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Put $p=(3x^2-2)/\sqrt{2}$ and use the orthonormal basis $\mathcal{B}=(1,\sqrt{3/2}x,p)$ from iii. The columns of the given matrix giveConsequentlyThe images of $\mathcal{B}$ are therefore $(p,\sqrt{3/2}x,-1)$, again an orthonormal basis. By i, $T$ is an orthogonal transformation. Its matrix in this orthonormal basis is

If $T$ is an orthogonal transformation and $e_1,e_2,e_3$ is an orthonormal basis, thenThe images form an orthonormal set, hence are linearly independent and form an orthonormal basis in dimension three.

Conversely, suppose the images $T{e}_1,T{e}_2,T{e}_3$ of one chosen orthonormal basis form an orthonormal basis. Write $u={\sum }_i{a}_i{e}_i$ and $v={\sum }_j{b}_j{e}_j$. By linearity of $T$ and bilinearity,Thus mapping one orthonormal basis to an orthonormal basis is equivalent to preserving every inner product. In the converse, a single chosen orthonormal basis suffices.

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.

For $X\in {\mathrm{SO}}_3(\mathbb{R})$, the defining identity $X^{T}X=I$ givesThus an orthogonal transformation preserves the Euclidean norm and sends the unit sphere to itself. Restricting the given group action to this invariant set preserves both action axioms.

If $XN=N$, the first column of $X$ is $N=e_1$. The other columns are unit vectors perpendicular to it, so the first row is also $(1,0,0)$. Hence $X=\mathrm{diag}(1,B)$ with $B^{T}B=I_2$ and $\det B=1$. Conversely, every such matrix lies in $G$ and fixes $N$. Since every element of the special orthogonal group in two dimensions has the planar rotation matrix form, the stabilizer subgroup isGeometrically these are exactly the rotations in three dimensions about the first coordinate axis.

Choose the first orthonormal basis vector to be ${\mathbf{e}}_1=\mathbf{a}$. If $\mathbf{b}$ is not parallel to $\mathbf{a}$, choose ${\mathbf{e}}_2$ along $\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$; otherwise choose any unit vector orthogonal to $\mathbf{a}$. Complete to an orthonormal basis with ${\mathbf{e}}_3$. The matrix with rows ${\mathbf{e}}_i^T$ is an orthogonal matrix, and expressing the three unit vectors in this basis givesThe last two components of ${\mathbf{c}}^{\prime }$ have polar coordinates of the displayed form; we may choose $w\ge 0$. If they vanish, $\theta$ is arbitrary.

An orthogonal transformation preserves dot products. Taking dot products with ${\mathbf{a}}^{\prime }$, and then using the unit Euclidean norms, givesThe remaining dot product givesAn orthogonal matrix has determinant $\pm 1$, so it preserves the absolute scalar triple product. In the new coordinates this scalar triple product is $yw\sin \theta$. ConsequentlyTaking the nonnegative square root provesNo division by $y$ or $w$ was used, so the proof includes parallel vectors. Equivalently, the expression under the square root is the Gram determinant of the three unit vectors.

Represent the quadratic form by the symmetric matrixThe spectral theorem for real symmetric matrices gives an orthogonal matrix $Q$ and real eigenvalues ${\lambda }_1,{\lambda }_2$ such that $Q^{T}MQ=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. In the orthogonal change of coordinates $\mathbf{x}=Q\mathbf{u}$, the equation becomes ${\lambda }_1{u}_1^2+{\lambda }_2{u}_2^2=1$.

If both eigenvalues are positive, this is an ellipse with semiaxes ${\lambda }_1^{-1/2}$ and ${\lambda }_2^{-1/2}$. Conversely, a zero eigenvalue makes the set either empty or unbounded along that coordinate, while two negative eigenvalues give the empty set. Opposite signs give an unbounded hyperbola. A negative eigenvalue together with a zero one also gives the empty set. None of those cases is an ellipse. Therefore the exact condition is that $M$ be a symmetric positive-definite matrix.

The sum and product of its eigenvalues are its matrix trace and determinant:Two real numbers are both positive precisely when their sum and product are positive. ThusAn orthogonal transformation preserves area. Using i gives the quadratic-form ellipse area

Let $\mathbf{r}(u,v)$ parametrize $S$. Then $A\mathbf{r}$ parametrizes its image, with tangent vectors $A{\mathbf{r}}_u$ and $A{\mathbf{r}}_v$. An orthogonal matrix preserves dot products, so the Lagrange identity for the cross product gives, for any $\mathbf{p},\mathbf{q}$,The area element of a surface is therefore the same pointwise before and after applying $A$. Integrating provesThis includes orthogonal transformations with determinant $-1$: reversal of orientation does not affect the Euclidean norm in the surface area formula.

