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I saw on wikipedia that the determinant of a rotation matrix is always one (possibly by definition?), but it doesn't say anything about the determinant of the Jacobian of such a matrix.

Since applying a rotation shouldn't change the integral of a rotationally invariant function over $\mathbb{R}^d$, if $(v_1,...,v_d) = \phi(u_1,...,u_d)$ is our rotation, then it should be the case that $d_{v_1}...d_{v_d} = |det(D \phi)(u)| d_{u_1}...d_{u_d} = d_{u_1}...d_{u_d}$ right (here $D$ stands for the Jacobian)?

I don't know enough about rotation matrices to make that justify that claim though.

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The Jacobian matrix of the map on $\mathbb{R}^n$ induced by a matrix $T$ is everywhere equal to $T$. Keep in mind that the Jacobian gives the best linear approximation of a function, and that this means equality if the function is linear. So when you are doing a change of variables with a linear map (like a rotation), you just multiply by the determinant of the map, in your case $1$.

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    @Jonas, yeah my question came from not really processing/comprehending what you said earlier. I think I get it now, though. $T$hanks for all the help! $A$nd you too, Andres!2011-01-30
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Let $A$ be any $n$x$n$ matrix. Let $h(\mathbb{x}) = A \cdot \mathbb{x}$. Now if you take some $S \subset \mathbb{R}^n$, then you have that:

$volume(h(S)) = |\det A| \cdot volume(S)$

which is just a direct application of a change of variables. Rotation preserves volume, hence if $A$ is your rotation matrix, then $|det A| = 1$. In general, the deteminant is negative in the case that the transformation is orientation reversing.

Now to consider the Jacobian: clearly $D h(\mathbb{x}) = D (A \cdot \mathbb{x}) = A$, hence $|\det \ D h| = 1$ as well.

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    Yes sorry my bad. I had written at first that it's negative if the transformation reverses orientation. I'll change it the way it was.2011-01-31