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Use signature $\eta =\mathrm{diag}(1,-1,-1,-1)$, the numerical matrices ${\sigma }_{\mu }=(I,{\sigma }_i)$ and ${\bar{\sigma }}_{\mu }=(I,-{\sigma }_i)$ given in the question, and raise genuine tensor indices with $\eta$. A real Minkowski spacetime vector corresponds to the Hermitian matrixFor SL(2,C) matrices, $X^{\prime }=AX{A}^{†}$ is Hermitian and has the same determinant. It therefore induces a real linear Lorentz transformation; composition of the congruence actions agrees with matrix multiplication. The kernel consists of $\pm I$: preservation of $X=I$ makes a kernel element unitary, and preservation of every Hermitian $X$ makes it scalar.

Polar decomposition of an invertible complex matrix connects every determinant-one complex matrix to its unitary factor, so the group is connected. The action gives the Lorentz spinor double cover of the Proper orthochronous Lorentz group. It covers rotations via $A_R=e^{-i\vartheta \mathbf{n}\cdot \mathbf{\sigma }/2}$ and boosts via the positive Hermitian matrices below. Every proper orthochronous transformation is a boost followed by a rotation, since one can first match its image of the future unit time vector and then use its rotation stabilizer. The congruence construction does not cover spatial parity or time reversal. In particular parity has determinant $-1$ as a four-vector transformation; all transformations continuously produced from SL(2,C) have determinant $+1$.

Set $c_{\theta }=\cosh (\theta /2)$, $s_{\theta }=\sinh (\theta /2)$ and $N=\mathbf{n}\cdot \mathbf{\sigma }$. The Pauli matrix multiplication law gives $N^2=I$. The proposed boost matrix is $A_B=c_{\theta }I+s_{\theta }N$, with eigenvalues $e^{\pm \theta /2}$, determinant one and $A_B^{†}=A_B$. Split $\mathbf{x}=x_{\parallel }\mathbf{n}+{\mathbf{x}}_{\perp }$. The perpendicular Pauli part anticommutes with $N$, so multiplication givesThis is an active Lorentz boost. The image of a rest worldline has velocity $\mathbf{v}=\tanh \theta \mathbf{n}$, fixing the velocity sign convention.

For two nonzero boosts, write $A^{\prime }=c^{\prime }I+s^{\prime }{\mathbf{n}}^{\prime }\cdot \mathbf{\sigma }$ and $A=cI+s\mathbf{n}\cdot \mathbf{\sigma }$. Their product isThe last term is anti-Hermitian. Thus $A^{\prime }A$ is not Hermitian unless the axes are collinear; a pure boost has Hermitian lifts $\pm A_B$, so it cannot give the same Lorentz transformation. The noncollinear boost obstruction from Pauli products is thereforeThat rotation is a Wigner rotation. A zero-rapidity factor is the trivial exception, regardless of the arbitrary axis assigned to it.

Define $J_i=\frac{1}{2}{ϵ}_{ijk}{M}_{jk}$ and $K_i=M_{0i}$. The printed Lorentz algebra brackets giveSet $L_i=(J_i+i{K}_i)/2$, $R_i=(J_i-i{K}_i)/2$. ThenThis is the chiral decomposition of the complex Lorentz algebra. The two copies are the complexifications of the SU(2) algebras conventionally labelled left and right. They are not two independent compact real subalgebras of the real Lorentz algebra: $L,R$ are complex linear combinations of the real generators. The distinction is needed for noncompact boosts.

To verify the infinitesimal two-component action, putIts trace vanishes by antisymmetry, so $\det A=1+O({\omega }^2)$. The Pauli matrix multiplication law impliesSince $a^{†}=\frac{1}{4}{\omega }^{\mu \nu }{\bar{\sigma }}_{\nu }{\sigma }_{\mu }$, antisymmetrizing this identity yieldsConsequently $AX{A}^{†}=X+{\sigma }_{\mu }{\omega }^{\mu }{}_{\nu }{x}^{\nu }+O({\omega }^2)$, exactly the claimed infinitesimal coordinate transformation with ${\omega }^{\mu }{}_{\nu }={\omega }^{\mu \rho }{\eta }_{\rho \nu }$.

