Is the Image of a Normal Operator the Same as Its Adjoint?

In summary: So in summary, if you accept the fact that the images of T and T^{*} are the same, then the kernels are also the same.
  • #1
Sepen
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!
 
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  • #2
Sepen said:

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
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
Sepen said:
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.
 
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  • #5
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
Sepen said:
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
Ah it would be Im(T) and it would be Im(T*) so thus they have to be equal?
 
  • #8
Sepen said:
Ah it would be Im(T) and it would be Im(T*) so thus they have to be equal?

Right!
 

1. What is an image of a normal operator?

An image of a normal operator is the set of all vectors that can be obtained by applying the operator to any vector in its domain.

2. How is the image of a normal operator related to its eigenvectors?

The image of a normal operator is spanned by its eigenvectors. This means that any vector in the image can be expressed as a linear combination of the operator's eigenvectors.

3. Can the image of a normal operator be the same as its domain?

Yes, the image of a normal operator can be equal to its domain if the operator is surjective, meaning that every vector in its range is mapped to by at least one vector in its domain.

4. How can the image of a normal operator be used to find its eigenvalues?

The eigenvalues of a normal operator can be found by examining the spectrum of the operator, which is the set of all values that the operator can map an eigenvector to. The image of the operator can help determine the spectrum and thus the eigenvalues.

5. Is the image of a normal operator always a subspace?

Yes, the image of a normal operator is always a subspace of its range. This is because it contains the zero vector, is closed under addition and scalar multiplication, and is a subset of the range.

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