Is the Image of a Normal Operator Equal to Its Adjoint's Image?

[kingwinner]

Q) Let V be an inner product space and T:V->V a linear operator. Prove that if T is normal, then T and T* have the same image. (i.e. imT=imT*)

My Attempt:
<T(v),T(v)>
=<T*T(v),v>
=<TT*(v),v>
=<T*(v),T*(v)>

=>||T(v)|| = || T*(v)||

But this doesn't seem to help...

Thanks!

 

[HallsofIvy]

You are trying to prove two sets (Im(T) and Im(T*)) are the same. Start by assuming v is in Im(T). Then there exist u such that v= T(u). You want to show that v= T*(w) for some w in V.

 

[kingwinner]

You are trying to prove two sets (Im(T) and Im(T*)) are the same. Start by assuming v is in Im(T). Then there exist u such that v= T(u). You want to show that v= T*(w) for some w in V.

I understand the definitions, but I have no idea how to prove this.
Can you give me more hints, please?

 

[kingwinner]

My attempt:
v=T(u) for some u E V
=> T*(v)=T*T(u)
=> T*(v)=TT*(u) since T is normal

And now I am stuck, how can I prove that v=T*(w) for some w E V?

 

[morphism]

This seems false unless you're given some other condition on V. Is it finite-dimensional or complete? If so, then from your observation that ||Tv||=||T*v|| we can conclude immediately that kerT=kerT*; on the other hand, it's easy to see that the orthogonal complement of imT* is kerT (i.e. (\text{im} T^*)^{\perp} = \ker T). Now use finite-dimensionality to place things together (in case of completeness, use direct sum decompositions).

 

[kingwinner]

We are assuming finite-dimensional vector space.

v=T(u) for some u E V
=> T*(v)=T*T(u)
=> T*(v)=TT*(u) since T is normal
I have no idea how to proceed from here...

How is it possible to prove without using (\text{im} T^*)^{\perp} = \ker T?

 

[morphism]

Why don't you want to use it?