Equivalent definitions of an orthogonal matrix. I wish to show that the following definitions of an $n \times n$ real matrix $Q$ are equivalent:

  
*
  
*$QQ^T=I$,
  
*$Qx\cdot Qx=x\cdot x$ for all $x\in \mathbb{R}^n$.
  

I found it easy to show that $(1) \Rightarrow (2)$: 
Suppose that $QQ^T=I$. Then, we have that $$Qx\cdot Qx=(Qx)^TQx=x^TQ^TQx=x^TIx=x^Tx=x\cdot x.$$
However, I am unsure about how to show that $(2) \Rightarrow (1)$. 
 A: Hint Polarizing the equation in (2) gives that
$$Qx \cdot Qy = x \cdot y$$
for all $x, y \in \Bbb R^n$, and using the same manipulation as in the proof of the other implication gives that this is equivalent to
$$y^T Q^T Q x = y^T x.$$
Now, evaluate both sides at $x = e_a$, $y = e_b$, where $(e_1, \ldots, e_n)$ is the standard basis.

 Evaluating the l.h.s. gives $e_b^T Q^T Q e_a$, which is the $(a, b)$ entry of the matrix $Q^T Q$.

A: As Travis pointed out, by polarizing we have that:
$\langle Qx, Qy \rangle$=$\langle x,y\rangle \quad \forall x,y$
Hence:
$\langle Q^TQx, y \rangle$=$\langle x,y\rangle \quad \forall x,y$
$\implies \langle Q^TQx-x, y\rangle=0 \quad \forall x,y$ 
Fix an arbitrary $x$. Put $y:=Q^TQx-x$. Therefore, $\langle Q^TQx-x, Q^TQx-x\rangle=0 \implies Q^TQx-x=0 \implies Q^T Q x=x$. Since $x$ is arbitrary, $Q^TQ$ is the identity.
A: Suppose
$Qx \cdot Qx = x \cdot x, \forall x \in \Bbb R^n; \tag{1}$
then
$x^TQ^TQx = Qx \cdot Qx = x \cdot x$
$=  x^T Ix, \forall x \in \Bbb R^n, \tag{2}$
whence,
$x^T(Q^TQ - I)x = 0, \forall x \in \Bbb R^n; \tag{3}$
we note that $Q^TQ - I$ is symmetric:
$(Q^TQ - I)^T =(Q^TQ)^T - I^T = Q^TQ - I. \tag{4}$
The desired result now follows with the aid of the following
Lemma:  Let $A$ be a real, symmetric, $n \times n$ matrix, $A^T = A$; if $x^TAx = 0$, $\forall x \in A$, then in fact $A = 0$.  
Proof:  we have, for $x, y \in \Bbb R^n$, 
$(x + y)^TA(x + y) = 0, \tag{5}$
or
$x^TAx + y^TAx + x^TAy + y^TAy = 0; \tag{6}$
now
$x^TAx= y^TAy = 0, \tag{7}$
so (6) yields
$y^TAx + x^TAy = 0; \tag{8}$
since $x^TAy$ is a scalar quantity, and $A^T = A$, we have
$x^TAy = (x^TAy)^T = y^TA^Tx = y^TAx; \tag{9}$
thus (8) becomes
$2y^TAx = 0 \Rightarrow y^TAx = 0; \tag{10}$
since (10) binds for all $y$, 
$Ax = 0, \forall x \in \Bbb R^n \Rightarrow A = 0. \tag{11}$
QED.
We now apply this lemma to find that
$Q^TQ - I = 0 \tag{12}$
by virtue of (3) and (4); but this is eqivalent to
$QQ^T = I \tag{13}$
since $Q$ is of finite size.  QED.
A: Consider the orthonormal basis $\{e_i\}_{i=1}^n$ and consider a matrix $Q=(Q_{ab})$ that satisfies $(2)$. Then, for $1\leq i,j\leq n$, we have
$$
Qe_i\cdot Qe_j = e_i^TQ^TQe_j= \sum_{1\leq a,b\leq n} Q_{ib}^TQ_{aj} = e_i\cdot e_j = \delta_{ij}.
$$
On the other hand, note that 
$$
\left(Q^TQ\right)_{ij} = \left(QQ^T\right)_{ij} = \sum_{1\leq a,b\leq n} Q_{ib}^TQ_{aj}=\delta_{ij}.
$$
