# No point spectrum in $l_2$

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 $$\lambda=0$$.

Question:

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

• 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... Commented May 20, 2018 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.