Q:

$\langle Tx,Ty\rangle=\langle x,y\rangle$ for every $x,y \in H$. implies $T^*=T^{-1}$

Let $T:H \to H$ be a linear operator. $T$ is called isometry if $\|T(x)\|=\|x\|$. Prove that the following are equivalent:

T is an isometry
$\langle Tx,Ty\rangle=\langle x,y\rangle$ for every $x,y \in H$.
$T^*=T^{-1}$
the columns of the matrix that represent $T$ are orthogonal vectors of $\mathbb{R}^n$.

(1) $\rightarrow$ (2): Assume that T is an isometry. We need to show $\langle Tx,Ty \rangle = \langle x,y\rangle$. Let $x,y \in H$. We know that T is called isometry if $\| T(x) \| =\| x\|$. We can use the definition again and say $\| T(y) \| =\| y\|$.
\begin{equation*}
\begin{split}
4\langle Tx,Ty\rangle & = \|Tx+Ty\|^2-\|Tx-Ty\|^2 \\
& = \|T(x+y)\|^2-\|T(x-y)\|^2 \\
& = \|x+y\|^2-\|x-y\|^2 \\
& = 4<x,y> \\
\end{split}
\end{equation*}
Hence $\langle Tx,Ty\rangle=\langle x,y\rangle$.
Done
(2) $\rightarrow$ (3): Assume that $\langle Tx,Ty\rangle=\langle x,y\rangle$. We need to show $T^*=T^{-1}$.
How do I do the proof?
(3) $\rightarrow$ (4): Assume that $T^*=T^{-1}$. We need to show the columns of the matrix that represent T are orthogonal vectors of $\mathbb{R}^n$. If $T^{-1}=T^T$, then $T^TT=I$. Hence each column has unit length and is perpendicular to every other column. (i.e. That is an orthogonal matrix).
Done
(4) $\rightarrow$ (1): Assume the colum of the matrix that represent $T$ are orthogonal vectors of $\mathbb{R}^n$. We need to show that T is an isometry.
How do I do the proof?

A:

Ok. I see my comment won't do to answer your questions. To see (2) implies (3) note that, $\langle Tx,Ty\rangle= \langle x,T^*Ty\rangle =\langle T^*Tx,y\rangle =\langle x,y\rangle$. Since the equalities hold for all $x,y\in H$, by uniqueness of the adjoint operator $T^*T=TT^*= I$.
Note that (3) implies (2) and (2) implies (1) are trivial so you can prove (4) iff (3) (remember that the associated matrix of the adjoint operator is the conjugate transpose of the  associated matrix).

