Unitary operators in Hilbert spaces Let $T$ be a bounded linear operator on a Hilbert space $H$. I have to show that the following are equivalent:
(i) $T$ is unitary
(ii) For every orthonormal basis $\{u_{\alpha}:\alpha\in \Lambda\}$, $\{T(u_{\alpha}):\alpha\in \Lambda\}$ is an orthonormal basis.
(iii) For some orthonormal basis $\{u_{\alpha}:\alpha\in \Lambda\}$, $\{T(u_{\alpha}):\alpha\in \Lambda\}$ is an orthonormal basis.
I have proved that (i)$\implies$ (ii). Also (ii)$\implies$ (iii) is obvious. 
How to show that (iii)$\implies$ (i)? Please suggest anything?
 A: It is enough to compute $T^*T(u_\alpha)$ for all $\alpha$, since the operator is determined by its values at a basis.
We can compute the coordinates of this vector in the basis $u_\alpha$. So, we do
$$T^*T(u_\alpha)\cdot u_{\beta}=T(u_\alpha)\cdot T(u_\beta)=\delta_{\alpha,\beta}$$
Therefore $T^*T(u_\alpha)=\sum_\beta(T^*T(u_\alpha)\cdot u_{\beta})u_{\beta}= u_{\alpha}$, from where we get that $T^*T$ is the identity.
This is the same computation we did above, but hiding things a little.
A: For $(iii) \Rightarrow (i)$, you want to show that
$$
\langle Tx,Ty\rangle = \langle x,y\rangle \qquad (\ast)
$$
for all $x,y\in H$. First note that $(\ast)$ is true if $x,y\in S := \{u_{\alpha}\}$. By sesqui-linearity, $(\ast)$ is true if $x,y\in \text{span}(S)$. However, $\text{span}(S)$ is dense in $H$, so if $x\in H,v\in \text{span}(S)$ and $\epsilon > 0$, then $\exists u\in \text{span}(S)$ such that $\|x-u\| < \epsilon$. Hence,
$$
|\langle Tx,Tv\rangle - \langle x,v\rangle| \leq |\langle Tx,Tv\rangle - \langle Tu,Tv\rangle| + |\langle Tu,Tv\rangle - \langle u,v\rangle| + |\langle u,v\rangle - \langle x,v\rangle|
$$
Since $T$ is bounded, you can make the right hand side as small as you want, proving that
$$
\langle Tx,Tv\rangle = \langle x,v\rangle
$$
Similarly, one can replace $v$ by an arbitrary $y\in H$, proving $(\ast)$ for all $x,y\in H$.

Edit: My apologies, in addition to proving $(\ast)$, one also needs to prove that $T$ is surjective. To see this, one simply notes that $R(T)$, the range of $T$ contains an orthonormal basis for $H$, so if $x\in H$, then write
$$
x = \sum a_{\alpha} T(u_{\alpha}) \qquad (\dagger)
$$
so that the series $y = \sum a_{\alpha} u_{\alpha}$ converges because $(\dagger)$ converges. Now $y\in H$ and $x = T(y)$ since $T$ is continuous. Hence, $R(T) = H$.
