Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

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 $\|w_1-w_2\|^2=0$.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.