Image of Normal Operator

  • Thread starter Sepen
  • Start date
  • #1
4
0

Homework Statement


Show that if T is a normal operator on a finite dimensional vector space than it has the same image as its adjoint.


Homework Equations


N/A


The Attempt at a Solution


I have been able to show that both T and [itex]T^{*}[/itex] have the same kernel. Thus, by using the finite dimension property and the rank nullity theoremit just suffices to show containment one way.

However, if you suppose that a vector v is in the Im(T) I haven't been able to find some representative w such that T*(w) = v.

Does anyone have any idea how to proceed?

Thanks!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619

Homework Statement


Show that if T is a normal operator on a finite dimensional vector space than it has the same image as its adjoint.


Homework Equations


N/A


The Attempt at a Solution


I have been able to show that both T and [itex]T^{*}[/itex] have the same kernel. Thus, by using the finite dimension property and the rank nullity theoremit just suffices to show containment one way.

However, if you suppose that a vector v is in the Im(T) I haven't been able to find some representative w such that T*(w) = v.

Does anyone have any idea how to proceed?

Thanks!
You should also be able to show that Im(T) is orthogonal to Ker(T). Use that.
 
  • #3
4
0
Is that necessarily true? And if it is (I know that you can show that Im(T) is orthogonal to Ker(T*) and vice versa), but how does that specifically help?
 
  • #4
22,089
3,293
Is that necessarily true? And if it is (I know that you can show that Im(T) is orthogonal to Ker(T*) and vice versa), but how does that specifically help?
The idea is to show both ##\textrm{im}(T)## and ##\textrm{im}(T^*)## to be orthogonal to ##\textrm{ker}(T)##. For former you will need normality, the latter should be easy.
 
  • Like
Likes 1 person
  • #5
4
0
I understand that but does that necessarily imply that they have to be equal? I recognize that they have to have the same dimensions in that case, but that doesn't get me anywhere.
 
  • #6
22,089
3,293
I understand that but does that necessarily imply that they have to be equal? I recognize that they have to have the same dimensions in that case, but that doesn't get me anywhere.
If you accept the fact in my last post to be true for now, then what would ##\textrm{ker}(T)^\bot## be?
 
  • #7
4
0
Ah it would be Im(T) and it would be Im(T*) so thus they have to be equal?
 
  • #8
22,089
3,293
Ah it would be Im(T) and it would be Im(T*) so thus they have to be equal?
Right!
 

Related Threads on Image of Normal Operator

Replies
1
Views
894
Replies
2
Views
7K
Replies
5
Views
930
Replies
1
Views
572
  • Last Post
Replies
11
Views
1K
Replies
0
Views
5K
  • Last Post
Replies
3
Views
506
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
3K
Top