# Kernel of the adjoint of a linear operator

## Homework Statement

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 kernel (T* is the adjoint of T).

## The Attempt at a Solution

Let us assume x is in the kernel of T. Then, TT*x =T*Tx = T*0= 0. So T*x is in the kernel of T but this doesn't mean T*x = 0 and I get stuck here. Any help would be greatly appreciated.

## Answers and Replies

Dick
Science Advisor
Homework Helper
You aren't using that T* is the adjoint. Look at x*(T*T)x and x*(TT*)x.

I think I found out how to do this. Are my steps right?
Let x be an element in the kernel of T. Since TT*x = T*Tx = T*0=0, (TT*x,x) = 0 where (,) indicates the inner product.
Then, (TT*x,x) = (T*x,T*x) = ||T*x||^2 = 0 so T*x = 0 and x is in the kernel of T*. The other way can be done similarly.

BTW, what does "x*" indicate in x*(T*T)x? I don't understand...

Dick
Science Advisor
Homework Helper
x* is the adjoint of the vector x. So x*(TT*)x is (x,(TT*)x). Yes, that's it. Show (Tx,Tx)=(T*x,T*x)=||Tx||^2=||T*x||^2.

O.K., thank you.