Sphere is exhausted by orthogonal transformations of fixed vector

Question

For each $ \ n \in \mathbb{N} \ $ let $ \, O(n+1) \, $ be the set of orthogonal linear operators $ \ \mathbb{R}^{n+1} \to \mathbb{R}^{n+1} \, $, $ \ e_{n+1} = (1,0,0,...,0) \in \mathbb{R}^{n+1} \ $ and $ \ S^n = \big\{ x \in \mathbb{R}^{n+1}: \Vert x \Vert = 1 \big\} \, $.

Prove that, for all $ \ y \in S^n$, there exists $ \ Q_y \in O(n+1) \ $ such that $ \ Q_y (e_{n+1}) = y \ $.

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I could prove it easily for $ \ n=0 \, $, since $ \ S^0 = \{ -1 , 1 \} \, $, $ \ e_1 = 1 \ $ and $ \ O(1) = \big\{ id_{\mathbb{R}} , -id_{\mathbb{R}} \big\} \ $.

For $ \ n=1 \ $ I used the theorem 8.7, item (d), of Walter Rudin's "Principles of mathematical analysis" 3rd edition, at page 183. For fixed $ \ y \in S^1$, there exists an unique $ \ t \in [0, 2 \pi[ \ $ such that $ \ E(it)=y \ $. Then I used the functions $ \ C,S: \mathbb{C} \to \mathbb{C} \ $ defined in this same section of the book to pick the restrictions $ \ c=C|_{[0, 2 \pi[} \ $ and $ \ s = S|_{[0, 2 \pi[} \ $ and build the function $ \ Q_y : \mathbb{R}^2 \to \mathbb{R}^2 \ $ such that $ \ Q_y (a,b) = \big( c_t a - s_t b , s_t a + c_t b \big) \, $, $ \, \forall (a,b) \in \mathbb{R}^2 \, $. Where $ \ c_t = c(t) \ $ and $ \ s_t = s(t) \ $. It is straightforward that $ \, Q_y \, $ is a linear transformation of $ \, \mathbb{R}^2$. A simple computation gives $ \ [C(z)]^2 + [S(z)]^2 =1 \, $, $ \, \forall z \in \mathbb{C} \, $. Using this we see that $ \ \Vert Q_y (a,b) \Vert = \Vert (a,b) \Vert \, $, $ \, \forall (a,b) \in \mathbb{R}^2 \, $. So $ \ Q_y \in O(2) \ $. And we got $$Q_y(e_2) = Q_y (1,0) = (c_t,s_t) = c_t + i \, s_t = C(t) + i \, S(t) = E(it) = y \ \ . $$

So I tried to apply some induction argument for $ \ n \geq 2 \, $, but I failed. Can anyone help?

Thanks

Answer 1

The vectors $e_{n+1}$ and $y$ span a two-dimensional space $W$. If you can find a $n+1 - 2$ dimensional space $V$ such that every vector in $V$ is orthogonal to every vector in $W$ then you are done: you can just take $Q_y$ to be the map that acts as the $Q_y$ you constructed in your post on $W \cong \mathbb{R}^2$ and acts as the identity on $V$.

Does this help?

Answer 2

It happened that I managed to answer that question in an unexpected way.

For each $ \ k \in \{ 2, 3, ... , n+1 \} \ $ let $ \ e_{(n+1,k)} = \big( \delta^1_k , \delta^2_k , ... , \delta^{n+1}_k \big) \ $ be the $k$-th vector of the canonical basis of $ \, \mathbb{R}^{n+1}$. That is, $\delta^i_j$ is the Kronecker delta. So $ \ C = \big\{ e_{n+1} , e_{(n+1,2)}, ..., e_{(n+1,n+1)} \big\} \ $ is the canonical basis of $\mathbb{R}^{n+1}$. For fixed $ \ y \in S^n$, there is a basis $B'$ of $ \, \mathbb{R}^{n+1}$ that contains it. By Gram-Schmidt process, orthonormalising $B'$, fixing $y$ as the first vector, there exists an orthonormal basis for $ \, \mathbb{R}^{n+1}$ that contains $y \, $, say $ \ B = \big\{ y , y_2 , y_3 , ... , y_{n+1} \big\} \ $. Define $ \ Q_y : \mathbb{R}^{n+1} \to \mathbb{R}^{n+1} \ $ as the unique linear extension such that $ \ Q_y (e_{n+1}) = y \ $ and $ \ Q_y \big( e_{(n+1,k)} \big) = y_k \ $, $ \, \forall k \in \{ 2,3,...,n+1 \} \ $. Since $B$ and $C$ are both orthonormal basis for $ \, \mathbb{R}^{n+1}$, we have $ \ Q_y \in O(n+1) \ $.