Showing operator has no eigenvalues

Question

Let $T:\ell_2 \to \ell_2$ be defined by $(T(x))_n = \dfrac{x_{n-1}}{n}$ for $n \gt 1$ and $(T(x))_1=0$.

I have showed that $T$ is a compact operator.

Now I need to show that $T$ has no eigenvalues.

So assume $\lambda \ne 0$ is such, we have $T(x) = \lambda x$ so $\dfrac{x_n-1}{n} = \lambda x_n$ for all $n\gt 1$ which gives : $$x_n = \dfrac{1}{n! \lambda^{n-1}}x_1 $$

I'm not sure how to continue.

Thanks for helping!

Answer 1

You just forgot the condition $(Tx)_1=0$. If $Tx=\lambda x$ with $\lambda \neq 0$ then $0=(Tx)_1=\lambda x_1$ so $x_1=0$ . Your calculation now shows $x_n=0$ for all $n$. Can you check if $0$ is an eigen value?

Answer 2

Since $T(x)_1=0$ and ince $T(x)_1=\lambda x_1$, you get that $x_1=0$. Then you have $T(x)_2=\frac{x_1}2$ and therefore $T(x_2)=0$. And so on…

Answer 3

Write $x =\sum _{n\geq 1} x_n e_n $ where $\{ e_n :n \in \mathbb N \} $ is the standard orthonormal basis of $\mathcal l ^2$.

Then if $\lambda$ is an eigen-value with eigenvector $x$ we have $$\lambda x_n = \frac {x_{n-1} }{n} \forall n \in \mathbb N$$

This gives $ x_1=0$ and hence $ x_n=0 \forall n \in \mathbb N $