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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

A map $f:A\to {\mathbb{R}}^m$ is differentiable at $p\in A$ when there is a linear map $Df(p):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe linear map $Df(p)$ is the Fréchet derivative of $f$ at $p$.

The multivariable chain rule states that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, thenIndeed, write $f(p+h)=f(p)+Ah+r(h)$ and $g(f(p)+k)=g(f(p))+Bk+s(k)$, where $r(h)=o(|h|)$ and $s(k)=o(|k|)$. Since $k=Ah+r(h)=O(|h|)$,and the last two terms are $o(|h|)$. This proves the formula.

For matrix inversion near the identity matrix,The last term is $O(\Vert h{\Vert }^2)$ because inversion is bounded near $I$. Therefore the inversion map is differentiable at $I$ and

Inner-product preservation between programs for $U$ and $V$ givesfor all $|\alpha ⟩,|\beta ⟩$. If $⟨P_U|P_V⟩\ne 0$, this identity forces every matrix element of $U^{†}V$ to be the same scalar multiple of the corresponding identity matrix element. Hence $U^{†}V=e^{i\theta }I$, contrary to physical distinctness. Therefore

An elementary row operation is one of three operations: interchange two rows, multiply one row by a nonzero scalar, or add a scalar multiple of one row to a different row. Each operation is reversed by another operation of the same kind.

Performing such an operation on an $m\times n$ matrix $A$ is equivalent to left multiplication by the $m\times m$ elementary matrix obtained by applying that operation to the identity matrix. Each elementary matrix is invertible, with inverse corresponding to the reverse operation. Thus a sequence of operations givesFor square matrices, if $A$ were invertible, then $B=PA$ would have inverse $A^{-1}{P}^{-1}$. Hence if $B$ is singular, $A$ must be singular as well. In fact the two are invertible together, because $A=P^{-1}B$ gives the converse.

A real square matrix $A$ is an orthogonal matrix whenHere $A^T$ is the transpose and $I$ the identity matrix; equivalently, its columns form an orthonormal basis. If this holds, the squared Euclidean norm obeysso the linear transformation preserves length.

Conversely, suppose $|A\mathbf{v}|=|\mathbf{v}|$ for every vector $\mathbf{v}$. Put $B=A^{T}A-I$, a symmetric matrix. Then ${\mathbf{v}}^{T}B\mathbf{v}=0$ for every vector. Taking $\mathbf{v}={\mathbf{e}}_i$, the standard basis vectors, gives $B_{ii}=0$. Taking $\mathbf{v}={\mathbf{e}}_i+{\mathbf{e}}_j$ for $i\ne j$ gives $2B_{ij}=0$. Thus $B=0$, proving that the linear transformation preserves length if and only if $A$ is an orthogonal matrix.

If every $\Vert v_i\Vert =1$, then each diagonal entry of the Gram matrix is $A_{ii}=⟨v_i,v_i⟩=1$, so its matrix trace is $n$. Using iii and the permitted arithmetic-geometric mean inequality,The bound is attained by any orthonormal family, whose Gram matrix is the identity matrix.

Perform the elementary row operations $R_2\leftarrow R_2+2R_1$ and $R_3\leftarrow R_3-R_1$, then $R_2\leftarrow R_2/2$, $R_1\leftarrow R_1-R_2$, and finally $R_1\leftarrow R_1+R_3$, $R_2\leftarrow R_2-3R_3$:Consequently the reduced row echelon form is $I_3$. Applying the same operations to the augmented column $b=(b_1,b_2,b_3{)}^T$ givesThus every $b\in {\mathbb{R}}^3$ admits exactly one solution. This follows because the elementary row operations preserve the solution set of the augmented linear system, and the reduced coefficient matrix is the identity matrix.

For an $m\times n$ matrix, the corresponding elementary matrices are $m\times m$, because the operation is implemented by left matrix multiplication. Apply the desired elementary row operation to the identity matrix $I_m$. Writing $E_{ij}$ for the matrix unit with its only nonzero entry, $1$, in position $(i,j)$ gives:Indeed, these matrix products interchange rows $i,j$, scale row $i$, and add $c$ times row $j$ to row $i$, respectively. The displayed inverses undo precisely those operations. Algebraically, $E_{ij}^2=0$ when $i\ne j$, so $(I_m+c{E}_{ij})(I_m-c{E}_{ij})=I_m$; the other two inverse identities follow immediately from their row actions. Thus every elementary matrix is an invertible matrix.

The identity is the identity matrixLeft or right matrix multiplication by it preserves every entry, as is also immediate from the product formula above.

Parametrize the matrices byDirect matrix multiplication givesThis proves closure. The identity matrix is $h(0,0,0)$, and the formulagives both a left and a right inverse, as substitution in the product formula verifies. Thus each element is an invertible matrix, and the subgroup test shows that $H$ is a subgroup of ${\mathrm{GL}}_3({\mathbb{Z}}_3)$. It is the unitriangular group ${\mathrm{UT}}_3({\mathbb{F}}_3)$ and has $3^3=27$ elements.

The identity matrix has entries given by the Kronecker delta, soEach of the $n$ diagonal entries is one.

For $\mathbf{u}=(1,1{)}^T$ and $\mathbf{v}=(1,-1{)}^T$, the outer products areTheir inner product is zero and both squared norms are two. Thus $P_u=\mathbf{u}{\mathbf{u}}^T/2$ and $P_v=\mathbf{v}{\mathbf{v}}^T/2$ are orthogonal spectral projectors, with $P_u+P_v=I$. Since $A\mathbf{u}=5\mathbf{u}$ and $A\mathbf{v}=\mathbf{v}$, the eigenvalues are $\lambda =5$, $\mu =1$, andMultiplying these decompositions uses $P_u{P}_v=0$ and $P_u^2=P_u$, $P_v^2=P_v$, giving the identity matrix.

Use Gaussian elimination. The last equation gives $y=c-z$, and the second then gives $x=b-4c-2z$. Substitute these into the first equation to obtain $3b-10c-z=a$. Back-substitution yieldsThe linear system has matrix formThe coefficients of the arbitrary right-hand side in the solved expressions are the rows of the matrix inverse. ThereforeMultiplying the original matrix by this candidate gives the identity matrix; its determinant is $-1$, so the solution is unique for every $(a,b,c)$.