# Prove $R(a \times b) = Ra \times Rb$ given $R \in \mathcal{SO}(3)$ and $a,b \in \mathbb{R}^3$.

Here, $$R$$ is a proper rotation matrix, and $$(\times)$$ is the cross-product.

I already found 3 ways of proving this, all have problems, and I am requesting an elegant approach.

(1) Direct calculation, $$R = R_{\mathbf{u},\theta}$$ has an explicit expression, where $$\mathbf{u}$$ is the vector about which the rotation of $$\theta$$ is carried out. It is essentially possible to calculate both sides and compare. Inelegant.

(2) Using antisymmetric matrices: $$Ra \times Rb=S(Ra)Rb=RS(a)R^\top Rb=RS(a)b=R(a\times b)$$.
My issue with this is that the equality I am trying to prove, is used to prove $$RS(a)R^\top=S(Ra)$$. And so using this feels circular.

(3) $$Ra \times Rb=\|Ra\|\|Rb\|\sin(Ra,Rb)\mathbb{\hat{u}_1}$$ and $$a \times b=\|a\|\|b\|\sin(a,b)\mathbb{\hat{u}_2}$$.
Here, essentially $$\|Ra\|$$ should equal $$\|a\|$$ since $$R$$ only affects orientation.
Because the relative orientation does not change, the $$\sin(\cdot)$$ term should be equal.
Likewise, $$\mathbb{\hat{u}_1}$$ and $$\mathbb{\hat{u}_2}$$ intuitively I know are equal but I am having a hard time expressing it.
Lastly, I have no idea how to bridge $$(a \times b)$$ to $$R(a \times b)$$.
I intuitively see it, and perhaps $$\det R = 1$$ might be useful, but I feel it is hard to write.

Please, a fourth approach is welcome, and insight is always appreciated.

Recall that the cross product $$a\times b$$ is characterized by the property that

$$\det(x,a,b)=\langle x,a\times b\rangle, \qquad \forall x\in\mathbb{R}^3.$$

Now let $$R\in\mathcal{SO}(3)$$. Then by using the fact that $$R^{\mathsf{T}} = R^{-1}$$, we get

$$\langle x, R(a \times b) \rangle = \langle R^{\mathsf{T}}x, a \times b \rangle = \langle R^{-1}x, a \times b \rangle = \det(R^{-1}x, a, b).$$

Then, utilizing the assumption $$\det(R) = 1$$,

$$= \det(R) \det(R^{-1}x, a, b) = \det(x, Ra, Rb) = \langle x, Ra \times Rb \rangle.$$

Finally, since $$\langle x, R(a \times b) \rangle = \langle x, Ra \times Rb \rangle$$ holds for any $$x\in\mathbb{R}^3$$, the desired identity follows.

Addendum. A similar argument shows that, for any invertible $$3\times 3$$ real matrix $$T$$,

$$T(a \times b) = \frac{1}{\det T}(T T^{\mathsf{T}})( Ta \times Tb).$$

• Very interesting addendum, thank you very much. Sep 29, 2019 at 7:05

The simplest approach I see is the following, $$\langle R(a\times b), Ra\rangle = (Ra)^TR(a\times b)$$ $$=a^T(a\times b)=\langle a\times b, a\rangle =0$$ Similarly, $$\langle R(a\times b), Rb\rangle = 0$$ Hence, it follows that $$R(a\times b)= k( Ra\times Rb)$$ for some non-zero real number $$k$$. Now, observe that, $$\frac{1}{k}||R(a\times b)||^2$$ $$\langle R(a\times b), Ra\times Rb \rangle = det (R(a\times b), Ra,Rb)$$ $$=(det R)(det(a\times b,a,b))$$ $$=||a\times b||^2$$ $$=||R(a\times b)||^2$$ Required result follows.

• In the first step you prove that $Ra\times rb$ is orthogonal to $R(a\times b)$. That proves that $R(a\times b)$ and $Ra\times Rb$ are linearly dependent. I don't see how that already implies that they also have the same length. Sep 29, 2019 at 8:30
• @celtschk, yeah, additional argument is required. Corrected now. Sep 29, 2019 at 9:42

Here's a component-wise proof. Note that I'm using the Einstein convention: If an index appears twice in a product, it is summed over. For example $$(Rv)_i = R_{ij}v_j$$ actually means $$(Rv)_i = \sum_{j=1}^3 R_{ij} v_j$$ because $$j$$ appears twice in $$R_{ij}v_j$$.

