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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?

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Since A and B are both orthonormal bases, which means P preserves inner product, i.e. P is orthogonal transformation, so det(P)=1 or -1. –  chaohuang Jul 1 '12 at 21:35
also, since P is orthogonal,P^T = P^(-1) –  chaohuang Jul 1 '12 at 21:36

2 Answers 2

up vote 1 down vote accepted

$$\begin{eqnarray*} (P^T G_A P)_{il} &=& (P^T)_{ij} (G_A)_{jk} P_{kl} \\ &=& P_{ji} \langle x_k,x_j\rangle P_{kl} \\ &=& \langle P_{kl} x_k, P_{ji} x_j\rangle \\ &=& \langle (P^T x)_l, (P^T x)_i\rangle \\ &=& (G_B)_{il} \end{eqnarray*}$$

$\def\det{\mathrm{det}\,}$ 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 $\det P = \pm 1$.

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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).

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