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  "body": "\n## Proof\n\nWe argue by induction on $n$. A hyperplane of $V$ is a subspace of dimension $n - 1$.\n\n1. Let $V$ be a real inner product space with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$ and norm $\\norm{\\cdot}$. For $u \\in V$ with $u \\ne 0$, set $H_u = \\set{x \\in V : \\ip{x}{u} = 0}$. Over $\\R$, the inner product is symmetric.\n2. $H_u$ is the [[orthogonal complement|079D68S3AW]] of the line $\\R u$. If $\\dim V = n$, then $\\dim H_u = n - 1$ by the [[dimension formula|0E2NYPVSP9]]. So $H_u$ is a hyperplane.\n3. Claim: the [[reflection|025158VEYZ]] in $H_u$ is $R_u(x) = x - 2 \\frac{\\ip{x}{u}}{\\ip{u}{u}} u$. Set $c = \\ip{x}{u} / \\ip{u}{u}$ and $p = x - c u$.\n4. Then $\\ip{p}{u} = \\ip{x}{u} - c \\ip{u}{u} = 0$, so $p \\in H_u$. For $h \\in H_u$, $\\ip{c u}{h} = c \\ip{h}{u} = 0$, so $c u \\in H_u^\\perp$.\n5. Since $x = p + c u$ and the [[decomposition|0SP7P0G7KQ]] $V = H_u \\oplus H_u^\\perp$ is unique, the [[orthogonal projection|0Q8HZTJC7G]] is $P_{H_u}(x) = p$. So the reflection is $2p - x = x - 2 c u = R_u(x)$.\n6. From step 3, $R_u(u) = -u$ and $R_u(h) = h$ for $h \\in H_u$. Since $x = p + cu$, we get $R_u(R_u(x)) = R_u(p - cu) = p + cu = x$. So $R_u \\circ R_u = \\id$.\n7. For $n \\ge 0$, let $P(n)$ say: every linear isometry of an $n$-dimensional real inner product space is a composition of at most $n$ reflections in hyperplanes. Let $V$ have dimension $n$ and let $T \\colon V \\to V$ be a [[linear isometry|0717XXPWM6]].\n8. If $T = \\id$, then $T$ is the composition of zero reflections. This covers $n = 0$, where $V = \\set{0}$. Now let $n \\ge 1$, assume $P(n - 1)$, and let $T \\ne \\id$.\n9. Choose $v \\in V$ with $T(v) \\ne v$, and set $u = T(v) - v \\ne 0$. By [[preservation of inner products|0R7VXTYHYZ]], $\\ip{T(v)}{T(v)} = \\ip{v}{v}$. Write $a = \\ip{v}{v}$ and $b = \\ip{T(v)}{v}$.\n10. Then $\\ip{u}{u} = a - 2b + a = 2(a - b)$ and $\\ip{T(v)}{u} = a - b$. Since $u \\ne 0$, $a - b \\ne 0$.\n11. By step 3, $R_u(T(v)) = T(v) - 2 \\frac{a - b}{2(a - b)} u = T(v) - u = v$.\n12. Let $S = R_u \\circ T$. A reflection keeps norms, since $\\norm{p - cu}^2 = \\norm{p}^2 + c^2 \\norm{u}^2 = \\norm{p + cu}^2$ when $p \\perp u$. So $S$ is a linear isometry with $S(v) = v$.\n13. Since $T(0) = 0$, we have $v \\ne 0$. Let $E = H_v$, a hyperplane by step 2. For $x \\in E$, preservation of inner products by $S$ gives $\\ip{S(x)}{v} = \\ip{S(x)}{S(v)} = \\ip{x}{v} = 0$. So $S(E) \\subseteq E$.\n14. The restriction $S|_E \\colon E \\to E$ is a linear isometry of the real inner product space $E$, of dimension $n - 1$.\n15. By $P(n-1)$, $S|_E = R'_1 \\circ \\cdots \\circ R'_k$ with $k \\le n - 1$, where $R'_j$ is the reflection of $E$ in a hyperplane $G_j$ of $E$.\n16. Fix $j$. The orthogonal complement of $G_j$ in $E$ has dimension $(n - 1) - (n - 2) = 1$ by the dimension formula. Choose $w_j \\ne 0$ in it.\n17. Then $G_j \\subseteq G'_j = \\set{x \\in E : \\ip{x}{w_j} = 0}$, and $\\dim G'_j = n - 2$ by step 2 in $E$. A [[subspace of full dimension|0WGAXEK99E]] is the whole space, so $G_j = G'_j$.\n18. By steps 3 to 5 applied in $E$, $R'_j(x) = x - 2 \\frac{\\ip{x}{w_j}}{\\ip{w_j}{w_j}} w_j$ for $x \\in E$.\n19. Let $R_j = R_{w_j}$ be the reflection of $V$ in $H_{w_j}$, a hyperplane of $V$. By step 3 and step 18, $R_j(x) = R'_j(x)$ for $x \\in E$. Also $R_j(v) = v$, since $\\ip{v}{w_j} = 0$.\n20. Each $R_j$ maps $E$ into $E$. So $R_1 \\circ \\cdots \\circ R_k$ agrees with $S$ on $E$, and both fix $v$.\n21. By the [[decomposition|0SP7P0G7KQ]] for the line $\\R v$, each $x \\in V$ is $t v + e$ with $t \\in \\R$ and $e \\in E$, as $E = (\\R v)^\\perp$. Two linear maps that agree on $v$ and on $E$ are equal.\n22. Hence $S = R_1 \\circ \\cdots \\circ R_k$. By step 6, $T = R_u \\circ R_u \\circ T = R_u \\circ S = R_u \\circ R_1 \\circ \\cdots \\circ R_k$.\n23. This is a composition of $k + 1 \\le n$ reflections in hyperplanes of $V$. This proves $P(n)$, and the theorem follows by induction.\n\n## Idea\n\nIf $T$ moves a vector $v$, the reflection in the hyperplane orthogonal to $T(v) - v$ swaps $T(v)$ and $v$, because they have the same length. Composing with it gives an isometry that fixes $v$, and hence maps the hyperplane orthogonal to $v$ to itself. Induction on that hyperplane gives at most $n - 1$ more reflections.\n",
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