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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.