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The interior consists exactly of strictly copositive matrices. If $x^{T}Ax$ has positive minimum $\delta$ on the compact nonnegative unit sphere, perturbations of operator norm less than $\delta$ preserve this positive lower bound there. Conversely, if a unit nonnegative vector has zero quadratic form, $A-\epsilon I$ leaves the copositive cone for every $\epsilon >0$. Thus such a matrix is not interior. The identity matrix is a convenient interior point.

The Kronecker sum of square matrices combines their Kronecker products with identity matrices. If $Ax=\lambda x$ and $By=\mu y$, then $x\otimes y$ is an eigenvector with eigenvalue $\lambda +\mu$. Tensor-product spatial grids therefore turn separated coordinate Laplacians into a Kronecker sum. Two Hermitian matrices give a Hermitian Kronecker sum, allowing an energy method without computing every eigenvector.

For isotropic incompressible flow of a Newtonian fluid, the stress tensor is the displayed expression, with pressure $p$, dynamic viscosity $\mu$, velocity $\mathbf{u}$, and identity matrix $I$. In Couette flow, its off-diagonal component is the shear stress.

An orthonormal set consists of vectors that have norm one and are pairwise orthogonal. Its Gram matrix is the identity matrix.

Write $n=bk$, let $I_n$ be the identity matrix, let $J_m$ be the $m$-by-$m$ all-ones matrix, and put $B=\mathrm{diag}(J_k,\dots ,J_k)$, with one diagonal block per experimental block. The variance-covariance matrix isIts orthogonal decomposition is obtained by first subtracting each block's sample mean, then subtracting the grand sample mean from the block means. More explicitly, writing $x_{jt}$ for coordinate $t$ in block $j$, the three invariant subspaces areThey are pairwise orthogonal, with dimensions $b(k-1)$, $b-1$ and $1$, and their direct sum is ${\mathbb{R}}^n$. On $W$, both $B$ and $J_n$ vanish. On $U$, $B=kI$ and $J_n=0$. On $G$, $B=kI$ and $J_n=nI$. Thus the eigenvalues and corresponding invariant subspaces areIf some eigenvalues coincide, their eigenspace is the direct sum of the listed subspaces with that value. Zero-dimensional rows are omitted when $b=1$ or $k=1$. For an admissible covariance matrix the eigenvalues on nonzero subspaces must be nonnegative; these conditions are also sufficient for positive semidefiniteness. The expectation parameters do not affect this calculation.

Work in ${\mathbb{C}}^N$, the domain of the matrix $A$; the printed ${\mathbb{C}}^n$ in the introduction is a dimension typo. For a fixed nonempty support of a vector $S$, write $u$ for the coordinates of $x$ in $S$. ThenThe Gram matrix $A_S^{\ast }{A}_S$ is a Hermitian matrix, so its difference from the identity matrix is also a Hermitian matrix. By the finite-dimensional spectral theorem, its matrix 2-norm is the largest absolute eigenvalue, equivalentlyIndeed, an expansion in an orthonormal eigenbasis bounds every Rayleigh quotient by the largest absolute eigenvalue, and an appropriate eigenvector attains that bound. The restricted isometry constant must bound this quantity for every $S$ of size at most $s$, and the maximum of these quantities suffices for all sparse vectors of order $s$. There are finitely many sets, so the maximum exists. The empty set contributes zero. Therefore

For one-column Gram matrices, unit Euclidean norm of the columns gives $A_S^{\ast }{A}_S=1$. The restricted isometry constant formula therefore gives ${\delta }_1=0$. For a two-column Gram matrix, put $c=⟨a_i,a_j⟩$. The difference from the identity matrix has the formup to the convention for the complex inner product. Its characteristic polynomial is ${\lambda }^2-|c{|}^2$, so the eigenvalues are $\pm |c|$ and its matrix 2-norm is $|c|$. Maximizing over pairs gives ${\delta }_2=\mu$, where $\mu$ is the mutual coherence.

Part (b), together with the minimality defining the restricted isometry constant, gives ${\delta }_s\le {\mu }_1(s-1)$. Every summand defining cumulative coherence is at most the mutual coherence, so ${\mu }_1(s-1)\le (s-1)\mu$. Thus, for normalized columns and $N\ge 2$,The upper range $s\le N$ matters because the printed definition of cumulative coherence stops at $N-1$. For $N=1$, only ${\delta }_1=0$ is needed; the maximum over pairs defining mutual coherence is otherwise empty.

