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Let $A\colon \ell^{1}\to \ell^{1}$ be defined by $$A(x)=(x_{2}+x_{3}+x_{4}+ \dots,x_1,x_2,x_3,\dots),$$ where $x\in\ell^1$ iff $\sum|x_k|<\infty$. Let $D$ be the closed unit disc in $\Bbb C$ and $\lambda_0=(1+\sqrt5)/2$. Show that $$\sigma(A)=D\cup\{\lambda_0\}.$$

I have managed to show that Eig$(A)=\{\lambda_0\}$ and $D\subset\sigma(a)$ so that $D\cup\{\lambda_0\}\subset\sigma(A)$.

How can I prove the converse?

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1 Answer 1

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Assume that $\lambda\neq \lambda_0$ and $|\lambda|>1$. Let $B:=A-\lambda I$. We have to show that $B$ is injective and surjective.

  • Let $x\in \ell^1$ such that $Bx=0$. We have to show that $x=0$. We have $x_j=\lambda x_{j+1}$ for $j\geq 1$ hence $x_1=\lambda^{n-1}x_n$ and $x_n=\lambda^{1-n}x_1$. Comparing the first coordinates, this gives $$\sum_{n=1}^{\infty}x_1\lambda^{-n}=\lambda x_1.$$ If $x_1=0$ then $x=0$ and we are done, and if $x_1\neq 0$ then $\sum_{n=1}^{\infty}\lambda^{-n-1}=1$ hence $\frac {\frac{1}{\lambda^2}}{1-\frac{1}{\lambda}}=1$, that is $\lambda^2-\lambda-1=0$. As $\lambda>1$, $\lambda$ would be equal to $\lambda_0$. A contradiction. Hence $x=0$ and since Kernel$(B)=\{0\}$, B is injective.

  • We have to show that $B$ is surjective. Fix $y\in \ell^1$. If $x$ works, we should have $x_j-\lambda x_{j+1}=y_{j+1}$, hence $$\sum_{j=2}^n\lambda^{j-2}y_j=x_1-\lambda^{n-1}x_{n},$$ which gives $$x_{n}=\frac{x_1}{\lambda^{n-1}}-\sum_{j=2}^{n}\frac {y_j}{\lambda^{n+1-j}}.$$ The first coordinates give $$x_1\left(\frac{1}{1-\lambda}-\lambda\right)=y_1+\sum_{n=2}^{\infty}\sum_{j=2}^{n}\frac {y_j}{\lambda^{n+1-j}}$$ As $|\lambda|>1$, using a change of index in the double summation we can see that $x\in \ell^1$.

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