Let $H_1, H_2$ be Hilbert spaces and $T\in B(H_1,H_2)$.
Show that the following conditions are equivalent:
A. $T$ is isometry.
B. $T^*T=I_{H_1}$.
C. $T$ preserves the inner product.
D. $T$ "copies/sends" every orthonormal set in $H_1$ to orthonormal set in $H_2$.
E. $T$ is injective and there exist a basis to $H_1$ that $T$ "copies" to an orthonormal set in $H_2$.
A $\Rightarrow$ B is already shown here Show that an isometric linear operator $T:H\to H$ satisfies $T^* T=I$, where $I$ is the identity operator on $H$.
B $\Rightarrow$ C:
Assume that $T^*T=I$ then, by definition
$\forall x,y\in H_1$
$\langle Tx,Ty\rangle = \langle T^*Tx,y\rangle = \langle x,y\rangle$.
C $\Rightarrow$ D:
We assume that $T$ preserves inner product.
Let $\{u_a\}_{a\in A} \in H_1$ be an orthonormal set.
For every $a\in A$ , $\|Tu_a\|=\|u_a\|$ (by assume).
And for every $a\neq b \in A$
$\langle Tu_a,Tu_b\rangle =\langle u_a,u_b\rangle =0$.
So $\{T(u_a)\}_{a\in A}$ is an orthonormal set in $H_2$.
How to do D $\Rightarrow$ E and E $\Rightarrow$ A ?
And is that enough for showing that the claims are equivalent?
Best Answer
D$\Rightarrow$E:
Every Hilbert space has an orthonormal basis, so let $\ \big\{u_a\big\}_{a\in A}\ $ be such a basis of $\ H_1\ $. If $\ \text{D}\ $ holds, then $\ \big\{Tu_a\big\}_{a\in A}\ $ must be an orthonormal set in $\ H_2\ $. Now let $\ x=\sum_\limits{a\in A}x_au_a\ $, $\ y=\sum_\limits{a\in A}y_au_a\ $ be arbitrary members of $\ H_1\ $ with $\ Tx=Ty\ $. Then for any $\ b\in A\ $, \begin{align} \ \big\langle Tx,Tu_b\big\rangle&=\big\langle T\sum_\limits{a\in A}x_au_a,Tu_b\big\rangle\\ &=\big\langle \sum_\limits{a\in A}x_aTu_a,Tu_b\big\rangle\\ &=x_b \end{align} by the orthonormality of $\ \big\{Tu_a\big\}_{a\in A}\ $. Likewise, $\ \big\langle Ty,Tu_b\big\rangle=y_b\ $. Thus, since $\ Tx=Ty\ $, $\ x_b=y_b\ $ for all $\ b\in A\ $ and hence $\ x=y\ $. Therefore $\ T\ $ is injective.
E$\Rightarrow$A:
Suppose $\ \text{E}\ $ holds. Then there exists a basis $\ \big\{u_a\big\}_{a\in A}\ $ of $\ H_1\ $ such that $\ \big\{Tu_a\big\}_{a\in A}\ $ is an orthonormal set in $\ H_2\ $. If $\ x=\sum_\limits{a\in A}x_au_a\ $ is an arbitrary member of $\ H_1\ $, then $\ \|x\|^2=\sum_\limits{a\in A}|x_a|^2\ $ and \begin{align} \|Tx\|^2&=\langle Tx,Tx\rangle\\ &=\Big\langle\sum_\limits{a\in A}x_aTu_a,\sum_\limits{a\in A}x_aTu_a\Big\rangle\\ &=\sum_\limits{a\in A}|x_a|^2\\ &=\|x\|^2\ . \end{align} Therefore $\ T\ $ is an isometry.