# Let $Q$ be a special orthogonal matrix. Show that $Q(u\times v)=Q(u)\times Q(v)$ for any vectors $u, v\in\mathbb R^3$.

Let $Q$ be a $3\times3$ special orthogonal matrix. Show that $Q(u\times v)=Q(u)\times Q(v)$ for any vectors $u, v\in\mathbb R^3$.

I have no idea how to start. I'm not sure if $Q(u)\cdot Q(V)=Q(u\cdot v)$ would helps. Please give me some help. Thanks.

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More generally, $M(u) \times M(v) = (\det M) M^{-T}(u \times v)$ as mentioned (but not proved) in Wikipedia. –  lhf Jul 18 '13 at 4:31

The cross-product $u\times v$ is the unique vector such that $$\det(u,v,w)=(u\times v)\cdot w\qquad \forall w$$ where $\det(u,v,w)$ is the determinant of the $3\times 3$ matrix whose columns are $u,v,w$ in this order, that is the determinant of the linear map that sends the canonical basis to $(u,v,w)$. That's a common definition of the cross-product. See below if needed.

Recall that $Q$ special orthogonal means $Q^T=Q^{-1}$ and $\det Q=1$.

We need to prove that $Q^T(Qu\times Qv)=u\times v$. So let us compute $$Q^T(Qu\times Qv)\cdot w=(Qu\times Qv)\cdot Qw=\det(Qu,Qv,Qw)=\det Q\det(u,v,w)=\det(u,v,w).$$ By the uniqueness defining $u\times v$, this proves $Q^T(Qu\times Qv)=u\times v$, i.e. $Qu\times Qv=Q(u\times v)$.

Note: the same argument shows more generally that, as mentioned by lhf and wikipedia, $$M^T(Mu\times Mv)=(\det M) u\times v\quad\Rightarrow \quad (Mu\times Mv)=(\det M) M^{-T}(u\times v)$$ for every invertible $3\times 3$ matrix $M$, where $M^{-T}=(M^{-1})^T=(M^T)^{-1}$. The formula on the left is true for every matrix $M$ and is just $0=0$ in the singular case, since we have $Mu\times Mv=0$ for every $u,v$ in this case.

The fact that the identity $\det(u,v,w)=(u\times v)\cdot w$ is satisfied by every $u,v,w$ can be checked directly, by computations, from the determinant definition of $u\times v$. Another way to see it is to note that the map $(u,v,w)\longmapsto (u\times v)\cdot w$ is multilinear, anti-symmetric (or alternating), and sends the canonical basis to $1$, whatever definition of the cross-product you might have. So it must be the determinant. Uniqueness of $u\times v$ satisfying the identity follows from $(\mathbb{R}^{3})^\perp=\{0\}$, as $w_1\cdot w=w_2\cdot w$ for every $w$ implies $(w_1-w_2)\cdot w=0$ for every $w$, in particular for $w=w_1-w_2$, whence $\|w_1-w_2\|^2=0$.

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How to tell $\det(Qu,Qv,Qw)=\det Q\det(u,v,w)$ ? –  ᴊ ᴀ s ᴏ ɴ Jul 18 '13 at 9:08
@JASON That's the usual $\det AB = \det A \det B$. Note that $\det(u,v,w)$ is by definition the determinant of the linear map that sends the canonical basis to $(u,v,w)$. So $\det(Qu,Qv,Qw)$ is the det of the composition of the latter with $Q$. –  1015 Jul 18 '13 at 14:33

Maybe it will be easiest to show this explicitly for the basis vectors $\{ (1,0,0) \cdots \}$ , and then the general case follows from linearity of all things involved. It will be useful to note that if $\vec{Q_1}, \vec{Q_2}, \vec{Q_3}$ are the column vectors of $Q$, then the fact that $\det(Q) = 1 = \vec{Q1} \cdot (\vec{Q_2} \times \vec{Q_3})$ gives the "right hand rule" that $\vec{Q_1} \times \vec{Q_2} = \vec{Q_3}$.

Please note, I have been very sloppy and did not check the signs and orders of things. You should check that all formulae are indeed right.

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