1. ## Adjoint Operation

Let $H$ be a Hilbert Space and $T$ be an operator in $H$.
Prove that the mapping $T\rightarrow T^*$ is an one to one and onto mapping of $B(H)$ into itself,where $T^*$ is the adjoint of $T$ and $B(H)$ is the set of bounded operator on $H$.

Since we know that $T$ uniquely determines the adjoint $T^*$ and by the fact that $T^*$ is an operator in $H$,hence,we can claim that the mapping from $T\rightarrow T^*$ is one to one and onto.

Is this the right way to prove?Can anyone help?

2. What you need is not that "T uniquely determines T*" but that "T* uniquely determines T". Otherwise it is possible that two different Ts determine the same T* and the mapping would not be one to one.

3. HallsofIvy,can you give me some hints in proving the one to one?

4. Use the definition of "adjoint"! If T* is the adjoint of T, the <Tu, v>= <u, T*v> for all u and v in H. < , > is the inner product in H. Suppose that, for a given T, there were two adjoints, $T_1*$ and $T_*$. Then we must have [tex]<Tu, v>= <u, T_1*v>[tex] and $= $ so that $= $ for all u and v in H. Can you show that $T_1*= T_2*$ from that?