Prove that if $\Vert{Qx}\Vert = \Vert{x}\Vert$ then $Q^{-1} = Q^{t}$ The case where you have to prove $(Q^{t}Q)_{ii} = 1 \ \forall \ 1 \le i \le n$ is simple (you can choose $e_{i}$ as your $x$) but I am not able to show that  $(Q^{t}Q)_{ij} = 0 \ \forall \ 1 \le i, j \le n, \ i \neq j$. Any help will be appreciated.
 A: Note that
$$
\left\|Q\mathbf{x}\right\|=\left\|\mathbf{x}\right\|\iff\left\|Q\mathbf{x}\right\|^2=\left\|\mathbf{x}\right\|^2\iff\mathbf{x}^{\top}Q^{\top}Q\mathbf{x}=\mathbf{x}^{\top}\mathbf{x}
$$
holds for all $\mathbf{x}\in\mathbb{R}^n$.
For one thing, take $\mathbf{x}=\mathbf{e}_j$, and the equality implies that
$$
\mathbf{e}_j^{\top}Q^{\top}Q\mathbf{e}_j=\mathbf{e}_j^{\top}\mathbf{e}_j=1
$$
holds for all $j=1,2,\cdots,n$.
For another, take $\mathbf{x}=\mathbf{e}_j+\mathbf{e}_k$, and the equality implies that
$$
\mathbf{e}_j^{\top}Q^{\top}Q\mathbf{e}_j+\mathbf{e}_j^{\top}Q^{\top}Q\mathbf{e}_k+\mathbf{e}_k^{\top}Q^{\top}Q\mathbf{e}_j+\mathbf{e}_k^{\top}Q^{\top}Q\mathbf{e}_k=\mathbf{e}_j^{\top}\mathbf{e}_j+\mathbf{e}_j^{\top}\mathbf{e}_k+\mathbf{e}_k^{\top}\mathbf{e}_j+\mathbf{e}_k^{\top}\mathbf{e}_k,
$$
or using the last equality we just figured out,
$$
1+\mathbf{e}_j^{\top}Q^{\top}Q\mathbf{e}_k+\mathbf{e}_k^{\top}Q^{\top}Q\mathbf{e}_j+1=1+\mathbf{e}_j^{\top}\mathbf{e}_k+\mathbf{e}_k^{\top}\mathbf{e}_j+1.
$$
This reduces to
$$
\mathbf{e}_j^{\top}Q^{\top}Q\mathbf{e}_k+\mathbf{e}_k^{\top}Q^{\top}Q\mathbf{e}_j=\mathbf{e}_j^{\top}\mathbf{e}_k+\mathbf{e}_k^{\top}\mathbf{e}_j.
$$
Note that
$$
\mathbf{e}_j^{\top}\mathbf{e}_k\in\mathbb{R}\Longrightarrow\mathbf{e}_j^{\top}\mathbf{e}_k=\left(\mathbf{e}_j^{\top}\mathbf{e}_k\right)^{\top}=\mathbf{e}_k^{\top}\mathbf{e}_j,
$$
and that, likewise,
$$
\mathbf{e}_j^{\top}Q^{\top}Q\mathbf{e}_k=\mathbf{e}_k^{\top}Q^{\top}Q\mathbf{e}_j.
$$
Consequently, we conclude that
$$
\mathbf{e}_j^{\top}Q^{\top}Q\mathbf{e}_k=\mathbf{e}_j^{\top}\mathbf{e}_k
$$
holds for all $j,k=1,2,\cdots,n$.
Finally, recall that
$$
\mathbf{e}_j^{\top}A\mathbf{e}_k
$$
returns exactly the $\left(j,k\right)$-th entry of the square matrix $A$. Thus
$$
\left(j,k\right)\text{-th entry of }Q^{\top}Q=\mathbf{e}_j^{\top}\mathbf{e}_k=\delta_{jk}.
$$
This immediately leads to
$$
Q^{\top}Q=I_n\iff Q^{\top}=Q^{-1}.
$$
A: \begin{align*}
\left<Q^{\ast}Qx,x\right>&=\left<Qx,Qx\right>\\
&=\|Qx\|^{2}\\
&=\|x\|^{2}\\
&=\left<x,x\right>\\
&=\left<Ix,x\right>,
\end{align*}
so $Q^{\ast}Q=I$. Now use the fact that $\|Q^{\ast}x\|=\|x\|$ and conclude that $QQ^{\ast}=I$.
A: $(Qx)^TQx = x^TQ^TQx = x^Tx \ \forall x \rightarrow \ Q^TQ = I \rightarrow Q^{-1} = Q^T$ (the fact that $Q$ is invertible can be concluded from: $rank(Q^TQ) = rank(Q) = rank(I)$ ) 
A: If
$\Vert Qx \Vert = \Vert x \Vert, \tag 1$
then
$\langle Qx, Qx \rangle = \Vert Qx \Vert^2 = \Vert x \Vert^2 = \langle x, x \rangle; \tag 2$
thus,
$\langle x, Q^TQ x \rangle = \langle Qx, Qx \rangle = \langle x, x \rangle; \tag 3$
we are given that (1) holds for all $x$; hence so do (2) and (3).  Now let for arbitrary vectors $y$ and $z$ let
$x = y + z; \tag 4$
then from (3),
$\langle y + z, Q^TQ(y + z) \rangle = \langle y + z, y + z \rangle; \tag 5$
we now have
$\langle y + z, Q^TQ(y + z) \rangle = \langle y, Q^TQ y \rangle + \langle y, Q^TQz \rangle + \langle z, Q^TQ y \rangle + \langle z, Q^TQ z \rangle, \tag 6$
and
$\langle y + z, y + z \rangle = \langle y, y \rangle + \langle y, z \rangle + \langle z, y \rangle + \langle z, z \rangle; \tag 7$
substituting (6) and (7) into (5) yields
$\langle y, Q^TQ y \rangle + \langle y, Q^TQz \rangle + \langle z, Q^TQ y \rangle + \langle z, Q^TQ z \rangle$
$= \langle y, y \rangle + \langle y, z \rangle + \langle z, y \rangle + \langle z, z \rangle, \tag 8$
whence, since (3) binds for all $x$,
$\langle y, Q^TQz \rangle + \langle z, Q^TQ y \rangle = \langle y, z \rangle + \langle z, y \rangle; \tag 9$
furthermore,
$\langle y, z \rangle = \langle z, y \rangle, \tag{10}$
and
$\langle y, Q^TQz \rangle = \langle Qy, Qz \rangle = \langle Q^TQy, z \rangle = \langle z, Q^TQy \rangle; \tag{11}$
(9) thus becomes
$2\langle z, Q^TQy \rangle = 2\langle z, y \rangle, \tag{12}$
or
$\langle z, Q^TQy \rangle = \langle z, y \rangle; \tag{13}$
since this holds for all $z$ we find
$Q^TQy = y = Iy \tag{14}$
for all $y$.  Therefore 
$Q^TQ = I, \tag{15}$
and from this we of course conclude that
$(Q^TQ)_{ii} = 1, \; 1 \le i \le n, \tag{16}$
and 
$(Q^TQ)_{ij} = 0, \; 1 \le i, j \le n; \tag{17}$
finally, (15) directly yields
$Q^{-1} = Q^T. \tag{18}$
