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Use the standard implicit assumption that $p$ is prime, so ${\mathbb{F}}_p$ is a finite field. The matrices in question form $G{L}_2({\mathbb{F}}_p)$, the general linear group over a finite field. Products stay in the set because determinants multiply, the identity has determinant one, andhas entries in the finite field and nonzero determinant. Together with associativity this verifies the group axioms. The multiplication of matrices with column vectors is a group action because $Iv=v$ and $(gh)v=g(hv)$.

Let $C=⟨g⟩$ be the order-$p$ cyclic subgroup. By the orbit-stabilizer theorem, its group orbits on ${\mathbb{F}}_p^2$ have size either one or $p$. If $f$ is the number of fixed vectors, counting all $p^2$ vectors gives $f\equiv p^2\equiv 0(\mathrm{mod}p)$. Since the zero vector is fixed, $f\ge 1$ and hence $f\ge p$. There is a nonzero fixed vector $v$.

Extend $v$ to a basis $(v,w)$. In this basis $g$ has matrix $\left(\begin{array}{cc}1& b\\ 0& d\end{array}\right)$. From $g^p=I$ we get $d^p=1$; in the prime finite field, $d^p=d$, so $d=1$. Since $g$ has order $p>1$, it is not the identity, and $b\ne 0$. Replace $w$ by $b^{-1}w$. Then $g(b^{-1}w)=v+b^{-1}w$, and in the new basisThus all the order-p matrices in GL2 over the prime field form one conjugacy class. This proof also works at $p=2$. Primality is essential to the first assertion: for modulus four, $\mathrm{diag}(2,1)$ has nonzero determinant but no inverse, so the set defined using merely nonzero determinants would not be a group.

The chain rule yieldsIf $f$ is a canonical transformation, the symplectic matrix identity $AJ{A}^T=J$ makes this expression $\{F,G{\}}_{Q,P}\circ f$. Thus preservation of the symplectic form implies preservation of every Poisson bracket, for all $C^1$ functions.

Conversely, assume this Poisson bracket identity for every $F,G$. Take $F(Z)=u\cdot Z$ and $G(Z)=v\cdot Z$, for arbitrary constant vectors $u,v\in {\mathbb{R}}^{2n}$. It gives $u^{T}AJ{A}^{T}v=u^{T}Jv$. Since this holds for every $u,v$, $AJ{A}^T=J$, hence $A^{T}JA=J$. Therefore preservation of all Poisson brackets is equivalent to canonicity. Only first derivatives of $f,F,G$ enter this argument.

For a Cartesian change of basis represented by an orthogonal matrix $Q$, vectors transform as $E_i^{\prime }=Q_{ia}{E}_a$ and $B_i^{\prime }=Q_{ia}{B}_a$. Their squared norms are invariant and $Q_{ia}{Q}_{jb}{\delta }_{ab}={\delta }_{ij}$. Substitution into the given expression therefore givesthe transformation law of a Cartesian second-rank tensor. The magnetic field's extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.

Write the Maxwell stress tensor as $T=\mathbf{EE}+\mathbf{BB}-\frac{1}{2}(E^2+B^2)I$. Its divergence isThe identity $(\mathbf{A}\cdot \nabla )\mathbf{A}-\frac{1}{2}\nabla A^2=-\mathbf{A}\times (\nabla \times \mathbf{A})$ follows by contracting two Levi-Civita symbols, or directly by differentiating components. Using Maxwell's equations consequently yieldsHence the local conservation of electromagnetic momentum isThe two subtracted terms are the Lorentz force density, while $\mathbf{E}\times \mathbf{B}$ is the electromagnetic momentum density in the units used here.

Let $S=\mathrm{supp}x$ be the support of a vector $x$, with $|S|\le s$. Suppose the null space property holds. Every other feasible vector is $z=x+v$, where $0\ne v\in \ker A$. Splitting its L1 norm over $S$ and $S^c$ and using the triangle inequality givesThus the sparse vector $x$ is the unique minimizer in basis pursuit. Notice that the argument works for complex coordinates: it uses the absolute value inequality, rather than a real sign function.

