Proof of Alignment of Orthonormal Families by Plane Rotations

Throughout, $\sqrt{\,\cdot\,}$ is the positive square root, and we use repeatedly that finite compositions of plane rotations preserve dot products and orthonormal families, by Plane Rotations Preserve the Dot Product.

Step 1 (single-vector alignment). Let $w\in\mathbb{R}^{n}$ with $w\cdot w=1$. Then there are a finite composition $h$ of plane rotations and $\varepsilon\in\{1,-1\}$ with $h(w)=\varepsilon\,e_1$; and if $n\ge2$, one may take $\varepsilon=1$.

If $n=1$, then $w=(w_1)$ with $w_1^{2}=1$, so $(w_1-1)(w_1+1)=0$ and $w_1=\pm1$; take $h$ the empty composition and $\varepsilon=w_1$. Let $n\ge2$. Process the coordinates $k=2,\dots,n$ in order, maintaining a current vector $u$, initially $w$; each applied rotation preserves $u\cdot u=1$ by Plane Rotations Preserve the Dot Product. At stage $k$: if $u_k=0$, apply no rotation; otherwise set $r=\sqrt{u_1^{2}+u_k^{2}}>0$ and apply the plane rotation in coordinates $(1,k)$ with parameters $a=u_1/r$, $b=u_k/r$ (so $a^{2}+b^{2}=1$): the new first coordinate is $a\,u_1+b\,u_k=(u_1^{2}+u_k^{2})/r=r>0$, the new $k$-th coordinate is $-b\,u_1+a\,u_k=(-u_ku_1+u_1u_k)/r=0$, and all other coordinates are unchanged; in particular the coordinates $2,\dots,k-1$, already $0$, remain $0$. After stage $n$ the current vector is $(u_1,0,\dots,0)$ with $u_1^{2}=1$, so $u_1=\pm1$. If $u_1=-1$, apply one further plane rotation, in coordinates $(1,2)$ with parameters $(-1,0)$, which maps $(-1,0,\dots,0)$ to $(1,0,\dots,0)$. The resulting composition $h$ satisfies $h(w)=e_1$.

Step 2 (induction on $p$). We prove parts 1-3 by induction on $p$, for all $n$ simultaneously.

Base $p=1$. Part 1 holds since $n\ge1$. Parts 2 and 3 are exactly Step 1 (part 3 concerns $p<n$, which forces $n\ge2$).

Step from $p-1$ to $p$, $p\ge2$. First, $n\ge2$: if $n=1$, then $w_1=(w_{11})$ with $w_{11}^{2}=1$, and $w_1\cdot w_2=w_{11}w_{21}=0$ forces $w_{21}=0$, contradicting $w_2\cdot w_2=1$. By Step 1 with $n\ge2$ there is a finite composition $h_1$ with $h_1(w_1)=e_1$. Set $w'_k=h_1(w_k)$ for $1\le k\le p$; by Plane Rotations Preserve the Dot Product this is an orthonormal family, and for $k\ge2$ its first coordinate is $w'_k\cdot e_1=w'_k\cdot w'_1=w_k\cdot w_1=0$. Write $w'_k=(0,v_k)$ with $v_k\in\mathbb{R}^{n-1}$ for $2\le k\le p$; the pairwise dot products of $v_2,\dots,v_p$ equal those of $w'_2,\dots,w'_p$ (the omitted first coordinates vanish), so $v_2,\dots,v_p$ is an orthonormal family of $p-1$ vectors in $\mathbb{R}^{n-1}$.

By the induction hypothesis: $p-1\le n-1$, hence $p\le n$, proving part 1; and there are a finite composition $h'$ of plane rotations of $\mathbb{R}^{n-1}$ and $\varepsilon\in\{1,-1\}$ with $h'(v_k)=e'_{k-1}$ for $2\le k\le p-1$ and $h'(v_p)=\varepsilon\,e'_{p-1}$, where $e'_1,\dots,e'_{n-1}$ are the standard basis vectors of $\mathbb{R}^{n-1}$; and $\varepsilon=1$ may be arranged when $p-1<n-1$, that is, when $p<n$.

Embedding. For a plane rotation $g'$ of $\mathbb{R}^{n-1}$ in coordinates $(i,j)$ with parameters $(a,b)$, let $\widehat{g}$ be the plane rotation of $\mathbb{R}^{n}$ in coordinates $(i+1,j+1)$ with the same parameters. For every real $s$ and $v\in\mathbb{R}^{n-1}$,

since the first coordinate of $\mathbb{R}^{n}$ is untouched ($i+1\ge2$) and the remaining coordinates transform exactly as under $g'$. Let $\widehat{h}$ be the composition of the embedded rotations of $h'$, taken in the same order; iterating the display gives $\widehat{h}\bigl((s,v)\bigr)=(s,\,h'(v))$.

Conclusion. Set $h=\widehat{h}\circ h_1$. Since every plane rotation maps the zero vector to itself, $h'(0)=0$, so $h(w_1)=\widehat{h}\bigl((1,0)\bigr)=(1,h'(0))=e_1$. For $2\le k\le p-1$, $h(w_k)=\widehat{h}\bigl((0,v_k)\bigr)=(0,e'_{k-1})=e_k$; and $h(w_p)=(0,\varepsilon\,e'_{p-1})=\varepsilon\,e_p$, with $\varepsilon=1$ whenever $p<n$. This proves parts 2 and 3.

Step 3 (the sign obstruction). Let $n=2$ and define $D(u,v)=u_1v_2-u_2v_1$ for $u,v\in\mathbb{R}^{2}$. For a plane rotation $g$ with parameters $(a,b)$,

after expanding and cancelling the terms with coefficient $ab$. Hence $D$ is invariant under every finite composition $h$ of plane rotations. Since $D(e_1,-e_2)=-1$ and $D(e_1,e_2)=1$, no such $h$ maps the orthonormal family $(e_1,-e_2)$ to $(e_1,e_2)$. $\blacksquare$