# Thread: Operator has no eigenvalues

1. ## Operator has no eigenvalues

Hello everyone

Show: The operator T has no eigenvalues

$T:l^2(\mathbb{C}) \to l^2(\mathbb{C})$

$Tx:=\sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1}$ , where $x = (x_k)_{k \in \mathbb{N}}$

Proof:

Iff $\lambda \not= 0$ it follows that $x_1 = 0$ and $x_{k+1} = \frac{x_k}{\lambda(k+1)}$ if $k \ge 1$

$\Rightarrow x = 0$

I don't get it: how do we know that $x_1 = 0$ ?

Help would be much appreciated

Thanks
Rapha

2. Hi

I think it should say $(Tx)_1=0$.

You agree?

It is implicit in the definition of $T$ that

$Tx = 0\cdot e_1 + \sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1}$

3. Originally Posted by Rapha
Hello everyone

Show: The operator T has no eigenvalues

$T:l^2(\mathbb{C}) \to l^2(\mathbb{C})$

$Tx:=\sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1}$ , where $x = (x_k)_{k \in \mathbb{N}}$

Proof:

Iff $\lambda \not= 0$ it follows that $x_1 = 0$ and $x_{k+1} = \frac{x_k}{\lambda(k+1)}$ if $k \ge 1$

$\Rightarrow x = 0$

I don't get it: how do we know that $x_1 = 0$ ?
Look at the definition of T- the sum is $\frac{x_1}{2}e_2+ \frac{x_2}{3}e_3+ \cdot\cdot\cdot$. The crucial point is that there is NO " $e_1$" in that. In order for that to be a multiple of x, there must be NO " $e_1$" in x. That means that the coefficient of $e_1$, $x_1$, must be 0.

Help would be much appreciated

Thanks
Rapha

4. Hi!

Thank you so much! It would never have occurred to me. Your explanations made this excercise so easy.

Rapha