# Operator has no eigenvalues

• Feb 18th 2010, 08:31 AM
Rapha
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
• Feb 18th 2010, 11:25 AM
mabruka
Hi

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

You agree?

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

$\displaystyle Tx = 0\cdot e_1 + \sum^{\infty}_{k=1} \frac{x_k}{k+1}e_{k+1}$
• Feb 18th 2010, 12:11 PM
HallsofIvy
Quote:

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.

Quote:

Help would be much appreciated

Thanks
Rapha
• Feb 19th 2010, 07:20 AM
Rapha
Hi!

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

Rapha