[proofplan] We prove transitivity by taking arbitrary unit vectors $x,y\in S^{n-1}$ and constructing an orthogonal matrix carrying $x$ to $y$. The construction extends each of $x$ and $y$ to an [orthonormal basis](/page/Orthonormal%20Basis) of $\mathbb{R}^n$, then defines the [linear map](/page/Linear%20Map) sending the first basis to the second. The resulting matrix is orthogonal because its columns form an orthonormal basis, and its first-column action gives $Ax=y$. [/proofplan] [step:Extend each unit vector to an orthonormal basis] Let $x,y\in S^{n-1}$ be arbitrary. Since $|x|=|y|=1$, both vectors are nonzero. We use the following finite-dimensional inner-product fact: every unit vector in $\mathbb{R}^n$ extends to an orthonormal basis of $\mathbb{R}^n$. Applying this fact to $x$ and to $y$, choose orthonormal bases \begin{align*} (x_1,\dots,x_n) \end{align*} and \begin{align*} (y_1,\dots,y_n) \end{align*} of $\mathbb{R}^n$ such that $x_1=x$ and $y_1=y$. [guided] We need an orthogonal matrix that sends $x$ to $y$. A useful way to build such a matrix is not to prescribe only one vector, but to prescribe its action on a full basis. Because $x,y\in S^{n-1}$, the definition of the unit sphere gives \begin{align*} |x|=1 \end{align*} and \begin{align*} |y|=1. \end{align*} In particular, neither vector is zero. By the standard finite-dimensional inner-product fact that every unit vector in $\mathbb{R}^n$ extends to an orthonormal basis, there exist orthonormal bases \begin{align*} (x_1,\dots,x_n) \end{align*} and \begin{align*} (y_1,\dots,y_n) \end{align*} of $\mathbb{R}^n$ with first vectors \begin{align*} x_1=x \end{align*} and \begin{align*} y_1=y. \end{align*} This is the point where the geometry of the sphere enters: unit vectors are exactly the vectors that may appear as members of an orthonormal basis without rescaling. [/guided] [/step] [step:Build the matrix sending the first orthonormal basis to the second] Define matrices $P,Q\in M_n(\mathbb{R})$ by taking their columns to be the chosen basis vectors: \begin{align*} P=\begin{pmatrix} x_1 & \cdots & x_n \end{pmatrix} \end{align*} and \begin{align*} Q=\begin{pmatrix} y_1 & \cdots & y_n \end{pmatrix}. \end{align*} Since the columns of $P$ are orthonormal, for all $1\le i,j\le n$ we have \begin{align*} (P^\top P)_{ij}=\langle x_i,x_j\rangle. \end{align*} Thus $P^\top P=I_n$. Likewise, $Q^\top Q=I_n$. Define \begin{align*} A:=QP^\top\in M_n(\mathbb{R}). \end{align*} Then \begin{align*} AP=QP^\top P=Q. \end{align*} Therefore $A$ sends the $i$th column of $P$ to the $i$th column of $Q$ for every $1\le i\le n$. [/step] [step:Verify that the constructed matrix is orthogonal] Since $P^\top P=I_n$, the matrix $P$ is invertible with inverse $P^\top$, so $PP^\top=I_n$. Similarly, $Q^\top Q=I_n$. Hence \begin{align*} A^\top A=(QP^\top)^\top(QP^\top)=PQ^\top QP^\top. \end{align*} Using $Q^\top Q=I_n$ gives \begin{align*} A^\top A=P I_n P^\top=PP^\top=I_n. \end{align*} Therefore $A\in O(n)$. [/step] [step:Evaluate the matrix on the chosen point] Because $x=x_1$ is the first column of $P$ and $y=y_1$ is the first column of $Q$, the identity $AP=Q$ gives \begin{align*} Ax=Ax_1=y_1=y. \end{align*} Thus for arbitrary $x,y\in S^{n-1}$ there exists $A\in O(n)$ such that $Ax=y$. This is exactly transitivity of the action of $O(n)$ on $S^{n-1}$. [/step]