Prove that an operator preserves length if and only if it is orthogonal

Question

I want to show the following:

$T$ is orthogonal if and only if it preserves length, $i.e $ $||\alpha||=||T(\alpha)||, \forall$ $\alpha$ $\in$ $V$.

Answer 1

From $\langle \alpha,T^*T\alpha\rangle=\langle \alpha,\alpha\rangle$ you get $\langle \alpha,(I-T^*T)\alpha\rangle=0$ for all $\alpha$. Now using polarization you get $\langle \beta,(I-T^*T)\alpha\rangle=0$ for all $\alpha,\beta$. In particular you can take $\beta=(I-T^*T)\alpha$, to obtain $\|(I-T^*T)\alpha\|^2=0$.

Now comes the time to use some other hypothesis that you haven't posted. Because in an arbitrary Hilbert space, there are operators $T$ with $T^*T=I$ but $TT^*\ne I$.

Now if $T^*T=I$ and $T$ is surjective (which comes for free if $V$ is finite-dimensional), you get that $T$ is injective and surjective. Then $$ T^{-1}=IT^{-1}=T^*TT^{-1}=T^*, $$ and we are done.