Conversely, suppose basis pursuit uniquely recovers every sparse vector of order $s$. Fix $0\ne v\in \ker A$ and any $S$ with $|S|\le s$. Take $x=-v_S$ and $z=v_{S^c}$. These vectors have the same measurements, because $A{v}_S+A{v}_{S^c}=0$, and they are distinct since $z-x=v\ne 0$. The vector $x$ has at most $s$ nonzero coordinates, so uniqueness givesThis is the null space property for every such $S$. Uniform unique recovery by basis pursuit is equivalent to the order-$s$ null space property.

First the null space property implies sparse injectivity. If a nonzero $v\in \ker A$ had at most $2s$ nonzero coordinates, partition its support of a vector into disjoint $S,T$ with $|S|,|T|\le s$. Applying the null space property to each of these sets yieldswhich is impossible. Hence the null space contains no nonzero sparse vector of order $2s$.

The feasible vector $x$ has $\Vert x{\Vert }_0\le s$, where the L0 sparsity count counts its nonzero coordinates. Any different feasible $z$ with $\Vert z{\Vert }_0\le \Vert x{\Vert }_0$ would give a nonzero null space vector $z-x$ with at most $2s$ nonzero coordinates, contrary to sparse injectivity. Consequently every different feasible $z$ has strictly larger L0 sparsity count. The unique sparsest feasible vector is $x$:This proof does not treat the L0 sparsity count as a genuine norm; no triangle inequality for it is needed.

Use the stated characterization by the Lq null space property. Raising an Lq quasi-norm comparison to its positive exponent preserves its order, so the hypothesis at $q$ says that, for every nonzero $v\in \ker A$ and every $|S|\le s$,We prove the corresponding $p$ inequality; this is monotonicity of uniform sparse recovery in the exponent.

Fix a nonzero null space vector and arrange its coordinate magnitudes as $a_1\ge \cdots \ge a_N\ge 0$. Assume $s\ge 1$. The Lq null space property rules out a nonzero null space vector supported on at most $s$ coordinates, so $s<N$ and $t=a_s>0$. Since $p-q<0$, the factors $a_j^{p-q}$ are at most $t^{p-q}$ for $j\le s$ and at least $t^{p-q}$ for $j>s$ with $a_j>0$. Terms with $a_j=0$ contribute zero and require no negative power of zero. ThusThe largest $s$ coordinates maximize the $p$-power sum on any set of size at most $s$. Its complement therefore has the smallest complementary $p$-power sum. The displayed strict inequality proves the Lq null space property at exponent $p$ for every allowed support of a vector. The stated recovery characterization now applies to the Lq quasi-norm at $p$.

If $\ker A=\{0\}$, the measurement constraint already singles out $x$, for every objective; if $s=0$, only the zero sparse vector needs recovery. These cases do not need a positive threshold. Uniform recovery at $q$ implies uniform recovery at every $0<p<q$:

Let $\sigma =\mathrm{sgn}(x_S)$, using the real sign function on the support of a vector $x$. The fixed-sign null space condition in the PDF uses $S^c$ on its right-hand side. This complement is essential. For any real coordinate $x_j\ne 0$, the supporting-line inequality for the absolute value isEvery distinct feasible vector is $x+v$ with $0\ne v\in \ker A$. Summing the coordinate inequalities on $S$ and adding the L1 norm on $S^c$ givesThe strict final inequality is precisely the fixed-sign null space condition. Hence $x$ is the unique basis pursuit minimizer. Unlike the null space property of Question 1, this condition concerns the particular sign function values of $x$, rather than every vector supported in $S$.

Assume $x$ is the unique basis pursuit minimizer. Fix $0\ne v\in \ker A$ and setThere is $\epsilon >0$ such that the sign function of $x_j+t{v}_j$ equals that of $x_j$ at every $j\in S$ whenever $|t|<\epsilon$. To choose it, take less than the minimum of $|x_j|/|v_j|$ over the nonzero $v_j$ in $S$; if that set is empty, any positive $\epsilon$ works. On this interval the L1 norm has the exact expressionFor $0<t<\epsilon$, both $x+tv$ and $x-tv$ are distinct feasible vectors. Uniqueness forces their objective differences to be strictly positive, so $\alpha +\beta >0$ and $-\alpha +\beta >0$. ThereforeThe fixed-sign null space condition is necessary as well as sufficient. This argument also covers an empty support of a vector: then $\alpha =0$ and the nonzero null space vector has $\beta >0$. Strictness is indispensable: equality would make a sufficiently short feasible segment have the same objective as $x$.