The two Weyl spinor representations transform asBoth maps preserve group multiplication. For $\Lambda =e^{\omega }$, choose its spin lift and write the finite matrices asFor rotations these coincide as the SU(2) doublet; for boosts their generator signs are opposite. Any complex-linear intertwiner would commute with all rotations and hence be scalar by the Schur lemma, but a nonzero scalar cannot intertwine the opposite boost matrices. They are therefore inequivalent. Infinitesimally their $K_i$ matrices are $-i{\sigma }_i/2$ and $+i{\sigma }_i/2$, while their $J_i$ matrices are both ${\sigma }_i/2$. Thus they have chiral labels $(1/2,0)$ and $(0,1/2)$. The finite matrices are representations of the spin cover; choosing $A$ or $-A$ matters for spinors even though their four-vector transformations coincide.

Using the supplied block gamma matrices, the generator of the Dirac spinor isThe antisymmetry of $\omega$ removes the symmetric Clifford part. Hence

There is a conjugation-order error in the last displayed gamma identity in the PDF. The Clifford algebra gives, with ${\gamma }^{\rho }={\eta }^{\rho \sigma }{\gamma }_{\sigma }$,Exponentiating this linear commutator action proves the inverse Lorentz action on gamma matricesThe second identity is the order required for the requested bilinears when $\psi \mapsto S\psi$. An explicit countercheck to the printed order is a positive boost along the third axis: ${\omega }^{03}=-\theta$ giveswhereas the printed right side has a plus sign. This cannot be repaired by dropping index raising; the same raising convention is needed in the preceding coordinate transformation.

The adjoint relation supplied in the question gives ${\Omega }^{†}{\gamma }^0=-{\gamma }^0\Omega$, hence the Dirac spinor pseudo-unitarity identity $S^{†}{\gamma }^{0}S={\gamma }^0$. The Dirac adjoint therefore transforms as $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$. It now follows that the Dirac scalar bilinear isthe vector isand the antisymmetric second-rank tensor isThese are respectively a Lorentz scalar, Lorentz four-vector and Lorentz tensor. The source's inconsistent gamma identity is replaced by its correct inverse/order pair; all three transformation laws then follow with the stated spinor transformation.

A Schwartz function is a smooth function for which every seminormis finite. These seminorms define the Fréchet topology of the Schwartz space $\mathcal{S}({\mathbb{R}}^n)$. The tempered distributions form its continuous linear dual ${\mathcal{S}}^{\prime }$. We use bilinear distribution pairing, with no complex conjugation on the test function.

Differentiation under the Fourier integral and integration by parts giveTherefore each Schwartz seminorm of $\hat{\phi }$ is bounded by a finite sum of $L^1$ norms of polynomially weighted derivatives of $\phi$. Insert the integrable weight $⟨x{⟩}^{-n-1}$ to bound those norms by finitely many Schwartz seminorms. This proves that $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

To justify inversion without a merely formal exchange of integrals, insert $e^{-\epsilon |\xi {|}^2/2}$ in the inverse integral. The Gaussian Fourier integral givesThe normalized Gaussian is an approximate identity, so the right side tends to $\phi (x)$; the left side converges by dominated convergence theorem since $\hat{\phi }$ is integrable. It follows thatReflection is continuous in Schwartz seminorms, so ${\mathcal{F}}^{-1}=(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$ is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the distributional Fourier transform byIt agrees with the ordinary integral transform whenever Fubini applies. Transposing the Schwartz inverse gives an inverse on ${\mathcal{S}}^{\prime }$, and the same squared-transform identity holds. For the usual strong dual topology, a seminorm is $p_B(u)={\sup }_{\phi \in B}|⟨u,\phi ⟩|$ for a bounded set $B\subset \mathcal{S}$. Since a continuous linear Schwartz map takes bounded sets to bounded sets, $p_B(\hat{u})=p_{\mathcal{F}B}(u)$ proves continuity, and similarly for the inverse. Continuity also holds in the weak dual topology. Thus the transform is a continuous isomorphism on tempered distributions, with the stated normalization.

For a real symmetric positive-definite matrix $G$, there is $c>0$ with $g(x)=x^{T}Gx\ge c|x{|}^2$. The reciprocal is locally integrable in dimension three: the radial factor near zero is $r^2{r}^{-2}dr$. At infinity, rapid decay of a Schwartz test function makes it integrable. More quantitatively,This single seminorm bound proves a tempered distribution. No principal-value extension is needed at the origin.

