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Every Euclidean isometry of ${\mathbb{R}}^n$ is a product of at most $n+1$ operations of reflection in a hyperplane. An isometry fixing zero is orthogonal: the polarization identity preserves inner products, and its values on an orthonormal basis determine its linear action. An orthogonal transformation is a product of at most $n$ linear reflections, by reflecting the image of the first basis vector back to that vector and inducting on its orthogonal complement. For an arbitrary isometry, first reflect its image of zero back to zero, if necessary, and apply the orthogonal result. Affine reflection hyperplanes, rather than only hyperplanes through zero, are essential here.

For complementary vector subspaces $U,W$ of a real vector space, the rule $T(u+w)=u-w$ is a well-defined linear map satisfying $T^2=I$. It fixes $U$ pointwise and reverses $W$. In a basis formed by adjoining bases of $U$ and $W$, its matrix is $\mathrm{diag}(I,-I)$. Conversely every real linear map with $T^2=I$ has this form, with $U=\ker (T-I)$ and $W=\ker (T+I)$, as follows from $v=(v+Tv)/2+(v-Tv)/2$.

In a Euclidean space, such a linear map is an orthogonal transformation if and only if $U$ and $W$ are perpendicular. Necessity follows from $⟨u,w⟩=⟨Tu,Tw⟩=-⟨u,w⟩$; sufficiency follows by expanding the squared Euclidean norm of $u+w$. In three dimensions, a fixed plane with reversed complementary line is a possibly oblique reflection, and a fixed line with reversed complementary plane is a possibly oblique axial symmetry. Their diagonal forms alone do not imply a Euclidean reflection in a hyperplane or a Euclidean half-turn.

Use the standard ordered basis $({\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3)$ for both maps. The reflection in a hyperplane interchanges the second and third coordinates, so its matrix isOrient the rotation axis along $\mathbf{n}=(1,1,1)/\sqrt{3}$ and use the right-hand convention. The rotation matrix for angle $2\pi /3$ sends ${\mathbf{e}}_1$ to ${\mathbf{e}}_2$, ${\mathbf{e}}_2$ to ${\mathbf{e}}_3$, and ${\mathbf{e}}_3$ to ${\mathbf{e}}_1$. For example, the Rodrigues rotation formula givesMultiplying the rotation matrix by the dilation factor givesDirect multiplication gives $A^2=I$, $B^2=4\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)$ and $B^3=8I$. Hence $B^3=8A^2$. If the opposite axis orientation is chosen, $B=2R^T$ also has cube $8I$. Independently of coordinates, two applications of the reflection give the identity, while three applications of the rotation-dilation give a full turn and dilation by $2^3$. Every basis represents these resulting maps by $I$ and $8I$. Thus the equality even holds when the two matrices are expressed in different bases.

Let $T$ be the isometry. Since $T(0)=0$, it preserves norms, and the polarization identity applied to distances gives $⟨T(x),T(y)⟩=⟨x,y⟩$. For an orthonormal basis $e_1,e_2,e_3$, the vectors $T(e_i)$ are again an orthonormal basis, andExpanding in that basis proves $T(x)={\sum }_i⟨x,e_i⟩T(e_i)$. Hence $T$ is a linear map represented by an orthogonal matrix $Q$.

A reflection in a hyperplane with nonzero normal $v$ is $H_v(x)=x-2⟨x,v⟩v/⟨v,v⟩$. If $Q{e}_1\ne e_1$, choose $v=Q{e}_1-e_1$. Because $\Vert Q{e}_1\Vert =\Vert e_1\Vert$, substitution shows $H_{v}Q{e}_1=e_1$. If they agree, no reflection is needed. The resulting orthogonal transformation fixes $e_1$ and preserves its orthogonal complement. Apply the same construction there to fix $e_2$; the second normal is perpendicular to $e_1$, so the first vector remains fixed. Finally the restriction to the remaining line is either the identity or negation; in the latter case a third reflection, normal to $e_3$, fixes it. Reversing these steps writes $Q$ as a product of at most three reflections, each in a plane through the origin. This proves the three-dimensional case of the Cartan–Dieudonné theorem.

Three reflections are sometimes necessary. The map $-I$ is the product of the three coordinate-plane reflections. Its determinant is $-1$, excluding zero or two reflections. A single plane reflection fixes a two-dimensional vector subspace, whereas $-I$ fixes only zero, excluding one. Thus the bound is sharp.

An orthogonal matrix satisfies $R^{T}R=I$, so it preserves the Euclidean norm. For a nonzero real eigenvector $v$,Therefore $\boxed{\lambda =\pm 1}$. Geometrically, eigenvalue $1$ fixes every vector on the linear span of $v$; eigenvalue $-1$ reverses every vector on that line. In either case the line is preserved as a set.

For $M$, solving $(M-I)v=0$ gives $y=x$, $z=y$, $x=z$. For $N$, the equations are $-2y-2z=0$ and $-2z=0$, hence $y=z=0$. For $P$, let $a=(1,1,1{)}^T$ and $J=a{a}^T$, so that $P=I-\frac{2}{3}J$. Then $(P-I)v=0$ is equivalent to $x+y+z=0$. The resulting eigenspaces and their bases areSolving all the equations shows that these lists contain the maximum possible numbers of independent eigenvectors with eigenvalue $1$.

