TITLE: Adjoint Operator and Inverse
QUESTION [10 upvotes]: I am solving the following question and I am not really sure about the way I approach
Question 1: Assume that $T:U\rightarrow U$ is invertible map. Prove that $(T^*)^{-1}=(T^{-1})^*$
Here is my answer:
Notice that $\langle Tv,u\rangle = \langle v, T^*u \rangle$ for all $u,v\in U$
Then  $\langle T^*(T^{-1})^*v,u\rangle = \langle (T^{-1})^* v, Tu \rangle = \langle v,T^{-1}(Tu)\rangle = \langle u,v\rangle$
$T^* (T^{-1})^* = 1 $ so $(T^{-1})^* $ is the inverse of $T^*$. Hence, $(T^{-1})^*=(T^*)^{-1}$. 
Question 2: Prove that for every operator $T$ the operators $T^*T$ and $T T^*$ are self-adjoint
Assumption: I have to prove $T^*T = T$ and $T T^* = T $
Suppose $T\in L (V,W)$. Fix $ w\in W$ .
$\langle Tv,u\rangle = \langle v, T^*u \rangle$ for all $u,v\in U$
Then  $\langle TT^*v,u\rangle = \langle ((Tv)T^*,u \rangle = \langle Tv,Tu\rangle $ ??
$\langle T^*Tv,u\rangle = \langle (T^* (Tv),u \rangle = \langle Tv,Tu\rangle$ ???
I am not sure whether I am right or wrong for the above problem so please help me correct the above problem and understand adjoint and self-adjoint.

REPLY [11 votes]: Another way to solve the Question 1 is: 
We have $ (T^*)^{-1}T^* = 1 $ then 
 $ ((T^*)^{-1}T^*)^* = 1^*$ 
Since $1^* =1 $ and $(AB) = B^*A^*$ for every operators A and B then
$ T^{**}((T^*)^{-1})^* = 1 $ 
But $T^{**} = T$ so $((T^*)^{-1})^* = T^{-1}$ and then $(T^*)^{-1}=(T^{-1})^*$