Operator has no eigenvalues

**Rapha** · February 18th 2010, 08:31 AM

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

**mabruka** · February 18th 2010, 11:25 AM

Hi

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

You agree?

It is implicit in the definition of $T$ that

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

**HallsofIvy** · February 18th 2010, 12:11 PM

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

**Rapha** · February 19th 2010, 07:20 AM

Hi!

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

Rapha

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