The columns of $M$ form an orthonormal basis, so $M$ is an orthogonal matrix. Also $\det M=1$. Its fixed axis is the line through $a$, and on the perpendicular plane it acts as a two-dimensional rotation. If $\theta$ denotes the rotation angle, its matrix trace is $1+2\cos \theta$. Since $\mathrm{tr}M=0$, we obtain $\cos \theta =-1/2$. To fix the orientation relative to $u=a/\sqrt{3}$, note that $M{e}_1=e_3$ andThe component of $e_1$ perpendicular to $u$ has squared length $2/3$, and this scalar triple product equals $(2/3)\sin \theta$. Hence $\sin \theta =-\sqrt{3}/2$. ThusEquivalently, its unsigned angle is $2\pi /3$, or its positive angle is $2\pi /3$ about the opposite oriented axis.

For $P$, the reflection in a hyperplane formula givesIt fixes the plane $a\cdot v=0$ pointwise and maps $a$ to $-a$, so $\boxed{P\text{ reflects in }x+y+z=0}$. These actions also show that $P$ is an orthogonal matrix. In contrast, the second column of $N$ has squared length $5$, so $N$ is not orthogonal and cannot represent either of these Euclidean motions.

A diagonalizable matrix must have a basis of eigenvectors over the field in question. For $M$, a direct determinant computation gives ${\chi }_M(t)=1-t^3$. Its eigenvalues are $1,\omega ,{\omega }^2$, where $\omega =e^{2\pi i/3}$. Only one is real, with one-dimensional eigenspace, so $M$ is not diagonalizable over $\mathbb{R}$. Over $\mathbb{C}$ the three eigenvalues are distinct, and the vectors $(1,\lambda ,{\lambda }^2{)}^T$ for $\lambda =1,\omega ,{\omega }^2$ give an eigenbasis. The upper triangular matrix $N$ has only eigenvalue $1$, of algebraic multiplicity $3$, but its eigenspace has dimension $1$ over either field. Hence $N$ is not diagonalizable over either field. For $P$, the two real plane vectors displayed above, together with $a$, form an eigenbasis, with eigenvalues $1,1,-1$. They remain independent over $\mathbb{C}$. Therefore

For unit $u$, put $A=I-2u{u}^T$. Then $A^T=A$ and $A^2=I$, so $A$ is an orthogonal matrix. The affine reflection in a hyperplane is $R(x)=Ax+2cu$. Hence $R(x)-R(y)=A(x-y)$ preserves the Euclidean norm and distance, and $R(x)=x$ whenever $u\cdot x=c$. Thus it is an Euclidean isometry fixing the hyperplane pointwise.

For $p\ne q$, any reflection exchanging them must have normal parallel to $p-q$, and its fixed hyperplane must contain their midpoint. These requirements determine the perpendicular bisector uniquely:Changing both signs of $u,c$ leaves the same reflection. It belongs to the orthogonal group $O(n)$ exactly when its translation term vanishes, that is, $c=0$. Since $c=(\Vert p{\Vert }^2-\Vert q{\Vert }^2)/(2\Vert p-q\Vert )$,To prove the finite reflection decomposition of a Euclidean isometry, first note that a distance-preserving map $g$ fixing zero preserves inner products by the polarization identity. Its values on an orthonormal basis $e_i$ form an orthonormal basis, and $g(x)\cdot g(e_i)=x_i$ shows $g(x)={\sum }_i{x}_{i}g(e_i)$. Thus $g$ is orthogonal.

Every orthogonal transformation is a product of at most $n$ linear reflections: if $g(e_1)\ne e_1$, reflect $g(e_1)$ to $e_1$ using the hyperplane normal $g(e_1)-e_1$, which passes through zero. The resulting orthogonal transformation fixes $e_1$, and its restriction to $e_1^{\perp }$ is handled inductively in dimension $n-1$. If $g(e_1)=e_1$, no initial reflection is necessary. Reflections on the complement extend by fixing $e_1$; the induction starts in dimension zero.

For an arbitrary isometry $f$, if $f(0)\ne 0$, first reflect $f(0)$ to zero across its perpendicular bisector with zero. Composing this one affine reflection with $f$ gives an orthogonal transformation. If $f(0)=0$, skip the initial reflection. ThereforeThe glide reflection $(x,y)\mapsto (x+1,-y)$ cannot use fewer than three: its determinant sign is negative, excluding zero or two reflections, while absence of fixed points excludes a single reflection. Three suffice by reflecting successively in $x=0$, $x=1/2$, and $y=0$.

Write $a=(a_1,a_2,a_3,a_4{)}^T$. The reflection in a hyperplane orthogonal to $a$ sendssoIt acts as $-1$ on $\mathrm{span}\{a\}$ and as $1$ on the three-dimensional hyperplane $a^{\perp }$. Therefore

Let ${\Pi }_{\mathcal{G}}$ be the orthogonal projection onto the good linear subspace and writewhere the two displayed states are normalized projections into $\mathcal{G}$ and ${\mathcal{G}}^{\perp }$. Define the reflectionsThe amplitude amplification iterate $Q=R_{\psi }{R}_G$ preserves the good-bad plane, and its $k$th iterate satisfiesThus repeated reflections rotate amplitude toward the good subspace, reaching constant success probability after $O(1/\sin \theta )$ iterations.

Set $|\Psi ⟩=U|0^m⟩$. Unitary conjugation givesThe operator negates $|\Psi ⟩$ and fixes every vector in $\mathcal{S}$ orthogonal to it. It is therefore the reflection in a hyperplane whose normal is $|\Psi ⟩$.

In the ordered orthonormal basisof $\mathcal{S}$, the Pauli Z gate acts asIt fixes the line $|g{⟩}^{\perp }\cap \mathcal{S}$ and negates $|g⟩$, so it is the reflection in a hyperplane orthogonal to the good state.

If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.