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

Since we know that $\displaystyle T$ uniquely determines the adjoint $\displaystyle T^*$ and by the fact that $\displaystyle T^*$ is an operator in $\displaystyle H$,hence,we can claim that the mapping from $\displaystyle T\rightarrow T^*$ 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, $\displaystyle T_1*$ and $\displaystyle T_*$. Then we must have [tex]<Tu, v>= <u, T_1*v>[tex] and $\displaystyle <Tu, v>= <u, T_2*v>$ so that $\displaystyle <u, T_1*v>= <u, T_2*v>$ for all u and v in H. Can you show that $\displaystyle T_1*= T_2*$ from that?