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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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    Does your first statement say that, using the same notation as above (where $D$ is for the Jacobian), that $D(Tx) = Tx \forall x \in \mathbb{R}^d$? I am probably being really dumb, but what does that tell you about the determinant though?2011-01-30
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    @user1736: So the determinant of the Jacobian is identical to the determinant of the map you started with. Do you mean, why is the determinant of a rotation matrix 1?2011-01-30
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    @user1736: It tells you the determinant of the Jacobian is 1. Two minor points. Some authors use "Jacobian" to mean the determinant of the (square) matrix of first partials of change of variables mapping, and other authors use it to mean the matrix (as you have evidently done here). Also, be aware that a reflection is like a rotation in preserving distances between points, but differs in having determinant -1.2011-01-30
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    I guess my confusion is stemming from the fact that I have very poor intuition about what a determinant represents. So the reason why the determinants of the two matrices are equal is because the two matrices themselves are equal at every point. Is there an easy way to see why this property is true (since the two matrices themselves aren't equal)?2011-01-30
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    If "two matrices are equal at every point", they are equal! If "$A$ and $B$ are equal at every point" means that $Ax=Bx$ for all vectors $x$, then (by using the standard basis $e_1,e_2,\dots$) we see that the columns of $A$ and $B$ are equal ($Ae_1$ is the first column of $A$, etc).2011-01-30
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    @user1736: I'm not sure I understand. Equal matrices have equal determinants. Suppose $T$ is a real matrix, and $f:\mathbb{R}^n\to\mathbb{R}^n$ is the linear map induced by $T$ acting on column vectors, $x\mapsto Tx$. Then $f$ has a Jacobian matrix of partial derivatives $Df$, and $Df$ can be thought of as a function from $\mathbb{R}^n$ to $n$-by-$n$ matrices. In this case it is a constant function: $Df\vert_x = T$ for all $x$. Thus the Jacobian determinant is also a constant function, everywhere equal to $\det T$.2011-01-30
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    Anyway, the determinant of a matrix $A$ measures the "volume" of the function that maps each $x$ to $Ax$. This is why the Jacobian appears in the change of variable formulas, because it is telling you how volume changes as you change variables. (I say "volume" for two reasons. One is that, of course, if $n\ne 3$, instead of volume we have length, or area, or an appropriate notion of hyper-volume. But also, because we measure orientation, so some determinants are negative.)2011-01-30
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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. Thanks for all the help! And 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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    Dear milcak, Note that a rotation matrix always preserves orientation, and so in fact its determinant is equal to $1$.2011-01-31
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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