# Math Help - Hilbert Space

1. ## Hilbert Space

Show that an operator $T$ on a Hilbert Space is unitary iff $T(\{e_i\})$ is a complete orthonormal set whenever $\{e_i\}$ is.

2. What is your definition of unitary operator?

3. Unitary here means an isomorphism in Hilbert space(linear,one to one and onto)

4. Originally Posted by problem
Unitary here means an isomorphism in Hilbert space(linear,one to one and onto)
I'm going to assume that it's also an isometry (otherwise take $\mathbb{R} ^2$ and $T(e_1)=e_1$ and $T(e_2)=e_1+e_2)$.

$\Rightarrow$ ) Let $T:H_1 \rightarrow H_2$ be an isometric isomorphism then (by the polarization identity) $T$ preserves the interior product, and so if $(e_a)_{a \in A}$ is an orthonormal basis for $H_1$ then $(T(e_a))_{a\in A}$ is also an orthonormal set in $H_2$. Take $y=T(x)$ since $(e_a)$ is a basis we have $(x_n)_{n \in \mathbb{N} }$ $\subset$ $span((e_a)_{a \in A})$ such that $x_n \rightarrow x$ then $(T(x_n))_{n\in \mathbb{N} } \subset span((T(e_a))_{a\in A})$ and $T(x_n) \rightarrow T(x)=y$ (Since T, being an isometry, belongs to $\mathcal{B} (H_1,H_2)$). Since $T$ is onto, we have that $(T(e_a))_{a\in A}$ is a basis for $H_2$.

$\Leftarrow$ ) Let $(e_a)_{a\in A}$ be an orthonormal basis in $H_1$ and $(T(e_a)){a\in A}$ one for $H_2$. Let $L: H_1 \rightarrow H_2$ be such that $L(e_a)=T(e_a)=b_a$ for all $a \in A$ and if $x=\sum_{a\in A} \ e_a \in H_1$ $L(x):= \sum_{a\in A} \ b_a$. It follows from Bessel's inequality that $L$ is well defined and it's clearly linear and one-one. If $y= \sum_{a\in A} \ b_a \in H_2$ we define $x= \sum_{a\in A} \ e_a \in H_1$ (again by Bessel) and by definition $L(x)=y$ and so $L$ is onto.

If $y=L(x)$ then $\sum_{a\in A} \ b_a = \sum_{a\in A} \ b_a$ then $=$ for all $a\in A$ and we have (Note that this previous argument proves injectivity):

$\Vert L(x) \Vert ^2 = \Vert y \Vert ^2 = \sum_{a\in A} \ \vert \vert ^2 = \sum_{a\in A} \ \vert \vert ^2 =\Vert x \Vert ^2$ and so $L$ is an isometry.

Since $L$ and $T$ coincide on a l.i set, they are equal on the whole linear span and (assuming $T$ $\in \mathcal{B} (H_1,H_2)$ which I think is a necessary hypothesis) hence they agree on the closure of the linear span which is $H_2$. So $L \equiv T$ and this finishes the proof.