# 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.

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 $$Q^{\top}Q=I_n\iff Q^{\top}=Q^{-1}.$$

\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 $QQ^{\ast}=I$.

$(Qx)^TQx = x^TQ^TQx = x^Tx \ \forall x \rightarrow \ Q^TQ = I \rightarrow Q^{-1} = Q^T$ (the fact that $Q$ is invertible can be concluded from: $rank(Q^TQ) = rank(Q) = rank(I)$ )

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

$Q^{-1} = Q^T. \tag{18}$