Showing properties of unitary operator

Question

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.

1. 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 $T^{-1}=T^*$

2. Show that $T$ is unitary iff $T^*$ is.

3. 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.

Answer 1

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.

Answer 2

So just for completion 1c->1d
First assume that $T$ is bounded. Then $T^*$ is defined and bounded. Letting $\{e_i\}_{i\in I}$ be an orthonormal basis mapped by $T$ to some orthnormal basis, and proceeding as above shows that $T^* T(e_i) = e_i$ for all $i\in I$ and this proves that $T^* T = \text{id}_{H_1}$. Note that the assumed injectivity is not even used, but follows from this conclusion. Swapping the roles of the orthonormal bases $\{e_i\}_{i\in I}$ and $\{T(e_i)\}_{i\in I}$ we conclude that $T T^*=\text{id}_{H_2}$ which completes the proof of 1d.

Now what if we don't assume that $T$ is bounded? Then it's flat out wrong that $T$ has to be unitary. Let $H$ be a separable infinite dimensional Hilbert space with orthonormal basis $\mathcal{A}\subset H$. Let $\mathcal{B}\subset H$ be a family of vectors completing $\mathcal{A}$ to a Hamel basis of $H$. In other words $\mathcal{A}\cap\mathcal{B}=\emptyset$ and $\mathcal{A}\cup\mathcal{B}$ is a linear basis of $H$. Note that $\mathcal{A}$ is countable but $|\mathcal{B}|=2^{2^{\aleph_0}}$ Let $y$ be any element of $\mathcal{B}$ and define an operator $T:H\rightarrow H$ by setting $$T(y)=2y$$ and $$T(x)=x$$ for all $x\in\mathcal{A}\cup\mathcal{B}$ such that $x\neq y$. Then

1. $T$ can be extended uniquely to a linear operator with domain $H$.
2. $T$ is injective and in fact bijective because it maps a linear basis to a linear basis.
3. $T$ maps the orthonormal basis $\mathcal{A}$ to an orthnormal basis (itself.)
4. $T$ is not unitary because it does not preserve the norm of $y$.

It follows as a conclustion that $T$ cannot be bounded because if $T$ were bounded then the proof that it is unitary would apply. This means for instance that $T$ maps unit vectors to vectors of arbitrarily large norm even though this is not evident from its definition.
So unlike the other criteria, in 1c the boundedness of $T$ needs to be assumed, and the assumption of injectivity on the other hand is unnecessary.