Let $H_1, H_2$ be Hilbert spaces and $T:H_1\to H_2$.
We say that $T$ is unitary if it preserves the inner product and unto.
- Show that the following claims are equivalent:
A. $T$ is unitary.
B. T copies every orthonormal basis of H to an orthonormal basis of H.
C. T is injective and there exist an orthonormal basis of $H_1$ such that $T$ copies to an orthonormal basis of $H_2$.
D. T is invertible and $T^{-1}=T^*$
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Show that $T$ is unitary iff $T^*$ is.
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if $H_1=H_2$.
Show that $T$ is unitary iff $T$ preserves the inner product and is normal.
For 1:
A=>B:
Let T be an unitary operator, i.e it preserves the inner product.
Let $(u_a)\in H$ be a hilbert basis of H (for every hilbert space there's an orthonormal basis), then
$<Tu_a,Tu_b>=<u_a,u_b>=0$ for all $a\neq b$ and
$<Tu_a,Tu_a>=<u_a,u_a>=1$.
Thus, T copies the orthonormal basis $(u_a)$ to an orthonormal basis $(Tu_a)$.
D=>A:
Let T be invertible and $T^*=T^{-1}$.
Then, $<Tx,Ty>=<x,T^*Ty>=<x,y>$ so T is unitary by definition.
For 2:
Using that A <=>D
T is unitary iff it is invertible and $T^{-1}=T^*$.
If T is unitary then $<T^*x,T^*y>=<x,TT^*y>=<x,Iy>=<x,y>$, we get that $T^*$ is unitry.
In the second way, if $T^*$ is unitary, then
$<Tx,Ty>=<x,T^*Ty>=<x,y>$, so T is unitary.
For 3:
If T is unitary then T preservess the inner product (by def), and using A <=> D,
$T^*T=T^{-1}T=I$
$TT^*=TT^{-1}=I$
Therefore T is normal.
For the inverse, let T be norml and preseres the inner product,
$<Tx,Ty>=<x,T^*Ty>=<x,TT^*y>=<x,y>$, so $T^*T=TT^*=I$, so T is invertible and $T^*=T^{-1}$, thus T is unitary (by A <=>D).
Is what i did fine?
I did not get the idea in the rest => in 1, so will appreciate your help.
Best Answer
For 1b->1c
Well, first I hope you believe $H_1$ has some orthonormal basis $\{e_i\}_{i\in I}$ (it's a matter of axiom of choice). Then since in 1b you assume $T$ maps any orthnormal basis of $H_1$ to an orthonormal basis of $H_2$, then specifically $\{T(e_i)\}_{i\in I}$ must be an orthonormal basis of $H_2$. Now, for $i, j$ we have $\langle e_i, T^{*} T(e_j) \rangle = \langle T(e_i), T (e_j) \rangle = \delta^i_j$ which proves that $T^{*} T(e_i) = e_i$ for all $i \in I$. To show injectivity, assume $T(x) = 0$. Then $\langle x, e_i \rangle =\langle x, T^{*}T(e_i) \rangle = \langle T(x), T(e_i) \rangle = 0$ for all $i \in I$. Since $\{e_i\}_{i\in I}$ is an orthonormal basis this implies $x = 0$.
Note: I did not justify why $T^{*}$ exists. The question doesn't state what we know about $T$. I assume it's linear and defined on all of $H_1$. Even if we don't assume that $T$ is bounded, from the fact that it maps any orthonormal basis to an orthonormal basis and since any unit vector is part of an orthonormal basis, $T$ maps the unit ball into the unit ball so is bounded.