Find a rotation matrix that sends $v$ to $u$

Question

I know we have closed formulas for $\mathbb{R}^3$. I am looking for arbitrary $N$.

Given $\mathbb{R}^N$, find an orthogonal matrix $U$ that sends a unit-vector $u$ to unit-vector $v$.

Is there a closed form expression?

Answer 1

Let $w\in\mathbb{R}^n\setminus\{0\}$. The linear application associated to the matrix $$ U_{w}=I-2\frac{w w^T}{\|w\|^2}$$ maps $w$ into $-w$ and every element of $w^{\perp}$ into itself. $U_w$ is obviously an orthogonal matrix.
Check the linear application associated to $U_{u-v}$ to prove it maps $u$ into $v$ and vice-versa.
Note: it is crucial for $u$ and $v$ to have the same length; otherwise they cannot be mapped one into another by an isometry.

Keyword: Householder transformation.

Answer 2

Geometric Algebra Rotors represent rotations in n-dimensional Euclidean space. They can be seen as a generalization of Quaternions. But actually more powerful since they can represent rotations in Non-Euclidean geometries too (see Conformal Geometric Algebra). In 3D Geometric Algebra they are isomorphic to Hamilton Quaternions, in 2D Geometric Algebra they are isomorphic to complex numbers. In nD Geometric Algebra they are just nD rotations.

You can easily construct a nD Geometric Algebra based on a nD real vector space. All the formulations valid for 3D Geometric Algebra are valid for nD Geometric Algebra. For instance, 3D Rotors define rotation in terms of a 3D bivector angle, analogous to that, the nD rotors define rotation in terms of a nD-bivector angle, which is an hyperplane of the nD space. etc.

Answer 3

Let $T:\mathbb R^n\longrightarrow \mathbb R^n$ be a linear transformation and $\beta=\{b_1,b_2,\dots,b_n\}$,$\gamma=\{c_1,c_2,\dots,c_n\}$ be bases for $\mathbb R^n$. The matrix $[T]_{\beta\leftarrow\gamma}$ that represents the transformation $T$ from $\mathbb R^n$ in base $\gamma$ to $\mathbb R^n$ in base $\beta$ can be obtained as follows:

$\qquad$The $i$-th column of $[T]_{\beta\leftarrow\gamma}$ is given by $Tc_i$ written in base $\beta$.

This representation has some pretty intuitive properties. For instance

$$[T_2]_{\xi\leftarrow\beta}[T_1]_{\beta\leftarrow\gamma}=[T_2T_1]_{\xi\leftarrow\gamma}$$

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Now, I suppose your matrix $U$ is meant to be interpreted as some matrix $[T]_{\delta\leftarrow\delta}$, where $\delta$ is the canonical basis of $\mathbb{R}^n$. To make things absolutely clear, $\delta=\{e_1,\dots,e_n\}$ and $e_i$ is the vector with $1$ in its $i$-th coordinate and $0$ in all others.

We can try something like

$$U=[T_2]_{\delta\leftarrow\alpha_2}[M]_{\alpha_2\leftarrow\alpha_1}[T_1]_{\alpha_1\leftarrow\delta},$$

where $\alpha_1$, $\alpha_2$, $T_1$, $M$ and $T_2$ are to be defined. That's a lot of moving parts, but as long as $Uu=v$ and $T_1$, $M$ and $T_2$ are all orthogonal, we should be good. Well, one super easy orthogonal transformation is the identity, so we'll try

$$U=[I]_{\delta\leftarrow\alpha_2}[M]_{\alpha_2\leftarrow\alpha_1}[I]_{\alpha_1\leftarrow\delta},$$

Now all that's left is to define $\alpha_1$, $M$ and $\alpha_2$ such that $Uu=v$. Clearly, this reduces to $Mu=v$.

We can hence do as follows. Choose $\alpha_1=\{u_1=u,u_2,\dots,u_n\}$ and $\alpha_2=\{v_1=v,v_2,\dots,v_n\}$ so that they are orthonormal bases of $\mathbb R^n$. Define $Mu_i=v_i$. This definition makes $M$ automatically orthogonal, and moreover notice that $[M]_{\alpha_2\leftarrow\alpha_1}$ is the identity (as a matrix of numbers).

We're hence left with

$$U=[I]_{\delta\leftarrow\alpha_2}[I]_{\alpha_1\leftarrow\delta},$$

which solves our problem. Specifically, $[I]_{\alpha_1\leftarrow\delta}$ is the matrix whose $i$-th column is $e_i$ written in base $\alpha_1$, and $[I]_{\delta\leftarrow\alpha_2}$ is the matrix whose $i$-th column is $v_i$ written in the canonical base $($base $\delta)$.