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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 a point $\mathbf{x}$ on the plane, the Cauchy-Schwarz inequality gives $|d|=|\mathbf{x}\cdot \mathbf{n}|\le |\mathbf{x}|$, because $\mathbf{n}$ is a unit vector. Equality is achieved at $\mathbf{x}=d\mathbf{n}$. Thus the distance from the origin is $|d|$.

Let $\delta =\mathbf{p}\cdot \mathbf{n}-d$ and let $\mathbf{q}=\mathbf{p}-\delta \mathbf{n}$ be the perpendicular projection of the sphere's center onto the plane. Every point of the plane has form $\mathbf{x}=\mathbf{q}+\mathbf{u}$ with $\mathbf{u}\cdot \mathbf{n}=0$. The Pythagorean theorem givesIf $|\delta |=r$, the sphere equation therefore forces $\mathbf{u}=0$. There is exactly one contact point, $\mathbf{q}$; it lies on both surfaces.

Write $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The nonzero scalar triple product proves that $\mathbf{a},\mathbf{b},\mathbf{c}$ are linearly independent, so they form a positively oriented basis of ${\mathbb{R}}^3$ and define a nondegenerate tetrahedron.

For the three faces through the origin, choose the inward unit vectorsTheir signs are correct because each has positive inner product with the opposite vertex vector. Write the center as $\mathbf{p}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. It lies on the interior side of every face. Tangency of the interior sphere therefore means signed distance $r$, not $-r$, from each of those three planes. Taking inner products givesSolving proves the center formulaThis is the sphere center from three tetrahedron face distances; the interior hypothesis selects these signs rather than an exterior center.

Finally set $\mathbf{N}=\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}$. The scalar triple product identities give $\mathbf{N}\cdot \mathbf{a}=\mathbf{N}\cdot \mathbf{b}=\mathbf{N}\cdot \mathbf{c}=\tau$. In particular $\mathbf{N}\ne 0$, and the plane through the three vertices isThe second formula follows by normalizing $\mathbf{N}$ to a unit vector and applying the first distance result.

Use suffix notation with indices ranging from $1$ to $3$, and sum over repeated indices. By the component definitions of the dot product and cross product,The Levi-Civita symbol is unchanged by a cyclic permutation of its three indices: the cycle is two transpositions and hence has positive sign. Thus ${ϵ}_{ijk}={ϵ}_{kij}$. Commuting the scalar components givesTherefore the scalar triple product obeysCyclically moving the vectors preserves the scalar triple product; interchanging just two reverses its sign.

The two displacement vectors $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ lie in the required plane. Because the three points are non-collinear, their cross productis nonzero and is a normal vector to that plane. A point with position vector $\mathbf{r}$ lies in the plane exactly whenExpanding the cross product givesThe terms $\mathbf{a}\cdot (\mathbf{a}\times \mathbf{b})$ and $\mathbf{a}\cdot (\mathbf{c}\times \mathbf{a})$ vanish, so the equation of a plane through three points becomesNon-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.

For the specified coordinates,Their sum is $(3,3,6)$, with Euclidean norm $3\sqrt{6}$. Consequently either orientation of the unit normal is valid:As a check, the scalar triple product is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. The normal direction is $(1,1,2)$.

The Poisson bracket is $\{f,g\}=\mathbf{r}\cdot (\nabla f\times \nabla g)$. Since $\nabla {\rho }^2=2\mathbf{r}$, the scalar triple product vanishes, andEquivalently, check the identity on $x,y,z$ and extend to every polynomial using the Leibniz rule. Thus ${\rho }^2$ is a Casimir function of a Poisson manifold of this rotational Lie-Poisson structure on R3.

The Hamilton's equations areLinearization about $(1,0,0)$ gives $\dot{\alpha }=0$, $\dot{\beta }=(C-A)\gamma$, $\dot{\gamma }=(A-B)\beta$. The two nonzero exponents satisfy ${\lambda }^2=(C-A)(A-B)$. ConsequentlyWhen $A$ is outside this interval the transverse motion is oscillatory. At equality the linearization is nonhyperbolic; if just one transverse coupling vanishes, the other permits secular linear growth, rather than an exponential instability. This is the familiar intermediate-axis instability described by the Euler equations for a torque-free rigid body.

The dot product and Euclidean norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_i{x}_i{y}_i$ and $|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}$. For every real $t$,Its quadratic discriminant is nonpositive, proving the Cauchy-Schwarz inequality $|\mathbf{x}\cdot \mathbf{y}|\le |\mathbf{x}||\mathbf{y}|$. In the inequality asked for, equality holds exactly when $\mathbf{x}$ is a positive scalar multiple of $\mathbf{y}$, and

Using the Levi-Civita symbol and its contraction identity,

Put $T=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, the scalar triple product. Cyclic symmetry gives $\mathbf{m}\cdot \mathbf{a}=\mathbf{m}\cdot \mathbf{b}=\mathbf{m}\cdot \mathbf{c}=T$, so each requested angle $\varphi$ satisfies $\cos \varphi =T/|\mathbf{m}|$. If every pairwise angle is $\theta$ and $t=\cos \theta$, the preceding identity givesand therefore $|\mathbf{m}|=\sqrt{3}(1-\cos \theta )$. This is $0$ when $\theta =0$; for a right-handed orthonormal basis and $\theta =\pi /2$, $\mathbf{m}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ has norm $\sqrt{3}$.

