Orthogonal Matrices Scalar Product Proof I'm trying to figure this out: we're dealing with real numbers only here, with the standard scalar product defined as $<\mathbf{v},\mathbf{w}> = \mathbf{v}^T\mathbf{w}$, and we are told, unusually perhaps, that orthogonal matrices are defined as those for which $<R\mathbf{v},R\mathbf{w}> = <\mathbf{v},\mathbf{w}>$, and I need to show from this that the matrix R can be characterised by $R^TR=\mathbf{1}$ (and also that $det(R)=(+/-)1$ but this is fine).
Using the definition of the scalar product, it is not hard to get to $\mathbf{v}^TR^TR\mathbf{w}=\mathbf{v}^T\mathbf{w}$, which I feel is a good starting point, and seems to be close; whilst $R^TR=\mathbf{1}$ would make this work, it isn't the only matrix it can be. I also know $\mathbf{v}^TR^TR\mathbf{w}=\mathbf{w}^TR^TR\mathbf{v}$ but I'm not sure if that helps.
Thanks

Hint:
The relation $$\left\langle Rv,Rw\right\rangle
= \left\langle v,w\right\rangle$$ for an orthogonal matrix holds for any (!) $v,w\in\mathbb{R}^n$.
Hint 2:
It could be useful to look at the linear maps 
$$\psi_1: \mathbb{R}^n \to \mathbb{R}, w \mapsto \left(v^T R^TR-v^T\right)w$$
and then
$$\psi_2: \mathbb{R}^n \to \mathbb{R}^{1\times n}, v \mapsto v^T\left( R^TR - \mathrm{id}_{\mathbb{R}^n}\right).$$
