# Operator has no eigenvalues

• Feb 18th 2010, 09:31 AM
Rapha
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
• Feb 18th 2010, 12:25 PM
mabruka
Hi

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

You agree?

It is implicit in the definition of $T$ that

$Tx = 0\cdot e_1 + \sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1}$
• Feb 18th 2010, 01:11 PM
HallsofIvy
Quote:

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.

Quote:

Help would be much appreciated

Thanks
Rapha
• Feb 19th 2010, 08:20 AM
Rapha
Hi!

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

Rapha