Apr 13, 2020 at 22:28
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$\begingroup$ Okay, so if I've done this correctly, I now have the point spectrum $\sigma _{p}=\{1,-1\}$, because $\lambda ^{2}=1$. So I now just have to prove that it is the whole spectrum, correct? $\endgroup$ Apr 14, 2020 at 0:30
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$\begingroup$ Yes, indeed. As I said minimal polynomials have good use also in infinite dimension. In the present case, show that $1=P+Q$ where $P=(1-T)/2$ and $Q=(1+T)/2$ are projections onto the respective eigenspaces. $\endgroup$ Apr 14, 2020 at 0:45
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Only by definitions. Consider the equation $Tx=\lambda x$, i.e. $x_2-\lambda x_1=0$, $x_1-\lambda x_2=0$, $x_4-\lambda x_3=0$, $x_3-\lambda x_4=0$,... We get pairs of equations from which all $x_k$ can be found. For example: $\begin{cases} x_1-\lambda x_2=0,\\x_2-\lambda x_1=0\end{cases}$. This is a homogeneous system of linear equations and it has a nonzero solution if and only if its determinant is zero, i.e. $\begin{vmatrix} 1& -\lambda\\ -\lambda& 1 \end{vmatrix}=0$. From here we have $\lambda=\pm1$ -- this is a point spectrum (you can write out nonzero eigenfunctions corresponding to these numbers).
Further, let $\lambda\ne\pm1$. Consider the operator $T-\lambda I$ and show that this operator has a bounded inverse defined on the whole space. We need to solve equation $Tx-\lambda x=y$, $y\in l_1$. This equation splits into pairs of inhomogeneous systems that have unique solutions and can be found using the Cramer formulas: $$x=\left(\frac{y_2+\lambda y_1}{1-\lambda^2},\frac{y_1+\lambda y_2}{1-\lambda^2},\frac{y_4+\lambda y_3}{1-\lambda^2},\frac{y_3+\lambda y_4}{1-\lambda^2},...\right).$$ Because the $\|x\|\leq\dfrac{2\max\{1,|\lambda|\}}{|1-\lambda^2|}\|y\|$, operator $(T-\lambda I)^{-1}$ is bounded when $\lambda\ne\pm1$, so it's resovent of $T$.
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$\begingroup$ Thanks, I can see it now. I'm just not sure how you came to the inequality $||x||\le \frac{2\text{ max}\{1, |\lambda |\}}{|\lambda ^{2}-1|}||y||$ from $||x||=\frac{1}{|1-\lambda ^{2}|}\sum_{n=1}^{\infty }\{|y_{2}+\lambda y_{1}|,\ldots \}$. $\endgroup$ Apr 14, 2020 at 2:07
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$\begingroup$ $1\cdot|y_2|+|\lambda|\cdot|y_1|\leq\max\{1,|\lambda|\}|y_2|+\max\{1,|\lambda|\}|y_1|$, and so for each pair of terms. $\endgroup$– thingApr 14, 2020 at 2:18