Show $Q(x)\cdot Q(y)=x\cdot y\Rightarrow Q^TQ=I$

Question

According to my class notes the following are two equivalent definitions of an orthogonal matrix:

1. $Q^TQ=I$
2. $Q(x)\cdot Q(y)=x\cdot y$

I've been able to show that $1\Rightarrow 2$, yet I do not know how to show $2\Rightarrow 1$. I'd appreciate any help.

Answer 1

$x^\top y=(Qx)^\top(Qy)=x^\top Q^\top Qy$ so $x^\top(Q^\top Q-I)y=0$.

If $e_i$ denote the standard basis vectors, then for any matrix $A$ we have $A_{ij}=e_i^\top Ae_j$. So allowing $x$, $y$ to vary over pairs of the standard basis vectors implies that $Q^\top Q-I=0$, as desired.

Answer 2

Denote $C_1, \ldots C_n$ the columns of $Q$.

1. First remark that : $$Q^{\mathrm{T}} Q = I_n \iff \forall i, j \in \{1, \ldots, n\}, C_i.C_j = \delta_{i, j}$$ Where $\delta_{i, j}$ is the symbol of Kroneker : $$\delta_{i, j} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$$
2. We know that : $$\forall i \in \{1, \ldots, n\}, C_i = Q (e_i)$$ Then : $$\forall i, j \in \{1, \ldots, n\}, C_i.C_j = Q(e_i).Q(e_j) = e_i.e_j = \delta_{i, j}$$