$S^n = \{x\in \mathbb{R}^{n+1} : \|x\| = 1\}.$
I am using this definition: isometry is a surjective function $f:M \rightarrow N$ between two metric spaces $(M,d)$ and $(N,\rho)$ such that $$\rho (f(x),f(y)) = d(x,y) \quad \forall x,y \in M $$
My attempt:
Let $v = a-b$. Since $\|a\| = \|b\|$, the following Householder transformation maps $a$ to $b$ and vice versa:
$$H = I -2\frac{v v^t}{v^tv}$$ We have: $$H \cdot a= a -2\frac{(a-b) (a-b)^ta}{(a-b)^t(a-b)} = \frac{a(a-b)^t(a-b)-2(a-b)(a-b)^ta}{(a-b)^t(a-b)}$$ $$= \frac{aa^ta - aa^tb - ab^ta + ab^tb-2aa^ta + 2ab^ta+2ba^ta -2bb^ta}{(a-b)^t(a-b)}$$ $$= \frac{aa^ta - 2ab^ta + ab^tb-2aa^ta + 2ab^ta+2ba^ta -2bb^ta}{(a-b)^t(a-b)}$$ $$= \frac{- 2ab^ta + 2ab^ta+2ba^ta -2bb^ta}{(a-b)^t(a-b)} = \frac{2ba^ta -2bb^ta}{(a-b)^t(a-b)}= \frac{ba^ta + bb^tb -2bb^ta}{(a-b)^t(a-b)}$$ $$= \frac{b(a-b)^t(a-b)}{(a-b)^t(a-b)} = b$$
Householder transformations are orthogonal, then $H$ preserves norm and distance. Thus $f(x) = H\cdot x$ maps $S^n$ to $S^n$ and preserves distance. If $f$ is a surjective function, then $f$ is an isometry.
My questions are:
1) Is the function $H: S^n \rightarrow S^n$ surjective ?
2) Is it possible to solve this problem using rotations instead of reflections? How could I find a rotation matrix $Q$ such that $Q\cdot a = b$?
Thanks in advance.
Since $H^t$ is also orthogonal then $w=H^tv\in S^n$ for every $v\in S^n$ then $Hw=HH^tv=v$. Thus $H:S^n\rightarrow S^n$ is surjective. The answer to the second question is YES. Consider the plane spanned by $a,b$ and the linear transformation that rotates $a$ to $b$ and restricted to the orthogonal complement of this plane is the identity.