[proofplan] We decompose $\mathbb{R}^n$ as the orthogonal direct sum $\operatorname{span}(u)\oplus u^\perp$. Every orthogonal matrix fixing $u$ preserves $u^\perp$, so restriction gives a homomorphism from the stabiliser $O(n)_u$ to the [orthogonal group](/page/Orthogonal%20Group) of $u^\perp$. Conversely, every orthogonal map of $u^\perp$ extends uniquely to an orthogonal map of $\mathbb{R}^n$ by fixing $u$. Finally, choosing an [orthonormal basis](/page/Orthonormal%20Basis) of $u^\perp$ identifies that orthogonal group with $O(n-1)$. [/proofplan]

[step:Decompose $\mathbb{R}^n$ into the line through $u$ and its orthogonal complement] Define \begin{align*} u^\perp:=\{v\in\mathbb{R}^n:\langle v,u\rangle=0\}. \end{align*} Since $u\in S^{n-1}$, we have $|u|=1$, hence $u\ne 0$. For each $x\in\mathbb{R}^n$, define \begin{align*} \alpha_x:=\langle x,u\rangle\in\mathbb{R} \end{align*} and \begin{align*} w_x:=x-\alpha_x u\in\mathbb{R}^n. \end{align*} Then \begin{align*} \langle w_x,u\rangle=\langle x,u\rangle-\alpha_x\langle u,u\rangle=\alpha_x-\alpha_x|u|^2=0, \end{align*} so $w_x\in u^\perp$, and \begin{align*} x=\alpha_xu+w_x. \end{align*} If $x=\alpha u+w$ with $\alpha\in\mathbb{R}$ and $w\in u^\perp$, then taking the [inner product](/page/Inner%20Product) with $u$ gives \begin{align*} \langle x,u\rangle=\alpha\langle u,u\rangle+\langle w,u\rangle=\alpha, \end{align*} so $\alpha=\alpha_x$ and $w=w_x$. Thus \begin{align*} \mathbb{R}^n=\operatorname{span}(u)\oplus u^\perp \end{align*} as an orthogonal direct sum. [/step]

[step:Restrict every stabilising orthogonal matrix to $u^\perp$]Let $O(u^\perp)$ denote the group of linear maps $B:u^\perp\to u^\perp$ satisfying \begin{align*} \langle Bw,Bz\rangle=\langle w,z\rangle \end{align*} for all $w,z\in u^\perp$. Define \begin{align*} \Psi:O(n)_u\to O(u^\perp) \end{align*} by \begin{align*} \Psi(A)=A|_{u^\perp}. \end{align*} We first check that $\Psi$ is well-defined. Let $A\in O(n)_u$ and let $w\in u^\perp$. Since $A\in O(n)$ preserves inner products by [citetheorem:8281] and since $Au=u$, we have \begin{align*} \langle Aw,u\rangle=\langle Aw,Au\rangle=\langle w,u\rangle=0. \end{align*} Thus $Aw\in u^\perp$, so $A|_{u^\perp}:u^\perp\to u^\perp$ is a [linear map](/page/Linear%20Map). For $w,z\in u^\perp$, again using preservation of inner products, \begin{align*} \langle A w,A z\rangle=\langle w,z\rangle. \end{align*} Therefore $A|_{u^\perp}\in O(u^\perp)$. The map $\Psi$ is a [group homomorphism](/page/Group%20Homomorphism). If $A,C\in O(n)_u$, then for every $w\in u^\perp$, \begin{align*} \Psi(AC)(w)=(AC)w=A(Cw)=\Psi(A)(\Psi(C)(w)). \end{align*} Hence \begin{align*} \Psi(AC)=\Psi(A)\Psi(C). \end{align*}[/step]

[guided]The point of passing to $u^\perp$ is that an orthogonal matrix fixing $u$ cannot mix vectors perpendicular to $u$ into the $u$-direction. We make that precise as follows. Define \begin{align*} u^\perp:=\{v\in\mathbb{R}^n:\langle v,u\rangle=0\}. \end{align*} Let $A\in O(n)_u$, meaning $A\in O(n)$ and $Au=u$. Let $w\in u^\perp$. To prove that $Aw$ is still in $u^\perp$, we must prove $\langle Aw,u\rangle=0$. Since $Au=u$, we may replace $u$ by $Au$, and since $A$ is orthogonal it preserves inner products by [citetheorem:8281]. Therefore \begin{align*} \langle Aw,u\rangle=\langle Aw,Au\rangle=\langle w,u\rangle=0. \end{align*} Thus $Aw\in u^\perp$. Now define $O(u^\perp)$ to be the group of linear maps $B:u^\perp\to u^\perp$ preserving the restricted Euclidean inner product: \begin{align*} \langle Bw,Bz\rangle=\langle w,z\rangle \end{align*} for all $w,z\in u^\perp$. The restriction $A|_{u^\perp}:u^\perp\to u^\perp$ is linear because $A:\mathbb{R}^n\to\mathbb{R}^n$ is linear, and it preserves the inner product because $A$ preserves the inner product on all of $\mathbb{R}^n$. Hence $A|_{u^\perp}\in O(u^\perp)$. This gives a map \begin{align*} \Psi:O(n)_u\to O(u^\perp) \end{align*} defined by \begin{align*} \Psi(A)=A|_{u^\perp}. \end{align*} It respects multiplication because restriction commutes with composition: for $A,C\in O(n)_u$ and $w\in u^\perp$, \begin{align*} \Psi(AC)(w)=(AC)w=A(Cw)=\Psi(A)(\Psi(C)(w)). \end{align*} So $\Psi(AC)=\Psi(A)\Psi(C)$, and $\Psi$ is a group homomorphism.[/guided]