The three planes have equations $\mathbf{a}\cdot \mathbf{x}=\mathbf{a}\cdot \mathbf{p}$, $\mathbf{b}\cdot \mathbf{x}=\mathbf{b}\cdot \mathbf{q}$, and $\mathbf{c}\cdot \mathbf{x}=\mathbf{c}\cdot \mathbf{r}$. They meet at one point exactly when $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$, meaning their normals are linearly independent vectors. The point is

Using the Levi-Civita symbol and the epsilon-delta identity,This proves the vector triple product identityApplying it inside the scalar triple product gives Lagrange's identity

Using Einstein notation andwe obtainThus

The scalar triple product isPut $\Delta =[a,b,c]$. When $\Delta \ne 0$, the three vectorsform the reciprocal basis to $a,b,c$, so $e_i^{\prime }\cdot e_j={\delta }_{ij}$. Their scalar triple product is $1/\Delta$. Thereforesince the cyclic permutation preserves orientation. The identity also holds when $\Delta =0$, either by continuity or directly because the cross products are then linearly dependent.

For the given basis $e_1,e_2,e_3$, the same identities giveIf ${\sum }_i{\alpha }_i{e}_i^{\prime }=0$, dotting with $e_j$ gives ${\alpha }_j=0$, so the $e_i^{\prime }$ are linearly independent and hence form a basis. Moreover, $e_i\cdot e_j^{\prime }={\delta }_{ij}$, so the original basis is reciprocal to the primed basis. Uniqueness of a reciprocal basis gives

Every vector $K$ has a unique expansion $K={\sum }_i{\kappa }_i{e}_i^{\prime }$, and thenThis is an integer for every integer triple $(n_1,n_2,n_3)$ exactly when every ${\kappa }_i$ is an integer. Hence all such points areThey form the reciprocal lattice in the convention without a factor of $2\pi$.

LetLinear independence implies that the scalar triple product $Delta$ is nonzero. Applying the vector triple-product formulas to $(A\times B)\times (C\times D)$ recovers the coefficients of $D$ in the basis $A,B,C$. Equivalently, the reciprocal basis to $A,B,C$ isTaking scalar products with $D$ therefore yields

At late time $f(k,{\tau }_0)\to H/\sqrt{2k^3}$ and $a=-1/(H\tau )$. Define the elementary symmetric polynomialsThen part ii reduces towhere a prime removes the momentum-conserving Dirac delta distribution andExpanding the product givesRepeated integration by parts reduces every negative power to the supplied logarithmic integral. The power divergences are real and disappear when the imaginary part is taken. Writingone obtainsConsequently the late-time parity-odd primordial trispectrum isIts factor of $i$ is required by reality of a momentum-space scalar correlator: reversing all momenta complex-conjugates the correlator, while the scalar triple product changes sign.

Let the Bravais lattice have primitive vectors $a_1,a_2,a_3$, and putThe reciprocal lattice has the basisThe scalar triple product and cross product identities giveConsequently every $q={\sum }_j{n}_j{b}_j$ satisfies $q\cdot l\in 2\pi \mathbb{Z}$ for every lattice point $l={\sum }_i{m}_i{a}_i$, and conversely these three conditions force the coefficients of $q$ in the reciprocal basis to be integers. This proves that the displayed vectors generate ${\Lambda }^{\ast }$.

In the Born approximation, suppose that the crystal potential is the sum of translates of one atomic potential $V_0$ over the finite set $S$ of lattice sites. Its Fourier transform at the momentum transfer $Q=k-k^{\prime }$ factors asThus the single-atom scattering amplitude is multiplied by the crystal lattice structure factor $\Delta (Q)$. Writingseparates the sum into three finite geometric series. For $l_i=-L_i/2,\dots ,L_i/2$ this givesWhen the $L_i$ are large, a factor has a sharp maximum when ${\alpha }_i=2\pi n_i$. All three factors are therefore simultaneously large precisely when $Q\in {\Lambda }^{\ast }$. These are the reciprocal-lattice peaks of a finite crystal.

For the stated body-centered cubic lattice basis,the reciprocal-basis formula givesEquivalently,The shortest nonzero reciprocal vectors have squared Euclidean norm $2(2\pi /a{)}^2$, so

In elastic scattering, $|k|=|k^{\prime }|=k$. If $q=k-k^{\prime }\in {\Lambda }^{\ast }$ and $\theta$ is the angle between $k$ and $k^{\prime }$, Euclidean geometry givesThe first possible diffraction peak therefore occurs atas $ka\to \infty$, by the small-angle approximation. This is the elastic Bragg scattering condition for the shortest reciprocal-lattice vector.