First calculate the isotropic transform. For $f_{\epsilon }(x)=e^{-\epsilon |x|}/|x{|}^2$ and $k=|\xi |>0$, spherical integration and the supplied sine-integral identity giveThe original functions converge in ${\mathcal{S}}^{\prime }$ to $|x{|}^{-2}$ by dominated convergence theorem against tests. The transforms are bounded by $2{\pi }^2/|\xi |$, which is locally integrable in three dimensions and integrable against Schwartz functions at infinity. HenceLet $y=G^{1/2}x$. The Jacobian is $(\det G{)}^{-1/2}$ and the dual vector is $G^{-1/2}\xi$. The Fourier transform of a reciprocal positive quadratic form is consequentlyThe frequency-origin value is understood distributionally; there is no additional delta term.

For the complex extension use a symmetric matrix $A=G+iB$ with real symmetric $B$ and positive real part $G$. Symmetry is natural for a quadratic form; a skew-symmetric part contributes nothing. The bound $|x^{T}Ax|\ge x^{T}Gx$ again gives a regular tempered reciprocal. Both sides of the prospective formula depend holomorphically on the matrix while its real part is positive: compact parameter sets give a common $|x{|}^{-2}$ bound for pairing and differentiated integrands. Continue from the real positive matrices along $A_z=G+izB$. In a complex neighborhood of $z=0$, imaginary $z$ gives real positive matrices, so the one-variable identity theorem supplies the equality; connected continuation along $0\le z\le 1$ reaches $A$.

The analytic determinant square root for accretive symmetric matrices must follow that continuation, rather than an arbitrary scalar principal root of $\det A$. Explicitly, put $C=G^{-1/2}B{G}^{-1/2}$ and let its real eigenvalues be $b_j$. Thenwhere each factor has positive real part. This is the determinant square root normalized positively on real positive matrices. AlsoThus the quadratic-form square root has the unambiguous branch with positive real part. The accretive complex quadratic reciprocal Fourier transform isUniform local integrability of the right side justifies its analytic continuation as a tempered distribution, not merely pointwise away from the origin.

For the particular form,Substitution gives the required normalized expressionFor every nonzero real $\xi$, the radicand has positive real part, so the specified square root exists uniquely. Both sides are regular tempered distributions, despite their locally integrable singularities at zero.

For a prime $p>2$, a diagonal matrix with two distinct nonzero eigenvalues exists. The same common-eigenspace argument makes the normalizer of a diagonal subgroup of GL2 precisely the monomial matrices. There are $(p-1{)}^2$ choices of nonzero entries for each of two permutation patterns, givingAt $p=2$, however, the invertible diagonal subgroup is trivial. Every invertible matrix normalizes it, soThe first factor chooses a nonzero first column and the second chooses a column outside its one-dimensional span. The exceptional answer is necessary because the printed question permits every prime.

The coordinate axes are exactly the simultaneous eigenspaces of all invertible diagonal matrices in $T$. A matrix normalizing $T$ must permute these common eigenspaces: if $tv=\chi (t)v$, then $t(gv)=g(g^{-1}tg)v$ is again a scalar multiple of $gv$. Thus a normalizer element has exactly one nonzero entry in each row and column, a monomial matrix. In dimension two it is either diagonal or diagonal times $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Conversely $J$ swaps the diagonal entries and normalizes $T$. ThereforeThis is the normalizer of a diagonal subgroup of GL2; the quotient records the permutation of the two axes.

For a full-column-rank matrix, a thin QR decomposition has $A=Q_{1}R$, where $Q_1^T{Q}_1=I_n$ and $R$ is nonsingular upper triangular. Extend $Q_1$ to an orthogonal matrix $Q=(Q_1,Q_2)$. Orthogonality givesThus the unique least-squares solution satisfies $R{x}^{\ast }=Q_1^{T}b$, with residual norm $\Vert Q_2^{T}b\Vert$. In a rank-deficient problem $R$ is singular and uniqueness needs separate treatment.

A Givens rotation is the identity except for the rows/columns $p,q$, whose block is $\left(\begin{array}{cc}c& s\\ -s& c\end{array}\right)$ with $c^2+s^2=1$. Its transpose times itself has that block equal to the identity, so it is orthogonal. Choose $c=4/5,s=3/5$ in rows two and three. ThenBack substitution in the first three rows givesThe fourth row of the rotated residual is the unavoidable value $-2$, while the other three rows vanish. Thus this rotation also directly exhibits the least-squares optimality.