Introduce a parameter $t$ to keep track of total degree. The power series for the matrix exponential givesSet $S=X+Y$ and $C=\frac{1}{2}[X,Y]$, where the commutator is $[X,Y]=XY-YX$. Expanding a second matrix exponential givesSincethe two expressions agree through total degree two. More explicitly, applying the local matrix logarithm to the first expansion gives $\log (e^{tX}{e}^{tY})=t(X+Y)+\frac{1}{2}{t}^2[X,Y]+O(t^3)$. This proves the displayed order of the Baker--Campbell--Hausdorff formula. The Lie algebra is closed under the commutator, so its displayed exponent lies in the same Lie algebra. The expansion is a formal power series identity, or a convergent identity sufficiently near the identity matrix; this argument does not assert unrestricted global convergence.

The next two terms of the Baker--Campbell--Hausdorff formula are the two cubic commutators:Here $O(4)$ means total degree at least four in $X,Y$. If one continues by one more degree, the quartic contribution is $-\frac{1}{24}[Y,[X,[X,Y]]]$.

For the Pauli matrices, use the Pauli matrix multiplication lawPut $N=\mathbf{n}\cdot \mathbf{\tau }$. The antisymmetric term disappears when contracted with $n_a{n}_b$, so $N^2=|\mathbf{n}{|}^{2}I=I$. Consequently every even power of $N$ equals $I$, and every odd power equals $N$. Splitting the matrix exponential into even and odd terms of its power series yieldsThe Pauli matrices are Hermitian matrices, and the components of $\mathbf{n}$ are real, so $N^{†}=N$. For real $\alpha$, the resulting matrix $U$ satisfiesAlso $\mathrm{tr}N=0$ and $N^2=I$, so its two eigenvalues are $1,-1$. The eigenvalues of $U$ are therefore $e^{i\alpha },e^{-i\alpha }$, whose product is one. Thus $U$ is a unitary matrix with determinant one: it belongs to the SU(2) group.

Multiplying the two exact Pauli matrix exponentials, and using ${\tau }_1{\tau }_2=i{\tau }_3$, givesIn particular, the ${\tau }_3$ term has a minus sign. Its quadratic Taylor expansion isTaking $X=i\alpha {\tau }_1$, $Y=i\beta {\tau }_2$, the Pauli matrix commutator identity gives $[X,Y]=-2i\alpha \beta {\tau }_3$. The quadratic exponent in the Baker--Campbell--Hausdorff formula is therefore $Z=i\alpha {\tau }_1+i\beta {\tau }_2-i\alpha \beta {\tau }_3$. In $e^Z=I+Z+\frac{1}{2}{Z}^2+O(3)$, only the linear part of $Z$ contributes to $Z^2$ through degree two. The anticommutator ${\tau }_1{\tau }_2+{\tau }_2{\tau }_1=0$ gives $Z^2=-({\alpha }^2+{\beta }^2)I+O(3)$, reproducing the exact product's quadratic expansion.

With rightmost-first composition, a permutation cycle satisfiesTracing each $a_j$ verifies the equality, and every other letter is fixed. Thus every cycle is a product of transpositions; decomposing a permutation into disjoint permutation cycles proves that every element of $S_n$ is such a product. The identity is the empty product.

To establish that parity is well defined, use the permutation matrix $P_{\sigma }$ defined by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Its determinant is $\pm 1$, and $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. A transposition swaps two columns of the identity matrix and has determinant $-1$. Therefore if $\sigma$ is expressed as a product of $r$ transpositions,The left side depends only on $\sigma$, so any two decompositions have the same parity. DefineMultiplicativity of the determinant makes this the sign homomorphism. Its kernel of a group homomorphism is the alternating groupIt is a normal subgroup consisting of the even permutations. For $n\ge 2$ the sign map is surjective because a transposition has sign $-1$, so $[S_n:A_n]=2$. For $n=1$ both groups are trivial.

The gamma matrices represent the spacetime Clifford algebra. With the Minkowski metric convention above,The braces denote the anticommutator. Consequently $({\gamma }^0{)}^2=I_4$, $({\gamma }^i{)}^2=-I_4$, and distinct gamma matrices anticommute. The identity matrix is essential: the right side is a matrix equality, not a scalar equality. These signs agree with the explicit Dirac representation of the gamma matrices in the next part.

Use the Clifford algebra to write ${\gamma }^{\nu }{\gamma }^{\mu }=2{\eta }^{\mu \nu }{I}_4-{\gamma }^{\mu }{\gamma }^{\nu }$. The Lorentz generator from gamma-matrix commutators is thereforeThis also checks the diagonal case $\mu =\nu$, where $S^{\mu \mu }=0$ because $({\gamma }^{\mu }{)}^2={\eta }^{\mu \mu }{I}_4$. The metric term in the printed expression is understood to multiply the identity matrix. The generators are antisymmetric in their two spacetime indices.

