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Is the action of $O(n,\mathbb{R})$ on ${S}^{n-1}$ transitive? I think this is true as orthogonal matrices are supposed to rotate and keep the length fixed, but how do I prove this?

EDIT: Based on Alexander's comment, if $x = (x_{1}, \ldots, x_{n}) \in S^{n-1}$, then I can construct a matrix $A$ whose first row is the vector $x$ and whose other rows can be obtained by completing it to an orthonormal basis. But does it ensure that columns are also orthogonal?

Vishal Gupta
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    Given a point $x \in S^{n-1}$, maybe try constructing a matrix that takes $x$ to some point $(1, 0,...,0)$. If you can show this for any such $x$, this proves transitivity. – Alexander Aug 11 '13 at 18:37
  • Answer to your edit question: That the columns of a matrix $A$ are orthogonal, is precisely the condition $AA^T=I$. Transposing both sides, we get $(AA^T)^T=AA^T=I$, so the answer to your question is yes. – Fredrik Meyer Aug 11 '13 at 18:51
  • @FredrickMeyer I think you mean that columns are orthogonal precisely means that $A^{T}A = I$ and rows are orthogonal means $AA^{T} = I$. So transposing the condition for rows gives the condition for columns. – Vishal Gupta Aug 11 '13 at 18:54
  • @Alexander How do we show that there is a matrix that takes $x$ to $(1,0,...,0)$?

    The answer below shows how to obtain a matrix in $O(n)$ that takes $(1,0,...,0)$ to $x \in S^{n-1}$, for any $x$.

     – Saikat Aug 25 '22 at 20:41
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    @Alexander Ok I think I got the answer to my question. Since $A(1,0,...,0)^t = x^t$, we have $A^tA(1,0,...,0)^t = (1,0,...,0)^t = A^t x^t$, and as $A^t \in O(n)$, we are done. – Saikat Aug 25 '22 at 20:52
  • @Saikat Sounds good! – Alexander Aug 25 '22 at 20:54

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A very closely related recent question, where it is shown that you can rotate any point to the South Pole. As you are not picky about the sign of the determinant let me summarize: Let $A$ and $B$ be two distinct points on the sphere. What happens if you reflect the sphere w.r.t. the (hyper)plane $H$ that has $\vec{AB}$ as its normal and bisects the line segment $AB$?

If $\vec{n}=\vec{AB}$, then the formula for the said reflection is $$ \vec{x}\mapsto \vec{x}-2\frac{\vec{x}\cdot\vec{n}}{\Vert\vec{n}\Vert^2}\,\vec{n}. $$ With $\vec{n}$ known finding the matrix of this reflection is straightforward.

Community
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Jyrki Lahtonen
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With the said alexander:Let $x$ be a arbitary point in $S^{n-1}$,we know there exist a orthonormal basis in $\mathbb{R^n},\alpha_1,\alpha_2,...,\alpha_n$ and $\alpha_1=x$,now consider the linear map $T:\mathbb{R^n}\to \mathbb{R^n}$ s,t $e_i\mapsto\alpha_i(e_i$is standard basis which is also orthonormal).

Call the matrix of $T$,$A$.$A$ is orthonormal and$A{(1,0,...,0)}^t=x^t$.here PROPOSITION 10C part(b)

R Salimi
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  • Do we also need to ensure that any $x \in S^n$ can also be mapped to $(1,0,...,0)$ via some orthogonal matrix? – Saikat Aug 25 '22 at 20:36