Express the position vector in the suggested orthonormal basis as $\mathbf{r}=X{\mathbf{e}}_1+Y{\mathbf{e}}_2+Z{\mathbf{e}}_3$. ThenThis orthogonal transformation preserves $x^2+y^2+z^2=X^2+Y^2+Z^2$ and has Jacobian determinant of absolute value one, so $dV=dXdYdZ$. The basis actually reverses orientation; the absolute value of the Jacobian determinant is why no minus sign enters the volume integral. Also $x-y=\sqrt{2}Z$, so the half-space becomes $Z\ge 1/2$.

At each fixed $Z\in [1/2,1]$, the section is the disk $X^2+Y^2\le 1-Z^2$, of area $\pi (1-Z^2)$. Its reflection symmetry in $X$ makes the integral of $X$ vanish. Hence the required triple integral isThe negative sign is also consistent with the cap lying in the direction $(1,-1,0)$, where its average $y$ coordinate is negative.

Tensor transformation. Under an orthogonal transformation $R$, the magnetic moment and magnetic field intensity have transformed components $M^{\prime }=RM$, $H^{\prime }=RH$ for a proper rotation. The physical relation $M=\chi H$ must retain its form, giving $RM={\chi }^{\prime }RH$, henceThis is precisely the law for a Cartesian second-rank tensor. Both magnetic quantities are axial vectors; under reflections they each acquire the same determinant sign, which cancels from the transformation of the magnetic susceptibility tensor.

Parallel directions. The condition $M=\lambda H$ is the eigenvalue equation $\chi H=\lambda H$. For the matrix in the paper, expansion of $\det (\chi -\lambda I)$ givesSolving the three resulting homogeneous systems givesThe three directions are mutually orthogonal. For the zero eigenvalue, the induced magnetic moment is zero: the proportionality equation still holds with factor zero, though there is then no nonzero moment direction to assign.

Energy extrema. The stipulated energy is $\mathcal{E}=H^T\chi H$. At fixed $H^{T}H=H_0^2>0$, a Lagrange multiplier calculation givesbecause $\chi$ is a real symmetric matrix. Thus all stationary directions are the eigenvectors, for which $M=\lambda H$. Equivalently write $H=q_0{e}_0+q_1{e}_1+q_6{e}_6$; thenThe Rayleigh quotient immediately givesThe intermediate direction $e_1$ is stationary with $\mathcal{E}=H_0^2$ but is not an extremum on the sphere. At the minimum $M=0$, so the precise extremum condition is proportionality, including a vanishing proportionality factor.

Constant-energy surface. In the original field components,For a fixed positive value $E$, $q_0$ is unrestricted and the cross-section satisfies $q_1^2/E+q_6^2/(E/6)=1$. The surface is therefore an elliptic cylinder, with axis along $(1,-1,2)$ and transverse semiaxes $\sqrt{E}$ along $e_1$ and $\sqrt{E/6}$ along $e_6$. This is an example of level sets of a positive semidefinite quadratic form. At $E=0$ the surface degenerates to the axis line; no negative-energy level exists.

For Cartesian tensors, let $R$ be the orthogonal matrix taking old vector components to new ones, $v_i^{\prime }=R_{ij}{v}_j$. A rank-$n$ tensor transforms by one copy of $R$ on each index:Repeated indices are summed according to Einstein notation. In general coordinates a contravariant tensor index instead carries the corresponding Jacobian matrix $\partial x^{\prime i}/\partial x^j$, and a covariant tensor index carries its inverse; in orthonormal bases of Cartesian axes their component laws coincide with the displayed form.

An isotropic tensor has identical components after every proper orthogonal transformation, so $T^{\prime }=T$ for every $R$ with $R^{T}R=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The Kronecker delta is invariant under all orthogonal transformations, while the Levi-Civita symbol is invariant under proper rotations and acquires a determinant sign under reflections.

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis ${\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3$ corresponding to $6,\sqrt{3},-\sqrt{3}$, with ${\mathbf{u}}_1=(1,1,1{)}^T/\sqrt{3}$. Write $P=[{\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3]$ and $\mathbf{x}=P\mathbf{\xi }$. This orthogonal change of coordinates givesThe plane constraint is therefore ${\xi }_1=0$. On that plane the quadratic form vanishes exactly whenConsequently the entire set isThe two direction vectors are nonzero and not parallel, since ${\mathbf{u}}_2,{\mathbf{u}}_3$ are linearly independent. An invertible orthogonal transformation maps each full line to a full line. Thus the set consists of precisely two infinite lines through the origin.

Use Haar sampling by Gaussian QR decomposition:

The matrix $G$ is invertible almost surely. Indeed, successively sampled Gaussian columns have probability zero of lying in the proper linear subspace spanned by the previous columns. Thus the diagonal signs used above exist with probability one. Equivalently, apply the Gram-Schmidt process to the columns, always dividing each residual by its positive Euclidean norm; this produces $M$ directly.

To prove uniformity, fix $H\in O(n)$. Each column of $G$ has density proportional to $e^{-\Vert x{\Vert }^2/2}$, so orthogonal transformations preserve its probability distribution. Column independence therefore gives $HG\stackrel{\mathrm{d}}{=}G$.