Linearity of Stokes flow and rotational symmetry, including reflection in mathematics invariance of the spherical geometry, determine the possible rigid motions before any calculation. Translation is a polar vector; its linear dependence on a single vector $\mathbf{A}$ must be $c\mathbf{A}$. A traceless second-rank tensor $\mathbf{B}$ cannot produce a polar vector by an isotropic linear map: tensor contraction with the identity matrix vanishes and tensor contraction with the totally antisymmetric tensor vanishes because $\mathbf{B}$ is a symmetric second-rank tensor. Angular velocity is an axial vector. Neither $\mathbf{A}$ nor a symmetric $\mathbf{B}$ can produce one by an isotropic linear map. Products such as $\mathbf{A}\times \mathbf{BA}$ are excluded by Linearity of Stokes flow. Thus $\mathbf{\Omega }=0$.

Use the Unscaled Papkovich–Neuber representation, with harmonic functions $\mathbf{\Phi }$ and $\chi$:Indeed $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$ because each potential is a harmonic function. Put $r=|\mathbf{x}|$ and $S=\mathbf{x}\cdot \mathbf{Bx}$. The decaying harmonic functions with the necessary angular dependence are $\mathbf{A}\cdot \mathbf{x}/r^3$, $\mathbf{Bx}/r^3$ and $S/r^5$. The last is a harmonic function precisely because $\mathrm{tr}\mathbf{B}=0$. A $1/r$ Stokeslet is excluded by the force-free condition, and a rotlet by the torque-free condition. TryThe resulting Stokes flow isAt $r=a$, the coefficient of $\mathbf{n}(\mathbf{A}\cdot \mathbf{n})$ fixes $c=a^3/3$; the remaining constant vector gives $\mathbf{U}=-2\mathbf{A}/3$. In the $\mathbf{B}$ mode, matching $\mathbf{Bn}-(\mathbf{n}\cdot \mathbf{Bn})\mathbf{n}$ gives $e=a^4/2$ and $d=-a^2/2$. Therefore the complete two-mode tensorial squirmer flow isThe boundary condition is satisfied in the laboratory frame, so the Stokes flow tends to zero at infinity. The potential dipole in the $\mathbf{A}$ mode gives an irrotational flow and decays as $r^{-3}$; the leading $\mathbf{B}$ mode is a stresslet, decaying as $r^{-2}$. Neither carries a net force or torque. Direct surface integral averaging with the surface slip velocity formula gives the same translation, providing an independent check. Uniqueness of Stokes flow then identifies this decaying solution.

A pointed cone has no nonzero line through the origin: equivalently $K\cap (-K)=\{0\}$. If $A\in K\cap (-K)$, its quadratic form vanishes on the nonnegative orthant. Testing the coordinate vectors gives $A_{ii}=0$. Testing $e_i+e_j$ then givesSince $A$ is a symmetric matrix, every entry is zero. Thus $\boxed{K\cap (-K)=\{0\}}$.

The identity matrix gives an interior point. For a symmetric perturbation $E$ with operator norm $\Vert E{\Vert }_2<1$, the Cauchy-Schwarz inequality givesThis open norm ball around $I$ lies even in the positive semidefinite cone, hence in $K$. Consequently $\boxed{I\in \mathrm{int}K}$ and the interior is nonempty.

More generally, the interior of the copositive cone consists exactly of strictly copositive matrices. Positivity on the compact nonnegative unit sphere has a positive minimum and persists under small perturbations; a zero there is destroyed by an arbitrarily small negative multiple of $I$.

For each coefficient index $k\ge 0$, let $H_k$ have constant diagonal blocks $f_{k}J$ and cross blocks $g_{k}J$. The preceding block-constant argument gives $H_k⪰0$. Define the Hadamard powers $X^{\circ k}$, withThis is the constant entrywise power, including at zero entries; it is not the identity matrix. Repeated use of the Schur product theorem shows $X^{\circ k}⪰0$ for every $k\ge 0$.

Applying the Schur product theorem again makes every summand $H_k\circ X^{\circ k}$ positive semidefinite. The partial sums are positive semidefinite, and their entrywise limit is exactly $Y$. In finite dimension this is a matrix-norm limit; the positive semidefinite cone is closed. ConsequentlyEndpoint convergence follows from the stated expansions on the full interval: $f_k\ge 0$ and $\sum f_k=f(1)<\infty$, while $\sum |g_k|\le \sum f_k$. Thus the expansions converge absolutely at every matrix entry. This is coefficient-dominated entrywise positivity.

The coefficient condition must include index zero. We interpret the PDF's $\mathbb{N}$ accordingly. If it means only positive integers and no condition is imposed on $f_0,g_0$, the assertion is false: take $n=m=1$, $X=0$, $f=-1$ and $g=0$. All positive-index inequalities hold, but $Y=-I_2$.

The Leibniz formula for determinants isEquivalently, the determinant is the unique alternating function of the rows that is multilinear and takes value one on the identity matrix. Multiplying one row by $\lambda$ therefore gives