The adjugate matrix is the transpose of the cofactor matrix: $\mathrm{adj}(A{)}_{ij}=(-1{)}^{i+j}\det A_{\hat{j},\hat{i}}$. Cofactor expansion gives the adjugate identityFor an invertible matrix this yields $A^{-1}=\mathrm{adj}(A)/\det A$. Integer entries of $A$ imply integer cofactors, so $\det A=\pm 1$ makes its inverse integral. Conversely, if $A$ and $A^{-1}$ both have integer entries, their determinants are integers and $(\det A)(\det A^{-1})=1$. The only integer units are $\pm 1$. Hence an invertible integer matrix has an integer inverse exactly when it is a unimodular matrix, with determinant $\pm 1$. Invertibility is implicit whenever $A^{-1}$ is used.

In a zero-sum game with the row player maximizing, a sufficient saddle certificate is that $p,q$ are probability vectors andEvery row mixture then gains at most $v$ against $q$, and every column mixture concedes at least $v$ against $p$. Their pairing must equal $v$, proving optimality.

For the cost-weighted finite search game, order Colin's six columns as $123,132,213,231,312,321$ and Rowena's rows as $1,2,3$. The payoff matrix isUnder the proposed search mixture, an alternative location $i$ precedes the hiding location $j$ with probability $(c_i+c_k/2)/c$, where $k$ is the third location. The expected cost conditional on hiding at $j$ is consequentlyIt is independent of $j$, giving the upper bound $\mathrm{value}\le v$.

Choose $p_i=c_i/c$. Against any order $i,j,k$, its expected cost isThis gives the matching lower bound. ThusBoth strategies equalize the opponent's pure actions, giving the stated saddle certificate.

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

Use the natural empty-sum convention ${\mu }_1(0)=0$ for cumulative coherence. This is needed for the printed case $s=1$. For $x=0$ the conclusion is immediate. Otherwise let $S$ be its support of a vector, with $k=|S|\le s$. The Gram matrix $G=A_S^{\ast }{A}_S$ has diagonal entries one because the columns have unit Euclidean norm. Each off-diagonal row sum is bounded byThe last inequality uses monotonicity of cumulative coherence: enlarging an index set only adds nonnegative summands. There are enough indices to enlarge it because $s\le N$.

The Gershgorin circle theorem places every eigenvalue of $G$ within distance ${\mu }_1(s-1)$ of one. Since $G$ is a Hermitian matrix, its eigenvalues are real, and the finite-dimensional spectral theorem givesThis is the cumulative coherence bound for restricted isometry. The lower bound remains valid when ${\mu }_1(s-1)>1$, although it is then nonpositive. No assumption that the matrix is already a near isometry is required. The distortion is bounded by ${\mu }_1(s-1)$ on every order-$s$ sparse vector.

A stability analysis asks whether perturbations in an evolution remain bounded on a fixed physical time interval by a constant independent of the mesh. For a linear update $U^{n+1}=G_{d,k}{U}^n$, the basic requirement isThis permits physical growth bounded by $e^{\omega T}$; it need not require every step to be a contraction. Consistency of a numerical method measures how well its stencil reproduces the differential equation, whereas stability of a numerical method controls the accumulated defects. For a well-posed linear problem, the Lax equivalence theorem connects stability and convergence for a consistent approximation in the specified norm.

On the whole line or a periodic grid, constant-coefficient stencils commute with translations. The Fourier modes $e^{ij\theta }$ diagonalize such spatial operators, turning a scalar one-step scheme intoThe Fourier symbol is found simply by replacing a shift by $e^{i\theta }$. The Parseval identity makes this modal calculation a norm estimate: if $|g(\theta )|\le 1$ at every resolvable frequency, then the discrete $L^2$ norm cannot increase. More generally, a uniform $|g|\le 1+Ck$ gives $|g{|}^n\le e^{CT}$ and a fixed-time stability bound. All frequencies matter, including those near the grid scale; a small-wavenumber expansion establishes consistency but cannot test the entire stability range.

For the heat equation, put $r=k/d^2$ and $s={\sin }^2(\theta /2)$. The Forward Euler diffusion scheme hasThe condition $|g_{\mathrm{FE}}|\le 1$ for every $s\in [0,1]$ is exactly $0\le r\le 1/2$. This illustrates a parabolic time-step restriction $k\le d^2/2$. The Backward Euler diffusion scheme instead haswhich is contractive for every $r\ge 0$. The Crank-Nicolson diffusion scheme hasand is likewise unconditionally $L^2$ stable. However, its high-frequency amplification approaches $-1$ for large $r$, so it need not damp rapidly decaying modes effectively and may produce alternating transients from nonsmooth data. An A-stable time integrator can stabilize a diffusion semidiscretization with left-half-plane spectrum, but damping and accuracy remain separate questions.

