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

[Sepen]

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 T^{*} 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!

 

[Dick]

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 T^{*} 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.

 

[Sepen]

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?

 

[micromass]

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.

 

[Sepen]

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.

 

[micromass]

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?

 

[Sepen]

Ah it would be Im(T) and it would be Im(T*) so thus they have to be equal?

 

[micromass]

Ah it would be Im(T) and it would be Im(T*) so thus they have to be equal?

Right!