Exercise: Consider $A:l_2\to l_2$ where $Ax=(0,x_1,\frac{1}{2}x_2,\frac{1}{3}x_3...)$

Show that the operator has no point spectrum.

Point spectrum definition:$\ker(\lambda I-A)\neq 0\\Im(\lambda I-A)=X\\R(\lambda,A)=(\lambda I-A)^{-1}$ where $X$ is a normed vector space.

I noticed that $Ax=\lambda x\implies \lambda x_1,\lambda x_2,\lambda x_3...)$. Then $\lambda x_1=0,\lambda x_2=x_1,\lambda x_3=\frac{1}{2}x_2$.

The problem rises once I do not know if I can infer $\lambda x_1=0\implies \lambda=0 \vee x_1=0$ which further implies that all other elements are zero following the previous construction. So my proof would refute the point spectrum definition in the sense that $\ker(\lambda I-A)= 0$, when $x=0$ or $Im(\lambda I-A)\neq l_2$ when $\lambda=0$.


Is the point spectrum definition right? Is the proof right?

Thanks in advance!

  • $\begingroup$ Writing $Ax=(0,x_1,\frac12 x_2,\dots)$ actually doesn't specify what $A$ is! Is $x=(x_1,x_2,\dots)$ or $x=(x_0,x_1,\dots)$? Makes a big difference here... $\endgroup$ – David C. Ullrich May 20 '18 at 12:47

Suppose that $\lambda$ is in the point spectrum of $A$. Then there exists a non-zero $x \in \ell^2$ such that $Ax = \lambda x$.

First consider the case where $\lambda \ne 0$. Since $\lambda x_1 = 0$ and $\lambda \ne 0$ we can now deduce that $x_1 = 0$. Following your argument we get $x_n = 0$ for all $n \in \mathbb{N}$ so that $x = 0$, contradicting the fact that $x$ is non-zero.

Now consider the case that $\lambda =0$. Then for $\lambda$ to be in the point spectrum we must have $Ax = 0$ for some non-zero $x$. In particular, this says $A$ is not injective. However, it is not too hard to see that if $Ax = Ay$ then we must have $x = y$, so $A$ is injective. Thus, $0$ is not in the point spectrum.


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