• Support PF! Buy your school textbooks, materials and every day products Here!

Calculate the spectrum of a linear operator

  • #1
2
0
<mod note: moved to homework>

Calculate the spectrum of the linear operator ##T## on ##B(\ell^1)##.
$$T(x_1,x_2,x_3,\dots)=(\sum_{n=2}^\infty x_n, x_1, x_2, x_3, \dots)$$


I think the way to do it is to find the point spectrums of ##T## and its adjoint ##T^*##. But I don't know how to calculate them.

Some progress:

Let ##T(x_n)=\lambda (x_n)##. Then we have ##x_n=\lambda^{-n+1}x_1## for ##n \geq 2## and ##x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}##. Then
$$x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}=\frac{\sum_{n=2}^\infty \lambda^{-n+1}x_1}{\lambda}=\sum_{n=2}^\infty \lambda^{-n}x_1$$
Therefore, ##(1-\sum_{n=2}^\infty \lambda^{-n})x_1=0##.

Hence, ##\sum_{n=2}^\infty \lambda^{-n}=1##. Then we can solve for ##\lambda## to get the point spectrum of ##T##
 
Last edited:

Answers and Replies

  • #2
WWGD
Science Advisor
Gold Member
2019 Award
5,163
2,398
<mod note: moved to homework>

Calculate the spectrum of the linear operator ##T## on ##B(\ell^1)##.
$$T(x_1,x_2,x_3,\dots)=(\sum_{n=2}^\infty x_n, x_1, x_2, x_3, \dots)$$


I think the way to do it is to find the point spectrums of ##T## and its adjoint ##T^*##. But I don't know how to calculate them.

Some progress:

Let ##T(x_n)=\lambda (x_n)##. Then we have ##x_n=\lambda^{-n+1}x_1## for ##n \geq 2## and ##x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}##. Then
$$x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}=\frac{\sum_{n=2}^\infty \lambda^{-n+1}x_1}{\lambda}=\sum_{n=2}^\infty \lambda^{-n}x_1$$
Therefore, ##(1-\sum_{n=2}^\infty \lambda^{-n})x_1=0##.

Hence, ##\sum_{n=2}^\infty \lambda^{-n}=1##. Then we can solve for ##\lambda## to get the point spectrum of ##T##
If ##|1/\lambda|<1## then this sum would converge to ## \frac {2}{1- \lambda} ##
 
Last edited:

Related Threads for: Calculate the spectrum of a linear operator

  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
12
Views
2K
Replies
0
Views
1K
Replies
5
Views
897
Replies
3
Views
4K
Replies
1
Views
340
  • Last Post
Replies
20
Views
3K
  • Last Post
Replies
1
Views
1K
Top