Is unitary $T^2$ if $T$ is unitary? If $T$ is unitary then $\|Tx\|=\|x\|$ then $T^2=T T$, then  $\|Tx\| \|Tx\| =\|x\| \|x\|$
$\|Tx|\|^2 =\|x\|^2|$ but $T^2$ is unitary because $T^2=T T$ and $T$ was unitary then $\|Tx\|=\|x\|$
Is true this?
Thank for all :D
Good day everyone 
 A: Your argument is not valid. You have never really attempted to prove $\|T^2x\|=\|x\|$.
With your definition of unitary which is actually called isometry, it is trivial since if $T$ is isometric then:
$$
\|T^2x\|=\|T(Tx)\|=\|Tx\|=\|x\|.
$$
So $T^2$ is isometric.
Note: $T:H\longrightarrow H$ ($H$ a Hilbert space) unitary means by definition
$$
T^*T=TT^*=I
$$
where $T^*$ is the adjoint.
In finite dimension, this is equivalent (proof follows at the end) to 
$$
\|Tx\|=\|x\|
$$
for all $x$. But this is not true in infinite dimension.
Now if $T$ is unitary, we have
$$
(T^2)^*T^2=T^*T^*TT=I=TTT^*T^*=T^2(T^2)^*.
$$
So indeed $T^2$ is unitary.
Proof of the equivalence: $T$ unitary $\Leftrightarrow$ $\|Tx\|=\|x\|$ for all $x\in H$ (one says $T$ is an isometry) in finite dimension.
First if $T$ is unitary, then
$$
\|Tx\|^2=(Tx,Tx)=(T^*Tx,x)=(x,x)=\|x\|^2
$$
so $T$ is an isometry.
Now if $T$ is an isometry, we will prove that it is unitary in the case of a real finite-dimensional Hilbert space.
$$
(T^*Tx,y)=(Tx,Ty)=\frac{1}{4}(\|T(x+y)\|^2-\|T(x-y)\|^2)
$$
$$
=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2)=(x,y).
$$
Since this is true for all $y$, this shows
$$
T^*Tx=x
$$
for all $x$, hence $T^*T=I$.
Now by the rank-nullity theorem, it follows that also
$$
TT^*=I.
$$
So $T$ is indeed unitary.
For the complex case, the proof is similat but involves a slightly more complicated polarization identity: http://en.wikipedia.org/wiki/Polarization_identity