LetThe vector triple product identity $(\mathbf{a}\wedge \mathbf{b})\wedge \mathbf{a}=|\mathbf{a}{|}^2\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$ shows that ${\mathbf{r}}_0\wedge \mathbf{a}=\mathbf{b}$. Moreover, $(\mathbf{r}-{\mathbf{r}}_0)\wedge \mathbf{a}=0$ exactly when $\mathbf{r}-{\mathbf{r}}_0$ is a scalar multiple of $\mathbf{a}$. Thus the given line isAlso $\mathbf{a}\wedge {\mathbf{r}}_0=-\mathbf{b}$, so $\mathbf{N}=\mathbf{a}\wedge (\mathbf{y}-{\mathbf{r}}_0)$. The assumption $\mathbf{N}\ne 0$ says precisely that $\mathbf{y}$ is outside $L$. Consequently $L$ and $\mathbf{y}$ determine a unique plane, with normal vector $\mathbf{N}$.

To identify its equation, use cyclic invariance of the scalar triple product:For $\mathbf{r}\in L$, this equals $\mathbf{y}\cdot \mathbf{b}$, while $\mathbf{r}\cdot \mathbf{b}=\mathbf{r}\cdot (\mathbf{r}\wedge \mathbf{a})=0$. Hence $\mathbf{r}\cdot \mathbf{N}=\mathbf{y}\cdot \mathbf{b}$ throughout $L$. The point $\mathbf{y}$ also satisfies that equation because $\mathbf{y}\cdot (\mathbf{a}\wedge \mathbf{y})=0$. Since $\mathbf{N}\ne 0$, the required plane is therefore exactlyThis also proves the reusable plane through a point and a line given by a cross product construction.

For vectors $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{R}}^3$, the scalar product, cross product and scalar triple product are respectivelyExpanding the dot product givesInterchanging $\mathbf{a}$ and $\mathbf{b}$ changes the sign of their cross product, so its dot product with $\mathbf{b}$ is also zero. Thus the cross product is orthogonal to both vectors, including when it is the zero vector.

The displayed scalar triple product is a determinant. A square matrix has nonzero determinant exactly when its rows are linearly independent. Hence

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.

Use the convention $[\mathbf{a},\mathbf{b},\mathbf{c}]=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ for the scalar triple product. With $\mathbf{q},\mathbf{w}$ as above, $\mathbf{q}=\mathbf{n}\times \mathbf{w}$. The vector triple product identity givesThe Lagrange identity for the cross product, $|\mathbf{a}\times \mathbf{b}{|}^2=|\mathbf{a}{|}^2|\mathbf{b}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2$, also givesTaking the dot product with the unit vector $\mathbf{n}$ proves

Write $\times$ for the paper's $\wedge$. In a right-handed Cartesian coordinate system, the cross product isIt is perpendicular to both vectors, has norm $|\mathbf{a}||\mathbf{b}|\sin \varphi$, where $\varphi$ is their angle, and has direction given by the right-hand rule when nonzero. Its norm is the area of the parallelogram spanned by the two vectors.

If $\mathbf{b}\times \mathbf{c}\ne 0$, use that parallelogram as the base of the parallelepiped. Its perpendicular height is $|\mathbf{a}\cdot \mathbf{n}|$, where $\mathbf{n}=(\mathbf{b}\times \mathbf{c})/|\mathbf{b}\times \mathbf{c}|$. Thus its volume isIf $\mathbf{b}\times \mathbf{c}=0$, both the volume and this scalar triple product vanish. If the cross product is nonzero, the scalar triple product vanishes exactly when $\mathbf{a}$ lies in the plane spanned by $\mathbf{b},\mathbf{c}$. If it is zero, $\mathbf{b},\mathbf{c}$ are linearly dependent, and adjoining $\mathbf{a}$ still spans at most a plane. Consequently the three vectors are coplanar if and only if $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$, including cases involving a zero vector.

Put $V=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$. The vector triple product identity givesTaking its dot product with $\mathbf{a}\times \mathbf{b}$ and using cyclic invariance of the scalar triple product yieldsThere was no division by $V$, so this also proves the identity for coplanar vectors.

Apply the squared scalar triple product from cyclic cross products proved in c:Hence the reciprocal basis has scalar triple product $\boxed{{\mathbf{a}}^{\prime }\cdot ({\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime })=1/V}$, with its sign retained.

Let $V=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. The scalar triple product is invariant under a cyclic permutation, so the definitions of the reciprocal basis give

Let $a$ be an arbitrary constant vector. By the cyclic identity for the scalar triple product,Apply Stokes theorem to $\mathbf{F}\times a$. Since $a$ is constant, the curl product identity givesConsequentlyFor precision, the operator in the last line has componentsThe partial derivatives act on $\mathbf{F}$, with $\mathbf{n}$ multiplying their values; they do not differentiate $\mathbf{n}$. Since the dot products agree for every constant $a$, the cross-product form of Stokes theorem follows:Here $\wedge$ in the paper denotes the cross product, and the printed surface integral has its area element $dS$ understood.