The Determinant of Transition Matrices Between Orthonormal Bases I am trying to show that if $A$ and $B$ are orthonormal bases of a real finite dimensional vector space $V$ and $P$ is the change of basis matrix between $A$ and $B$ then $\det(P) = \pm 1$. Here is my progress:
When working on this problem, I discovered the following fact: If $G_B$ denotes the Gram matrix of the inner product relative to $B$ and $G_A$ denotes the Gram matrix of the inner product relative to $A$ then the Gram matrices and the transition matrix $P$ are related through
$$
G_B = P^T G_A P.
$$
The original claim is then a corollary of this fact:
$$
G_B = P^T G_A P \implies \det(G_B) = \det(P^T) \det(G_A) \det(P) \implies \det(P) = \pm 1
$$
where I have used basic properties of the determinant and the fact that the determinant of the Gram matrix relative to an orthonormal basis is $1$.
So, I think this proves the original claim but I am now stuck with proving the more general claim that $G_B = P^T G_A P$. What would be a good way to approach this?
 A: Here is a possibly simpler approach:
Let $a_1,...,a_n$, $b_1,...,b_n$ be the elements of bases $A,B$ respectively.
By assumption, $b_i = P a_i$. By 'orthonormality', we have $\langle b_i, b_j \rangle = \langle P a_i, P a_j \rangle = \langle a_i, P^TP a_j \rangle  = \delta_{ij}$. It follows from this that $P^TP = I$, and that $(\det P)^2 = 1$, from which the result follows (since $\det P$ is real).
A: $$\begin{eqnarray*}
(P^T G_A P)_{il} &=& (P^T)_{ij} (G_A)_{jk} P_{kl} \\
&=& P_{ji} \langle x_k,x_j\rangle P_{kl} \\
&=&  \langle P_{kl} x_k, P_{ji} x_j\rangle  \\
&=& \langle (P^T x)_l, (P^T x)_i\rangle \\
&=& (G_B)_{il}
\end{eqnarray*}$$
$\def\det{\mathrm{det}\,}$
Addendum: On second reading I may have misinterpreted the meaning of "this" in the last sentence of your question. 
Let $X_{ij} = (x_i)_j$ be the matrix whose rows are the orthonormal basis vectors.
$X$ transforms like $X\to P^T X$. 
We have $X X^T = \mathbb{I}$, since the basis is orthonormal. 
Then 
$$(P^T X)(P^T X)^T = P^T X X^T P = P^T P = \mathbb{I},$$
since the new basis is orthonormal. 
Therefore, $\det(P^T P) = (\det P)^2 = 1$, so $\det P = \pm 1$. 