Let $u=A^{\ast }h$ be the strict dual certificate for basis pursuit supplied by condition (ii). For $0\ne v\in \ker A$, we haveHere the adjoint operator is the real transpose. The injectivity of $A_S$ ensures $v_{S^c}\ne 0$: otherwise $A_S{v}_S=0$ would force $v=0$. Thus at least one nonzero term lies outside the support of a vector $x$. Since $|u_l|<1$ at every such index,This proves the fixed-sign null space condition, so part (a) gives uniqueness in basis pursuit. If $S^c$ is empty, the injectivity of $A_S=A$ instead means there is no nonzero null space vector, and the feasible set is a singleton. Injective active columns and a strict dual certificate for basis pursuit ensure unique recovery. Both ingredients matter: strictness outside $S$ cannot detect a nonzero null space direction supported entirely inside $S$.

Write $\sigma =\mathrm{sgn}(x_S)$ and $G=A_S^{\ast }{A}_S$. The injectivity of $A_S$ makes this Gram matrix a positive-definite matrix, since $u^{\ast }Gu=\Vert A_{S}u{\Vert }_2^2>0$ for $u\ne 0$. Thus its matrix inverse exists. Construct the least-norm dual certificateOn the active coordinates, $A_S^{\ast }h=G{G}^{-1}\sigma =\sigma$. For $l\notin S$, symmetry of the real Gram matrix and its matrix inverse givesCondition (iii) makes the absolute value of this coordinate strictly less than one. Consequently $h$ is a strict dual certificate for basis pursuit, and part (c) applies. If $S$ is empty, take $h=0$; the zero vector uniquely minimizes the L1 norm on its feasible set. Condition (iii) supplies an explicit certificate and therefore unique recovery. There is no claim that this particular least-norm dual certificate is necessary: other valid certificates may exist when this one fails.

Interpret the angle as the directed subspace angle defined by the infimum of projected unit vectors; it is different from the smallest angle between two subspaces. WriteConsider the bounded linear operator $T:W⟶V^{\perp }$ given by $Tw=P_{V^{\perp }}w$. Its adjoint operator, between these two Hilbert spaces, is $T^{\ast }u=P_{W}u$: for $w\in W$ and $u\in V^{\perp }$, the orthogonal projections give $⟨Tw,u⟩=⟨w,u⟩=⟨w,P_{W}u⟩$. The two positive directed subspace angle cosines yieldThe first bound makes $T$ injective and gives a closed range. Explicitly, if $T{w}_j$ converges, then $\Vert w_j-w_k\Vert \le a^{-1}\Vert T{w}_j-T{w}_k\Vert$, so $(w_j)$ is a Cauchy sequence. The closed subspace of a Hilbert space $W$ is complete, and its limit maps to the proposed range limit. The second bound gives $\ker T^{\ast }=\{0\}$. A vector $u\in V^{\perp }$ orthogonal to the range has $T^{\ast }u=0$, so it must be zero. The range is therefore dense as well as closed in $V^{\perp }$, and $T$ is onto. This is the mechanism of invertibility from lower bounds on an operator and its adjoint.

For any $f\in H$, choose the unique $w\in W$ with $Tw=P_{V^{\perp }}f$. Then $P_{V^{\perp }}(f-w)=0$, so $f-w\in V$. Moreover, if $w\in W\cap V$, then $Tw=0$ and hence $w=0$. Every vector has a unique decomposition, andThe direct sum is a topological one as well: the component $w=T^{-1}{P}_{V^{\perp }}f$ depends boundedly on $f$, with operator norm at most $a^{-1}$.

For precision, the quoted equality of the norms of complementary oblique projections needs both summands nonzero. For example, with $H=\mathbb{R}$, $V=H$ and $W=\{0\}$, the oblique projection is $I$, so $\Vert I\Vert =1$ but $\Vert I-I\Vert =0$. The secant function has value one here, so the second equality in the quoted formula fails. With nonzero complementary summands its intended version is valid. The proof above does not use that formula. The angle itself is undefined on a zero source space because it has no unit vectors; expressing the hypotheses as the two lower bounds handles zero spaces without ambiguity.