[step:Extend every orthogonal map of $u^\perp$ by fixing $u$] For $B\in O(u^\perp)$, define a linear map \begin{align*} T_B:\mathbb{R}^n\to\mathbb{R}^n \end{align*} by the rule \begin{align*} T_B(\alpha u+w)=\alpha u+Bw \end{align*} for the unique decomposition $\alpha\in\mathbb{R}$ and $w\in u^\perp$ from the orthogonal direct sum above. Let $\alpha,\beta\in\mathbb{R}$ and $w,z\in u^\perp$. Since $w$ and $z$ are orthogonal to $u$, and since $B$ preserves the inner product on $u^\perp$, we have \begin{align*} \langle T_B(\alpha u+w),T_B(\beta u+z)\rangle=\langle \alpha u+Bw,\beta u+Bz\rangle. \end{align*} Expanding by bilinearity of the Euclidean inner product gives \begin{align*} \langle \alpha u+Bw,\beta u+Bz\rangle=\alpha\beta\langle u,u\rangle+\alpha\langle u,Bz\rangle+\beta\langle Bw,u\rangle+\langle Bw,Bz\rangle. \end{align*} Because $Bw,Bz\in u^\perp$ and $|u|=1$, this becomes \begin{align*} \langle T_B(\alpha u+w),T_B(\beta u+z)\rangle=\alpha\beta+\langle w,z\rangle. \end{align*} On the other hand, \begin{align*} \langle \alpha u+w,\beta u+z\rangle=\alpha\beta+\langle w,z\rangle. \end{align*} Thus $T_B$ preserves the Euclidean inner product on $\mathbb{R}^n$. By [citetheorem:8281], $T_B\in O(n)$. Also \begin{align*} T_Bu=T_B(1u+0)=u, \end{align*} so $T_B\in O(n)_u$. Define \begin{align*} \Theta:O(u^\perp)\to O(n)_u \end{align*} by \begin{align*} \Theta(B)=T_B. \end{align*} If $B,C\in O(u^\perp)$, then for every $\alpha\in\mathbb{R}$ and $w\in u^\perp$, \begin{align*} T_BT_C(\alpha u+w)=T_B(\alpha u+Cw)=\alpha u+B(Cw)=T_{BC}(\alpha u+w). \end{align*} Hence $\Theta(BC)=\Theta(B)\Theta(C)$, so $\Theta$ is a group homomorphism. [/step]

[step:Verify the restriction and extension maps are inverse isomorphisms] Let $A\in O(n)_u$. For every $x\in\mathbb{R}^n$, write $x=\alpha u+w$ with $\alpha\in\mathbb{R}$ and $w\in u^\perp$. Since $Au=u$ and $\Psi(A)=A|_{u^\perp}$, we have \begin{align*} (\Theta\circ\Psi)(A)(x)=T_{\Psi(A)}(\alpha u+w)=\alpha u+Aw=A(\alpha u+w)=Ax. \end{align*} Thus $(\Theta\circ\Psi)(A)=A$. Conversely, let $B\in O(u^\perp)$. For every $w\in u^\perp$, \begin{align*} (\Psi\circ\Theta)(B)(w)=\Psi(T_B)(w)=T_Bw=Bw. \end{align*} Thus $(\Psi\circ\Theta)(B)=B$. Therefore $\Psi$ and $\Theta$ are inverse group isomorphisms, and \begin{align*} O(n)_u\cong O(u^\perp). \end{align*} [/step]

[step:Identify $O(u^\perp)$ with $O(n-1)$ after choosing an orthonormal basis] Since $u^\perp$ is an $(n-1)$-dimensional real [inner product space](/page/Inner%20Product%20Space), choose an orthonormal basis $e_1,\dots,e_{n-1}$ of $u^\perp$. Define the linear isometry \begin{align*} E:\mathbb{R}^{n-1}\to u^\perp \end{align*} by \begin{align*} E(a_1,\dots,a_{n-1})=\sum_{i=1}^{n-1}a_i e_i. \end{align*} For $M\in O(n-1)$, define \begin{align*} \Gamma(M):=E M E^{-1}:u^\perp\to u^\perp. \end{align*} Because $E$ is a linear isometry and $M$ preserves the Euclidean inner product on $\mathbb{R}^{n-1}$, the map $\Gamma(M)$ preserves the inner product on $u^\perp$, so $\Gamma(M)\in O(u^\perp)$. The assignment $\Gamma:O(n-1)\to O(u^\perp)$ is a group isomorphism, with inverse $B\mapsto E^{-1}BE$. Combining the group isomorphisms \begin{align*} O(n)_u\cong O(u^\perp) \end{align*} and \begin{align*} O(u^\perp)\cong O(n-1), \end{align*} we obtain \begin{align*} O(n)_u\cong O(n-1). \end{align*} This proves the theorem. [/step]