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Spatial homogeneity says that the velocity of $B$ relative to $A$ depends only on their separation. Comparing both galaxies with $C$ givesThus $\mathbf{v}(\mathbf{x}-\mathbf{y})=\mathbf{v}(\mathbf{x})-\mathbf{v}(\mathbf{y})$, so continuity of the macroscopic velocity field turns this Cauchy functional equation into a linear map:Isotropy requires $RH=HR$ for every rotation matrix $R$. The only such rank-two tensor is a scalar multiple of the identity, and henceThis is the Hubble law. A comoving galaxy has physical position $\mathbf{r}(t)=a(t)\mathbf{x}$, so

The rotation matrix $e^{xA}$ commutes with $C$ and is orthogonal, soSince $C^n=\mathrm{diag}(1,(-1{)}^n,(-1{)}^n)$,

Let $\Pi$ be the orthogonal projection onto the good linear subspace $\mathcal{G}$, and let $|\psi ⟩$ be a prepared unit vector with $p=⟨\psi |\Pi |\psi ⟩$. For $0<p<1$, defineThe amplitude amplification theorem states that, if the reflection operators about the initial state and the good subspace can be implemented, the iterationsatisfiesTo prove it, the two reflections preserve the linear span of $|g⟩,|b⟩$. In this ordered orthonormal basis, direct multiplication givesMultiplying this rotation matrix by $(\sin \alpha ,\cos \alpha {)}^T$ replaces $\alpha$ by $\alpha +2\theta$, proving the formula by mathematical induction.

For known $0<p\le 1/2$, choose $j$ to be a nearest nonnegative integer to $\pi /(4\theta )-1/2$. Then $|(2j+1)\theta -\pi /2|\le \theta$, so measurement finds the good subspace with probability at least $1-p$, using $O(p^{-1/2})$ iterations. For $p>1/2$, measurement of the initial state already has constant success probability. If $p=1$ success is certain; if $p=0$ these reflections cannot create any good component. If $|\psi ⟩=A|0⟩$, the initial-state reflection is $A(2|0⟩⟨0|-I)A^{†}$, implemented using quantum state preparation and its inverse quantum circuit.

Because $H_n|0^n⟩=|{\psi }_0⟩$,which is the reflection operator in the line spanned by $|{\psi }_0⟩$. Meanwhile $I_G$ changes the sign of every good basis state and fixes every bad basis state; onit is the reflection in the $|{\psi }_B⟩$ axis. Both reflections preserve $\mathcal{P}$, so their product $Q$ does too.

WriteIn Euclidean geometry, the composition of reflections in two lines through the origin has a rotation matrix through twice the oriented angle between the lines. Hence $Q$ rotates the good-bad plane by $2\theta$ from $|{\psi }_B⟩$ toward $|{\psi }_G⟩$. In particular, the Grover rotation angle formula is

Write $\xi =V_1$, $\eta =\varphi$, and $u_j=d^{j}u/d{x}^j$. On the jet space of a scalar ordinary differential equation, definewhere the total derivative operator isThe order-$N$ prolongation of a vector field is

In the coordinates $(x,u)$, the given action of the special orthogonal group isDifferentiating at $s=0$ gives the infinitesimal generator of a Lie point symmetryso $\xi =u$ and $\eta =-x$. The recursive formula givesandTherefore the second prolongation of the rotation generator for plane graphs is

For ${\Delta }_0=u_{xx}$,which vanishes when ${\Delta }_0=0$. Forwe obtainwhich likewise vanishes on the equation. This is the infinitesimal invariance criterion for both equations.

Geometrically, $u_{xx}=0$ describes straight lines. The signed curvature of a plane graph isso the second equation says $\kappa =1$ and describes consistently oriented arcs of unit circles. Rotation preserves straight lines, circles, and signed curvature, explaining both invariances.

A plane reflection has determinant $-1$, while a rotation matrix has determinant $+1$. The determinant of their product, in either order, is therefore $-1$. Since every rotation has determinant $+1$, such a product can never be another rotation.

An anticlockwise rotation matrix and reflection in the $x$-axis areThe columns of $R$ are the images of the standard coordinate vectors, while $M(x,y{)}^T=(x,-y{)}^T$.

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

Put $\sqrt{p}=\sin \theta$ and $\sqrt{1-p}=\cos \theta$, where $0<\theta <\pi /2$. In the ordered basis $(|b⟩,|g⟩)$, the product of the two reflections isThe leading minus sign is a physically irrelevant global phase. The remaining rotation matrix rotates the good-bad plane through $2\theta$, whereEquivalently, one iterate changes the initial angle $\theta$ to $3\theta$.

A planar rotation about the origin is the linear map represented by the rotation matrix $R_{\theta }=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. It preserves inner products, lengths, oriented angles and area, with positive angles measured anticlockwise. Its inverse is $R_{-\theta }$ and its determinant is one.

Under an orthogonal coordinate transformation $R$, a Cartesian pseudotensor has the usual tensor transformation law multiplied by $\det R$. Thus proper rotation matrices act as on ordinary tensors, while an orientation-reversing change contributes an extra minus sign. Interpreted as a rank-three pseudotensor, the Levi-Civita symbol has the same numerical components in every orthonormal frame, because its ordinary three-index transformation already contributes a factor $\det R$.

An axial vector transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$ under an orthogonal matrix $Q$. It agrees with a polar vector under a proper rotation matrix and has the additional determinant sign under an orientation-reversing transformation. Angular momentum and angular velocity are axial vectors.

Under rotation by a rotation matrix, a symmetric second-rank tensor transforms by conjugation, so its traces of powers are invariant. For a three-dimensional traceless second-rank tensor, the Cayley-Hamilton theorem implies $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$. In two dimensions its eigenvalues are $s,-s$, so $\mathrm{Tr}{Q}^3=0$. These identities organize the Landau-de Gennes free energy.

An object has rotational symmetry when a nontrivial rotation matrix leaves it unchanged.

The determinant transformation formula for the Levi-Civita symbol shows that it is invariant under every proper rotation matrix, for which $\det R=1$. Under an orthogonal reflection its sign reverses. This distinguishes proper-rotation isotropy from the pseudotensor transformation convention when both orientations are included.

Every $g\in S{L}_2(\mathbb{R})$ factors into an upper triangular determinant-one matrix and a rotation matrix. This follows because the triangular subgroup acts transitively on the complex upper half-plane, a group orbit of the full group, while $SO(2)$ is the stabilizer subgroup of $i$. The intersection of the two subgroups is $\{\pm I\}$, giving exactly two matrix pairs $(h,k)$ and $(-h,-k)$. Requiring the diagonal of $h$ to be positive gives uniqueness. For $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $q=\sqrt{c^2+d^2}$, the positive branch is $h=\left(\begin{array}{cc}1/q& (ac+bd)/q\\ 0& q\end{array}\right)$ and $k=q^{-1}\left(\begin{array}{cc}d& -c\\ c& d\end{array}\right)$.