The QR decomposition with positive diagonal is unique: if $G=Q_1{R}_1=Q_2{R}_2$, then $Q_2^T{Q}_1=R_2{R}_1^{-1}$ is both an orthogonal matrix and an upper triangular matrix, hence diagonal with entries $\pm 1$. Its diagonal entries are positive, so it is the identity. It follows that the orthogonal factor of $HG=HMR$ is $HM$. ThereforeBy uniqueness of normalized Haar measure on the compact group $O(n)$, the returned matrix has exactly the required probability distribution. Reflection invariance gives each determinant sign probability $1/2$, so the algorithm samples both components of the orthogonal group. The positive-diagonal correction is essential when a numerical QR decomposition chooses its own signs.

Using the consistently normalized Jordan–Wigner transformation from part (c), write ${\sigma }_1^x{\sigma }_2^x+{\sigma }_1^y{\sigma }_2^y=2(s_1^{+}{s}_2^{-}+s_1^{-}{s}_2^{+})$. Since $f_1^{†}{P}_1=f_1^{†}$ and $f_1{P}_1=-f_1$, the parity string givesThereforeThe orthogonal transformation $b=(f_1+f_2)/\sqrt{2}$, $c=(f_1-f_2)/\sqrt{2}$ preserves the canonical anticommutation relations and diagonalizes this bilinear Hamiltonian:The Fock vacuum is $|\downarrow \downarrow ⟩$, with energy $2h$. Occupying $b$ alone gives the symmetric opposite-spin state with energy $-2J$; occupying $c$ alone gives the antisymmetric state with energy $2J$; occupying both gives $|\uparrow \uparrow ⟩$, up to a global phase, with energy $-2h$. This reproduces the entire spectrum in part (b).

Both the canonical kinetic term and $V({\varphi }^{\mathsf{T}}\varphi )$ are invariant under the orthogonal group $O(n)$. The stated nonzero vacuum expectation value has length $|v|$. Its stabilizer is the orthogonal group $O(n-1)$ acting on the perpendicular subspace. HenceThe two nonzero coordinates of this particular vacuum do not break more generators than an axis-aligned vacuum: the two choices are related by a common orthogonal transformation.

Explicitly, with $s={\varphi }^{\mathsf{T}}\varphi$,stationary point at nonzero $v$ requires $V^{\prime }(v^2)=0$. The $n-1$ transverse eigenvalues vanish, while the radial squared mass is $4v^2{V}^{\prime \prime }(v^2)$. If this curvature also vanishes, there is an additional accidental massless radial mode; the number of symmetry-protected Goldstone bosons remains $n-1$. If $v=0$, the symmetry is unbroken and none is required.

The primitive exterior square of a symplectic vector space is the kernel of its symplectic contraction of an exterior square. In dimension four, the invariant inverse-form bivector $\Omega$ has nonzero wedge square. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$; the symmetric bilinear form $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$ is nondegenerate on ${\Lambda }^{2}V$. The primitive subspace is ${\Omega }^{\perp }$, and therefore inherits a nondegenerate form of dimension five. The symplectic Lie algebra acts on it by infinitesimal orthogonal transformations.

An orientation-preserving orthogonal transformation. In three dimensions it has an invariant axis and acts as a planar rotation through angle $\theta$ perpendicular to that axis. Its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so $\mathrm{tr}R=1+2\cos \theta$. In higher dimensions a proper rotation can involve several mutually orthogonal planes and need not fix an axis.

For unit vectors $u,v$ and a random normal $a$ with independent standard normal distribution coordinates,The Gaussian distribution is invariant under orthogonal transformations. Projecting onto the plane spanned by $u,v$ therefore gives a uniformly distributed direction. If the angle between $u,v$ is $\theta$, the sign-disagreement directions form two sectors of total angle $2\theta$ out of $2\pi$. The endpoint cases $u=v$ and $u=-v$ give probabilities zero and one directly.

A product of $m$ reflections has $\mathrm{rank}(I-Q)\le m$, by the rank inequality applied successively to $I-AB=(I-A)+A(I-B)$. Conversely an orthogonal transformation fixes the orthogonal complement of its moved space; successive Householder reflections fix one new basis direction at each step in that moved space. At most its dimension $\mathrm{rank}(I-Q)$ are required. Thus central inversion in three dimensions needs exactly three reflections.

For standard Brownian motion started at zero in ${\mathbb{R}}^n$, every fixed orthogonal transformation preserves the process law. Its transformed increments remain independent centred Gaussian random vectors with covariance $(t-s)I$, and its paths remain continuous. Consequently its exit law from any centred sphere is normalized surface measure. Translations give the same assertion for spheres centred at its starting point.

A Gaussian random vector with independent normal coordinates of expectation zero and variance one. Its distribution is invariant under orthogonal transformations. In particular, its direction in a fixed two-dimensional plane is uniform, which underlies the Gaussian sign-correlation identity.