For a diffeomorphism, the differential and its inverse preserve the natural pairing of a vector with a covector. Consequently the mixed tensor pullback commutes with every tensor contraction $C$:Differentiating at zero gives ${\mathcal{L}}_X(CT)=C({\mathcal{L}}_{X}T)$. The mixed tensor pullback also preserves tensor products, soThe ordinary product rule for differentiation therefore yieldsThis proves the contraction and Leibniz rule properties for all smooth generators, without needing straightening coordinates.

The suggested flow-box theorem applies locally only where $X\ne 0$. For example $X=x{\partial }_x$ vanishes at $x=0$, so it cannot equal a coordinate basis vector there. Its local flow is nevertheless ${\varphi }_s(x)=e^{s}x$ and ${\varphi }_s^{\ast }dx=e^{s}dx$, giving ${\mathcal{L}}_{X}dx=dx$ even at that zero. This Lie derivative at a zero of its generator illustrates why the general flow proof is needed to cover every point.

Use the positive graph $z=g(x,y)=\sqrt{1+x^2+y^2}$. The printed PDF bounds are $\sqrt{2}\le z\le \sqrt{5}$, so its projection is the annulus $1\le r\le 2$. The surface is the corresponding band of the upper sheet of a two-sheeted hyperboloid, bounded by circles of radii $1$ and $2$ at heights $\sqrt{2}$ and $\sqrt{5}$.

For the parametrization $\mathbf{R}(x,y)=(x,y,g(x,y))$, its tangent vectors are $(1,0,x/z)$ and $(0,1,y/z)$. Their cross product gives the upward vector surface element of a graph:Its magnitude gives the scalar surface area of a graph:Both formulas apply on $1\le \sqrt{x^2+y^2}\le 2$. Reversing the orientation changes only the sign of the vector element.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.

The antisymmetric part need not be a tensor. Its contraction with every symmetric test is zero, so the premise imposes no transformation law on it. This is the blindness of symmetric contraction tests to antisymmetric arrays, now with a third free index.

For a concrete example, prescribe the same numerical array in every orthonormal frame, with onlynonzero, and take $d^s=0$. Every contraction $d_{ijk}^a{s}_{ij}$ vanishes, and the zero result is a genuine vector in every frame. But under $R=\mathrm{diag}(-1,-1,1)$ the rank-three tensor transformation law would send the $121$ component to $-1$, whereas our framewise prescription leaves it equal to $1$. Thus the array is not a tensor despite satisfying every stipulated contraction test. An antisymmetric part could be a tensor if an additional transformation law were supplied; it is simply not forced to be one.

Split the array in its first two slots:For every symmetric second-rank tensor $s$, antisymmetry gives $d_{ijk}^a{s}_{ij}=0$. Hence the stipulated vector equals $d_{ijk}^s{s}_{ij}$.

Let $R$ change the orthonormal frame, so $s_{ab}^{\prime }=R_{ai}{R}_{bj}{s}_{ij}$. The vector transformation law givesfor every symmetric test $s$. Both coefficient arrays in $i,j$ are symmetric, so equality against every symmetric matrix forces equality of the coefficients; one may test the individual diagonal entries and the symmetric off-diagonal basis matrices. Multiplying by $R_{ai}{R}_{bj}$ and using orthogonality yieldsThis is exactly the rank-three tensor law. It is the symmetric-test criterion for a third-rank Cartesian tensor, a form of the quotient theorem for Cartesian tensors. Equivalently one may test $s_{ij}=(u_i{v}_j+v_i{u}_j)/2$ for arbitrary vectors $u,v$ and apply the quotient theorem twice.

Away from the origin, differentiate $r^2=x_j{x}_j$ using Einstein notation:HenceThis gradient is the radial unit vector. Division by $r$ is legitimate on the punctured domain; the distance function is not differentiable at the origin.

