# Image of Normal Operator

1. Mar 16, 2014

### Sepen

1. The problem statement, all variables and given/known data
Show that if T is a normal operator on a finite dimensional vector space than it has the same image as its adjoint.

2. Relevant equations
N/A

3. 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!

2. Mar 16, 2014

### Dick

You should also be able to show that Im(T) is orthogonal to Ker(T). Use that.

3. Mar 16, 2014

### 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?

4. Mar 16, 2014

### micromass

Staff Emeritus
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.

5. Mar 16, 2014

### 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.

6. Mar 16, 2014

### micromass

Staff Emeritus
If you accept the fact in my last post to be true for now, then what would $\textrm{ker}(T)^\bot$ be?

7. Mar 16, 2014

### Sepen

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

8. Mar 16, 2014

### micromass

Staff Emeritus
Right!