Minimal rotation matrix for two vectors in $\mathbb{R}^n$

Question

Suppose we have two normal vectors $v, u \in \mathbb{R}^n, \|u\|_2=1, \|v\|_2=1$. We would like to find a rotation matrix $R\in\mathbb{R}^{n \times n}$ that satisfies $R u = v$. This clearly does not uniquely determine matrix $R$ because there are only $n$ equations while there are $n(n-1)/2$ degrees of freedom for $R$. However we can make it unique by forcing it to only rotate vectors in the $Span\{u,v\}$. In other words: $$\forall w \in Span\{u,v\}^\perp : R w = w$$ Intuitively, it only rotates vectors in the $Span\{u,v\}$ plane while leaving every other direction untouched. In this sense, the rotation matrix is minimal.

In the very simple case that $Span\{u,v\} = Span\{e_i,e_j\}$, i.e., the linear space spanned by $u,v$ is the same as the one spanned by $e_i, e_j$ we can compute $R$ as follows: $R = I_n - M$, in which $M\in \mathbb{R}^{n\times n}$ is only zeros except for its $i$-th and $j$-th rows and columns: $$M_{i, j\times i, j} = \begin{bmatrix} 1-\cos \theta & \sin \theta\\ -\sin\theta & 1-\cos\theta \\ \end{bmatrix}$$ in which $\theta = \cos^{-1} \langle u, v \rangle$

Is there a way to compute the rotation matrix for general normal vectors $u, v$?

Answer 1

Let $v_1,v_2 \in\mathbb{R}^n$ be linearly independent normal vectors, i.e. $\|v_1\|=\|v_2\|=1$ and $\operatorname{dim}\operatorname{Span}(v_1,v_2)=2$.

Let $\{r_3, r_4,\dots, r_n\}$ be an orthonormal basis for the orthogonal complement $\operatorname{Span}(v_1,v_2)^{\perp}$ of the plane spanned by $v_1$ and $v_2$.

First we find an orthonormal basis for $\operatorname{Span}(v_1,v_2)$, let it be $\{w_1,w_2 \}$. One can simply use Gram–Schmidt process to obtain this basis as $$ w_1 = v_1,\quad w_2 = \frac{\Big(v_2-(v_2^{\rm T}w_1)w_1\Big)}{\|v_2-(v_2^{\rm T}w_1)w_1\|} \;. $$

Then the desired minimal rotation matrix $R$ can be found as

$$ R = \underbrace{\begin{bmatrix} \rule[-1ex]{0.5pt}{6.6ex} & \rule[-1ex]{0.5pt}{6.5ex} & \rule[-1ex]{0.5pt}{6.6ex} & & \rule[-1ex]{0.5pt}{6.6ex} \\ w_1 & w_2 & r_3 & \dots & r_n \\ \rule[-1ex]{0.5pt}{6.6ex} & \rule[-1ex]{0.5pt}{6.6ex} & \rule[-1ex]{0.5pt}{6.5ex} & & \rule[-1ex]{0.5pt}{6.6ex}\end{bmatrix}}_{\Pi} \underbrace{\begin{bmatrix} \cos(\theta) & -\sin(\theta) & & \\ \sin(\theta) & \phantom{-}\cos(\theta) & & \\ & & 1 & \\ & & & \ddots & \\ & & & & 1\end{bmatrix}}_{S} \underbrace{\begin{bmatrix} \rule[-1ex]{0.5pt}{6.6ex} & \rule[-1ex]{0.5pt}{6.5ex} & \rule[-1ex]{0.5pt}{6.6ex} & & \rule[-1ex]{0.5pt}{6.6ex} \\ w_1 & w_2 & r_3 & \dots & r_n \\ \rule[-1ex]{0.5pt}{6.6ex} & \rule[-1ex]{0.5pt}{6.6ex} & \rule[-1ex]{0.5pt}{6.5ex} & & \rule[-1ex]{0.5pt}{6.6ex}\end{bmatrix}^{\rm T}}_{\Pi^{\rm T}}, $$

where $\theta=\arccos(v_1^{\rm T}v_2)$ if $\{w_1,w_2\}$ were obtatined with Gram-Schmidt process, but in general $\theta=\pm\arccos(v_1^{\rm T}v_2)$. The sign is "$+$" if the orientation of $\{w_1,w_2\}$ is the same as the orientation of $\{v_1,v_2\}$, and the sign is "$-$" otherwise. Note that Gram-Schmidt process preserves orientation and gives us a "$+$".

Note that $\Pi \in O(n)$ and $S \in SO(n)$ and thus $$R^{}=\Pi^{} S^{} \Pi^{\rm T} = \Pi^{} S^{} \Pi^{-1} \in S^{}O^{}(n)$$ is a rotation.

Let $\{e_i\}_{i=1}^{n}$ denote a standard basis in $\mathbb{R}^n$. One can see that

$$ R^{}r_i = \Pi^{} S^{} \Pi^{\rm T} r_i = \Pi^{} S^{} e_i = \Pi^{} e_i = r_i , $$

which means that $R$ acts as an identity map on $\operatorname{Span}(v_1,v_2)^{\perp}$.

Both $v_1$, $v_2$ are sent by $\Pi ^{\rm T}$ to $\operatorname{Span}(e_1,e_2)$, then rotated by $S$ through an angle $\theta$ and then sent back by $\Pi$. The fact that $\Pi$ is an orthogonal matrix guarantees that the angle between $\Pi ^{\rm T}v_1$ and $\Pi ^{\rm T}v_2$ is the same as the angle between $v_1$ and $v_2$ and we have

$$ S^{}\Pi ^{\rm T}v_1=\Pi ^{\rm T}v_2 $$

$$ v_2=\Pi^{}S^{}\Pi ^{\rm T}v_1 $$

$$ v_2=Rv_1 $$

Answer 2

Too long for a comment... Let $a^{\top}$ be the projection of $a$ onto the span of $u$ and $v$ such that $a=a^{\top}+a^{\perp}$. Then rotate $a^{\top}$ by $\theta$ and add $a^{\perp}$ again.

Alternatively: Extend $\{u,v\}$ to a basis. We want $u$ to be mapped to $v$ and $v$ mapped to $2\langle u,v\rangle v-u$. In respect to this basis the matrix of $R$ is the identity except for the upper left $2\times2$ entries; they read $$\begin{pmatrix}0&-1\\ 1&2\langle u,v\rangle\end{pmatrix}.$$

Even more alternatively: call $c=\langle u,v\rangle$. Now compute as usual — with a little help of the Gram-matrix of $u$ and $v$ — the tangential component $$a^{\top}=\frac{1}{1-c^2}\bigl((\langle u,a\rangle-c\langle v,a\rangle)u +(\langle v,a\rangle-c\langle u,a\rangle)v\bigr).$$

To obtain $Ra^{\top}$ change $u$ to $v$ and $v$ to $2cv-u$. Now compute $$Ra=a^{\perp}+Ra^{\top}=a-a^{\top}+Ra^{\top} =a-\frac{1}{1+c}\bigl(\langle u+v,a\rangle u+\langle v-(1+2c)u,a\rangle v\bigr).$$