I have the following problem,
"Suppose that $X=\ell^1$ and define the operator $T\in B(X)$ as follows:
$$Tx=\left(\frac12x_2,\frac13x_3,\frac14x_4,...\right)\,,\textit{where,}\,\,\, x=(x_1,x_2,x_3,...)$$
Find the eigenvalues of $T.$"
Here is my attempt so far:
Let $\lambda$ be an eigenvalue of $T$, then,
$$\implies\exists x\in \ell^1:Tx=\lambda x,\lambda\in\mathbb C$$ $$\implies\frac12x_2=\lambda x_1,\,\frac13x_3=\lambda x_2,\,\frac14x_4=\lambda x_3,...$$
We can write this in an equivalent manner as follows:
$$\implies \frac12x_2=\lambda x_1,\,\frac13x_3=2!\lambda^2x_1,\,\frac14x_4=3!\lambda^3x_1,\,...\,,\frac1nx_n=(n-1)!\lambda^{n-1}x_1,\,...$$
For $n\in\mathbb N/\{1\}$.
In order for the eigenvalue problem to be satisfied we require $x\ne1$, and so, if we have that $x_1=0\implies x=0$, which yields a contradiction. Thus, $x_1\neq0.$
This means that we then have,
$$x=(x_1,\lambda x_1,2!\lambda^2x_1,...)$$
We must also ensure that $x\in X=\ell^1$, so we consider,
$$x\in\ell^1\iff\sum_{k=1}^\infty|x_k|\lt \infty$$
$$\iff\sum_{k=1}^\infty|(k-1)!\lambda^{k-1}x_1|\lt \infty$$
$$\iff|x_1|\sum_{k=1}^\infty|(k-1)!\lambda^{k-1}|\lt \infty$$
And we know that $x_1\neq0$, so,
$$\iff\sum_{k=1}^\infty|(k-1)!\lambda^{k-1}|\lt \infty$$
I am unsure how to move on from here. I had thought to argue along the lines that if $(k-1)!$ grows at a rate faster than the $\lambda^{k-1}$ tends to zero, then the sum will not be finite. I think I have to try and show that $|\lambda|\lt1$, but am not sure how to get rid of the $(k-1)!$ in the above. Is the aforementioned line of thought the right way to go about this? Cheers!
The operator $T$ satisfies $$ \|Tx\| \le \frac{1}{2}\|x\|, \\ \|T^2x\| \le \frac{1}{2\cdot 3}\|x\|, \\ \|T^3x\| \le \frac{1}{2\cdot 3\cdot 4}\|x\|. $$ In general $\|T^{n}\| \le 1/(n+1)!$. If $Tx=\lambda x$ and $x \ne 0$, then $$ |\lambda|^n\|x\|=\|T^nx\| \le \frac{1}{(n+1)!}\|x\| $$ If $\lambda \ne 0$, then $$ \|x\| \le \frac{|\lambda|^{-n}}{(n+1)!}\|x\|,\;\;\; n \ge 1. $$ The right side tends to $0$ as $n\rightarrow\infty$ because it is a general term of an everywhere convergent exponential series. Hence, $x=0$, which is a contradiction. Therefore, the only possible eigenvalue of $T$ is $0$. And $0$ is an eigenvalue of $T$.