Also in case you don't know those symbols, $$\delta_{ij} = \begin{cases} 1 & i=j\\ 0 & \text{otherwise} \end{cases}$$ and $$\epsilon_{ijk} = \begin{cases} 1 & ijk \in \{123,231,312\}\\ -1 & ijk \in \{132,213,321\}\\ 0 & \text{otherwise} \end{cases}.$$ Then we can easily check that $$(a\times b)_i = \epsilon_{ijk}a_j b_k.$$ The equation we want to prove is $$R(a\times b)=Ra\times Rb.$$ Now since $$R$$ is invertible, we can multiply both sides with $$R^{-1}$$ and get an equivalent equation: $$a\times b=R^{-1}(Ra\times R_b)$$ Thus we get: \begin{aligned} &&(a\times b)_i &= (R^{-1}(Ra\times Rb))_i\\ \iff&&(a\times b)_i &= (R^{-1})_{ij}(Ra\times Rb)_j\\ \iff&&(a\times b)_i &= R_{ji}(Ra\times Rb)_j && \text{orthogonality of R}\\ \iff&&\epsilon_{ikl}a_kb_l &= R_{ji} \epsilon_{jmn}(Ra)_m(Rb)_n\\ \iff&&\epsilon_{ikl}a_kb_l &= R_{ji} \epsilon_{jmn}R_{mk}a_kR_{nl}b_l\\ \iff&&\epsilon_{ikl}a_kb_l &= (\epsilon_{jnm} R_{ji} R_{mk} R_{nl}) a_k b_l \end{aligned} Thus we have to prove that $$\epsilon_{ikl} = \epsilon_{jnm} R_{ji} R_{mk} R_{nl} \tag{*}$$ Now $$\epsilon$$ is completely characterized by the following two properties:

• $$\epsilon_{123} = 1$$

• Exchange of two indices changes the sign.

So let's check both properties for the right hand side of the above equation (note that the free indices are $$i$$, $$k$$, and $$l$$):

• $$\epsilon_{jmn} R_{j1} R_{m2} R_{n3} \stackrel?= 1$$

The left hand side is the determinant of $$R$$, which by assumption is $$1$$, so this equation is indeed true.

• Exchange of two indices: Let's without loss of generality exchange the last two indices, $$§k$$ and $$l$$.

Clearly due to the commutativity of multiplication, we have $$\epsilon_{jnm} R_{ji} R_{ml} R_{nk} = \epsilon_{jnm} R_{ji} R_{nk} R_{ml}$$ Now the indices of $$\epsilon$$ in this term are summed over, therefore we can just rename them without changing anything. In particular, we exchange the names $$m$$ and $$n$$: $$\epsilon_{jnm} R_{ji} R_{nk} R_{ml} = \epsilon_{jnm} R_{ji} R_{mk} R_{nl}$$ Now we exchange the last two indices of the $$\epsilon$$ factor: $$\epsilon_{jnm} = -\epsilon_{jmn}$$ Putting all together, we then get $$\epsilon_{jnm} R_{ji} R_{ml} R_{nk} = -\epsilon_{jmn} R_{ji} R_{mk} R_{nl}$$ Thus exchanging $$k$$ and $$l$$ indeed changes the sign.

Thus we see that equation $$(*)$$ indeed holds, and thus $$R(a\times b)=Ra\times Rb$$.

I don't know whether this is really different from what you had in mind, but I would explain this as follows. It has a bit of Lie algebra air to it.

Let $$R_{\vec{a},t}$$ be the right-handed rotation about the axis $$\vec{a}$$ by the angle $$t|\vec{a}|$$. These form a 1-parameter group under composition, basically just add the angles of rotation as the axis is constant. The Lie algebra trick is to study the derivatives of such 1-parameter groups at $$t=0$$. The reason why cross products often appear when studying rotations in $$\Bbb{R}^3$$ is the following.

For all vectors $$\vec{b}\in\Bbb{R}^3$$ we have $$\frac d{dt}R_{\vec{a},t}\vec{b}\big\vert_{t=0}=\vec{a}\times\vec{b}.$$ In other words, cross product by $$\vec{a}$$ is an infinitesimal generator for the group of rotations about the axis $$\vec{a}$$.

The other fact I would use is the observation that

For all rotations $$R$$ we have the conjugation relation $$R\circ R_{\vec{a},t}\circ R^{-1}=R_{R\vec{a},t}.$$ In other words, to rotate a vector $$\vec{b}$$ about the vector $$R\vec{a}$$ we can first apply $$R^{-1}$$, to move the scene to be about rotations about $$\vec{a}$$, then rotate about $$\vec{a}$$ by the prescribed angle, and then as the last step rotate the axis back to $$R\vec{a}$$.

We have the obvious identity $$(R\circ R_{\vec{a},t}\circ R^{-1})R\vec{b}=R(R_{\vec{a},t}\vec{b}).\qquad(*)$$ And your claim follows from equating the derivatives of both sides of $$(*)$$ at $$t=0$$.

Don't know whether this was buried in one of your proofs, or whether you see too much handwavium in it.

• I deleted a latter section. While it explained the first boxed fact when $\vec{a}$ is one of the standard basis vectors, I think that my claim of the first boxed fact following from that by the second leads to a circular argument. Sep 29, 2019 at 5:51
• I did not think of it from this perspective. It's new to me. Thank you so much for broadening the horizons, I will enjoy digging into this. Sep 29, 2019 at 7:07