Orthogonal transformations and cross product In the appendix of Pressley's Differential Geometry, he states:
For a $3\times 3$ orthogonal matrix $M$, for $u,v\in \mathbb{R}^3$, $Mu\times Mv=(\det M)(u\times v)$, i.e. $Mu\times Mv=\pm u\times v$
However, this contradicts my intuition that, if $u$ and $v$ are rotated by $M$, then $u\times v$ should be rotated by $M$.
Can anyone explain what's wrong with my intuition? Thanks.
Edit: Here's the excerpt:

 A: Let
$$
        M = \left[\begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{matrix}\right]
$$
This is an orthoogonal matrix. $M\hat{i}=\hat{i}$, $M\hat{j}=\hat{k}$ and $M\hat{k}=\hat{j}$. Then $(M\hat{i})\times (M\hat{j})=\hat{i}\times \hat{k}=-\hat{j}$. And $\hat{i}\times\hat{j}=\hat{k}$. Are you sure you copied the problem correctly?
A: About the "proof" at orthogonal matrices preserve cross product, it's cheating when going from the last but one equality to the last one: look at that matrix $A$ "disappearing" from the right-hand side.
All they are doing is correct till they get at
$$
\langle Ax \times Ay, z\rangle = \det (A) \cdot \langle A (x\times y), z \rangle \ .
$$
And now, they say:
$$
\det (A) \cdot \langle A (x\times y), z \rangle = \det (A) \cdot \langle  x\times y, z \rangle  \ ,
$$
with that $A$ misteriously disappearing from the right-hand side.
But we can take advantage from their work till the point it is correct, namely
$$
\langle Ax \times Ay, z\rangle = \det (A) \cdot \langle A (x\times y), z \rangle =  \langle \det(A) \cdot A (x\times y), z \rangle   \ .
$$
And apply their lemma: since this equality is true for all $z$, we must have indeed
$$
 Ax \times Ay = \det(A) \cdot A (x\times y) = \pm A(x\times y) \ .
$$
A: Yeah, that's not true at all.  The correct result can be proved with Hodge duality.
Let $\epsilon$ be a 3-vector; clifford multiplication by $\epsilon$ gives the Hodge dual of what it multiplies.
The cross product is a Hodge dual:  $u \times v = (u \wedge v) \epsilon^{-1}$.
The inverse of a linear map is related to its adjoint through the Hodge dual:
$$M^{-1}(a) = M^*(a \epsilon) \epsilon^{-1} \frac{1}{\det M}$$
where, if $a \epsilon = c \wedge d$, then $M^*(a \epsilon) = M^*(c) \wedge M^*(d)$.  This is how linear maps act upon 2-vectors.
Of course, for orthogonal maps, the inverse is equal to the adjoint, so we get
$$M(a) = M(a \epsilon) \epsilon^{-1} \frac{1}{\det M}$$
Plug in $u \times v$ to get
$$M(u \times v) = M([u \wedge v] \epsilon^{-1} \epsilon) \epsilon^{-1} \frac{1}{\det M}$$
Or rather poetically,
$$M([u \wedge v]\epsilon^{-1}) = M(u \wedge v) \epsilon^{-1} \frac{1}{\det M}$$
But the easier to parse result is
$$M(u \times v) = M(u) \times M(v) \frac{1}{\det M} = \pm M(u) \times M(v)$$
A: Looking at the Wikipedia article for an orthogonal matrix 

[...]an orthogonal matrix preserves the dot product of vectors, and therefore acts as an isometry of Euclidean space, such as a rotation or reflection.

Therefore, if your matrix $M\in O(3)$ is specifically $M\in SO(3)$ that is, a rotation the vector product will not change. But if $M\in (O(3)-SO(3))$, then you basically turn your coordinate system from right-handed to left-handed and get a sign for the vector-product.
aka: Hint: look at the Wiki articles for vector product and orthogonal matrix
