So the prompt is merely an existence proof--just find a $u$ and $v$ that work. Well, I'm unfortunately a little stuck on getting started.

I know that $Q \in SO_4(\mathbb R) \implies QQ^T = I \text{ and } \det(Q) = 1$.

I tried to solve $Qx = uxv$ for $u,v$ but I was not able to do so successfully. This is because I don't necessarily know if $x$ is has an inverse.

Here's what I managed to deduce successfully:

$$|Qx| = |uxv| = |u||x||v| = 1\cdot |x| \cdot 1 = |x|$$

which means that multiplying $x \in \mathbb H$ by $Q$ doesn't change its length.

Let $Q = [q_1 \,|\, q_2 \,|\, q_3 \,|\, q_4]$, where $q_i$ is the $i^{th}$ column of $Q$, and let $x = a + bi + cj + dk$. Then,

$$\begin{align}Qx & = Q(a + bi + cj + dk)\\& = aQ + bQi + cQj + dQk \\&= aQ + bQ\begin{pmatrix}0 \\ 1 \\ 0 \\ 0\end{pmatrix} + cQ\begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix} + d\begin{pmatrix}0 \\ 0 \\ 0 \\ 1\end{pmatrix} \\ & = aQ + bq_2 + cq_3 + dq_4\end{align}$$

And unfortunately I don't see where to go from here. I'm not entirely sure that I did the multiplication of a quaternion by a matrix correctly. If $a \in \mathbb R$, what does $aQ$ mean? Thus, I don't think that's right.

Likely, the solution will boil down to $v = u^{-1}$ or something. But I'm still not quite sure how to arrive there.

  • $\begingroup$ $x$ only fails to have an inverse if it's $0$, and in that case any $u,v$ works. So it's safe to assume that $x\neq 0$ and therefore has an inverse. $\endgroup$ – Arthur Sep 27 '16 at 11:42
  • $\begingroup$ @Arthur It appears that assuming $x$ has an inverse is not all that helpful--it ends up giving me $Qx = uxu^{-1} \implies Q = uxu^{-1}x^{-1} = ux(xu)^{-1}$ which I know of no properties that might help me isolate $u$ by itself. $\endgroup$ – Decaf-Math Sep 27 '16 at 11:59

Here's a proof that you may find completely unsatisfactory, depending on whether you know about bundles and things like that.

Consider the map $p : SO(4) \to \mathbb S^3 : [q_1, q_2, q_3, q_3] \mapsto q_1$ that selects from a matrix its first column.

This is a fibration, and its fiber ($p^{-1}(q_1)$for $q_1 = \begin{bmatrix}1\\0\\0\\0 \end{bmatrix}$, for example) is just the set of $4 \times 4 $ orthogonal matrices whose first row and column are (1,0,0,0). If you look at the lower right $3 \times 3$ matrix, it must therefore be in $SO(3)$, since all its columns are orthonormal, and a first-row-expansion on the $4 \times 4$ matrix shows that the determinant is $+1$ instead of $-1$.

So now we have $$ SO(3) \to SO(4) \to S^3 $$ as a sequence of maps, where the first is the injection of the fiber and the second is a fibration. That means that $SO(4)$ can be written as a bundle where we look at a trivial $SO(3)$ bundle over the top half of $S^3$ and a trivial $SO(3)$ bundle over the bottom half, and the "gluing map" along the equator is a map from $S^2 \to SO(3)$, i.e., an element of $\pi_2(SO(3)$. Since $SO(3)$ is a Lie group, we know that $\pi_2(SO(3) = 0$.

So the gluing map is trivial, and there's a decomposition of $SO(4)$ as $SO(3) \times S^3$. (I have a feeling that was rather the long way around, but so be it.)

So do this: let $Q$ be your matrix, and let $u$ be the first column of $Q$, and let $L_{u^{-1}}$ be the matrix that represents left quaternion multiplication by $u^{-1}$ (quaternion inverse!), so that $L_{u^{-1}} \cdot e_1 = u^{-1} e_1= u^{-1}$, for instance, and $L_{u^{-1}}u = u^{-1}u = \mathbf {1}$, the quaternion $1$, which corresponds to the vector $e_1$.

What is $L_{u^{-1}}Q$? Well, its first columns is $e_1$, so its $(1,1)$ entry is $1$, so its first row (which must be a unit vector!) is $(1,0,0,0)$. So $$ L_{u^{-1}}Q = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & & & \\ 0 & & A & \\ 0 & & & \end{bmatrix} $$ where $A$ is a $3 \times 3$ matrix that must in fact be in $SO(3)$.

Now every rotation of 3-space, considered as the space of pure-vector quaternions, can be expressed in the form $q \mapsto s^{-1} q s$ for some unit quaternion $s$. So we can write the matrix $A$ in the form $L_{s^-1} R_s$, where these are the matrices of the linear transformations "multiply on the left by $s^{-1}$" and "multiply on the right by $s$", respectively.

So we have \begin{align} L_{u^{-1}} Q &= L_{s^{-1}}R_s \\ Q &= L_u L_{s^{-1}}R_s \\ \end{align} i.e., the matrix $Q$ represents left multiplication by $us^{-1}$ followed by right multiplication by $s$. Picking the $u$ in your solution to be my $us^{-1}$ and $v$ to be my $s$, we're done.

(Apologies for rambling answer...but I got there in the end...)

  • $\begingroup$ Thanks. I'm still trying to parse this answer in my brain, but here's what it seems like this proof accomplishes: we show that for any $x \in \mathbb H$, we have that $Q \in SO_4(\mathbb R)$ can be expressed as linear transformations such that left-multiplying $x$ by $Q$ gives us the desired result: $Qx = u x v$ for unit quaternions $u, v$. These linear transformations accomplish the left multiplying to $x$ and right multiplying to $x$, which in my opinion is pretty genius. $\endgroup$ – Decaf-Math Sep 27 '16 at 12:32
  • $\begingroup$ Right. And the first part (before the "so do this") is really not important; the second part really shows the first part in a less fancy-language way. The idea that left-multiply-by-$q$ is a linear map from $H$ to $H$, and therefore has a matrix rep (as does right-multiply!) is pretty darned cool, I agree. $\endgroup$ – John Hughes Sep 27 '16 at 12:40

It is a consequence of the fact that any 4D rotation can be canonically decomposed into a left-isoclinic and a right-isoclinic rotation. As you can see in the reference this means that any $A \in SO_4(\mathbb{R})$ acts, on a vector $\vec x=(x_1,x_2,x_3,x_4) \in \mathbb{R}^4$ as: $$ A \vec x=UV \vec x=U \vec x V^T $$ where $U$ and $V^T$ are matrices of the form: $$ U=\begin{pmatrix} a&-b&-c&-d\\ b&a&-d&c\\ c&d&a&-b\\ d&c&b&a \end{pmatrix} \qquad U^T=\begin{pmatrix} p&q&r&s\\ -q&p&-s&r\\ -r&s&p&-q\\ -s&-r&q&p \end{pmatrix} $$ with:$a^2+b^2+c^2+d^2=p^2+q^2+r^2+s^2=1$ Now we can see that $U$ and $V^T$ are the matrix representation in $M_4(\mathbb{R})$ of two unit quaternions $u,v$ and the vector $\vec x$ can be represented by the quaternion $x=x_1+x_2\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{i}+x_4\mathbf{k}$, so the rotation can be represented as the product $uxv$ .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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