Is unitary $T^2$ if $T$ is unitary?

Question

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

Answer 1

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