Place the cube at the origin with face normals $\pm e_1,\pm e_2,\pm e_3$, and label opposite faces by $(1,2),(3,4),(5,6)$. Every cube symmetry is a signed permutation matrix: there are $2^{3}3!=48$ choices. Its group action on faces is faithful because fixing all faces fixes all their normal directions. Central inversion $-I$ exchanges every opposite pair and therefore acts as $h$. Since $-I$ commutes with every linear cube symmetry, the face-action image lies in $C_{S_6}(h)$. The preceding count gives $|C_{S_6}(h)|=48$, so the faithful face action is onto this centraliser:For the remaining isomorphism, let $R$ be the rotational symmetry group of a cube. Half the signed permutation matrices have determinant $1$, so $|R|=24$. Central inversion has determinant $-1$, and every orientation-reversing symmetry is $(-I)r$ for a unique $r\in R$. Since $-I$ is central and $R\cap \{I,-I\}=\{I\}$, multiplication gives a direct product of groups:The group action of $R$ on the four body diagonals is faithful. To prove this, choose direction vectors $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. They span ${\mathbb{R}}^3$, and their only linear relation is that their sum is zero. A rotation fixing all four diagonal lines sends each listed vector to itself or its negative. Applying the relation forces all four signs to agree. The all-negative choice is $-I$, which is not in $R$, so the rotation is the identity. Thus $R$ embeds in $S_4$; both have order $24$, giving $R\cong S_4$. Finally $C_{S_6}(g)\cong C_2\times S_4$ from part (i). ConsequentlyThe direct-product decomposition of the cube symmetry group supplies the second isomorphism, while the first comes from the specified face action.

The Möbius transformations are bijections, so they preserve which coordinates of a triple are equal. Conversely, part (i) implies transitivity within each equality pattern. For a pattern with fewer than three distinct entries, extend the chosen distinct points to a triple before applying part (i); this works also for $p=2$, when the projective line has exactly three points. There are therefore five orbits of a group action.

We first count $|G{L}_2({\mathbb{F}}_p)|$. Its first column can be any nonzero vector, giving $p^2-1$ choices. The second must be outside the first column's one-dimensional span, giving $p^2-p$ choices. HenceFor the all-equal pattern take $(\infty ,\infty ,\infty )$. Its stabilizer subgroup consists of upper triangular matricesof order $p(p-1{)}^2$. Its group orbit has order $p+1$.

The three exactly-two-equal patterns have representatives $(\infty ,\infty ,0)$, $(\infty ,0,\infty )$ and $(0,\infty ,\infty )$. Each stabilizer subgroup fixes $0$ and $\infty$ individually, so it is the diagonal subgroup of order $(p-1{)}^2$. Each group orbit has order $p(p+1)$; their repeated-coordinate positions keep these three orbits distinct.

For the all-distinct pattern use $(0,1,\infty )$. Its stabilizer subgroup is the scalar subgroup of order $p-1$, and the group orbit has order $(p+1)p(p-1)$. Stabilizers of arbitrary triples are conjugates of the displayed representative stabilizers. In summary the equality-pattern orbits of projective triples haveThe sizes add to $(p+1{)}^3$, and each orbit size times its stabilizer order is $|G|$, as in the orbit-stabilizer theorem.

For any vector $\mathbf{a}$, differentiating its components and the rotating basis gives the rotating-frame derivative formulaApply it twice to the position $\mathbf{x}$, with constant angular velocity $\mathbf{\omega }$. The inertial acceleration iswhere dots denote rotating-frame derivatives. Newton's second law therefore becomesThe last terms are the Coriolis force and the force corresponding to centrifugal acceleration.

For the particle on a uniformly rotating inclined plane, choose orthonormal vectors ${\mathbf{e}}_1$ along the horizontal $\xi$ axis and ${\mathbf{e}}_2$ up the slope, and set ${\mathbf{e}}_3={\mathbf{e}}_1\times {\mathbf{e}}_2$, the upward normal vector. Then ${\mathbf{e}}_z=\sin \theta {\mathbf{e}}_2+\cos \theta {\mathbf{e}}_3$ andHere $N$ is the normal reaction. The marble is modeled as the smooth sliding particle specified by the printed equations, without an additional rolling constraint. The rotating-frame Coriolis acceleration and centrifugal acceleration are respectivelyThe normal rotating-frame acceleration is zero, so the normal reaction is not generally $mg\cos \theta$; it satisfiesProjecting the same equation onto the two tangential axes gives exactly

Multiply the equations by $\dot{\xi }$ and $\dot{\eta }$ and add. The Coriolis force terms cancel, as expected because this force does no work relative to the rotating frame. The normal reaction also does no work along the plane. The conserved quantity is rotating-frame kinetic energy plus gravitational and centrifugal potential energy:This need not be the conserved inertial energy, since the externally driven rotating plane can exchange energy with the particle.