# Thread: Operator has no eigenvalues

1. ## Operator has no eigenvalues

Hello everyone

Show: The operator T has no eigenvalues

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

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

Proof:

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

$\displaystyle \Rightarrow x = 0$

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

Help would be much appreciated

Thanks
Rapha

2. Hi

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

You agree?

It is implicit in the definition of $\displaystyle T$ that

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

3. Originally Posted by Rapha
Hello everyone

Show: The operator T has no eigenvalues

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

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

Proof:

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

$\displaystyle \Rightarrow x = 0$

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