For transport $u_t+a{u}_x=0$, define the hyperbolic Courant number $\nu =ak/d$. Forward Euler with a centered spatial difference gives $g=1-i\nu \sin \theta$ and $|g{|}^2=1+{\nu }^2{\sin }^2\theta$. It is unstable under a fixed nonzero CFL ratio. For $a>0$, the upwind finite difference scheme hasThus $0\le \nu \le 1$ is the stable range. Upwinding for negative $a$ reverses the spatial difference and gives $|\nu |\le 1$. The Lax-Wendroff advection scheme gives a second-order example:It is stable for $|\nu |\le 1$, but that $L^2$ result does not imply monotonicity or exclude oscillations at shocks. The Courant–Friedrichs–Lewy condition has a domain-of-dependence interpretation for explicit hyperbolic schemes; the Fourier calculation supplies the sharper algebraic stability criterion for the particular stencil.

For multilevel methods, each frequency gives an amplification polynomial and a companion matrix. Every amplification root must lie in the closed unit disk, with unit-modulus roots simple. Uniformity across frequencies and meshes is essential: one also needs control of the associated matrix powers, for example through a uniformly bounded diagonalization or the uniform power bound from separated amplification roots. A repeated unit root generates a Jordan block and polynomial growth in the number of steps. The same issue occurs for systems: checking only eigenvalue moduli of a nonnormal Fourier amplification matrix is insufficient. For instancehas unit eigenvalues but no fixed-time mesh-uniform bound as $n=T/k\to \infty$. For a hyperbolic system with a constant coefficient matrix admitting a uniformly bounded characteristic diagonalization, scalar modal estimates for each characteristic speed do give a system estimate. A positive energy symmetrizer provides another route. Frozen-coefficient Fourier tests for variable coefficients are useful local diagnostics, but a global estimate must also account for coefficient variation and boundary terms.

Boundary conditions change the allowable modes and can change the conclusion. A periodic calculation is exact only for a periodic closure. On a finite interval with zero Dirichlet boundary conditions, the centered diffusion matrix is diagonalized instead by the discrete sine transform. On $M$ interior points its eigenvalues are $-4{\sin }^2(j\pi /[2(M+1)])$ before scaling by $d^{-2}$. Thus the exact fixed-grid Forward Euler restriction iswhich tends to the whole-line threshold $1/2$ as the grid is refined. Backward Euler and Crank-Nicolson remain stable for all nonnegative $r$. An appropriate Neumann boundary condition uses cosine-type modes and includes a constant mode with eigenvalue zero. The constant mode must be preserved, rather than forced to decay; depending on the boundary stencil, the natural discrete inner product may have endpoint weights.

For hyperbolic transport, the continuous and discrete boundary closures must respect the direction of propagation. If $a>0$ on $[0,L]$, prescribe the left inflow value; prescribing an independent outflow value can overdetermine the problem. The energy method exposes this directly:Homogeneous inflow is dissipative, while forced inflow contributes data to the estimate. A compatible upwind boundary update has the same one-sided information flow. For diffusion,The last term vanishes for homogeneous Dirichlet or Neumann conditions. With Robin boundary conditions of dissipative sign it contributes a further nonpositive term. A non-dissipative boundary can produce genuine PDE growth, which a valid stability estimate should represent rather than incorrectly suppress.

Even a perfectly stable interior symbol cannot detect every defective boundary closure of a difference scheme. As a concrete counterexample, leave a stable explicit heat stencil on all grid points away from the left endpoint, but replace its first interior row by $U_1^{n+1}=2U_1^n$, still keeping $U_0=0$. The interior Fourier symbol is unchanged, yet $U_1^n=2^n{U}_1^0$ makes the finite-interval update unstable. The faulty boundary-adjacent row creates a mode absent from the periodic interior calculation. Thus the entire finite update matrix, or an energy estimate including its boundary rows, must be examined.

A more systematic boundary stability of a finite-difference method analysis transforms time and any tangential spatial variables, then looks for modes $U_j^n=z^n{\kappa }^j$ with $|z|>1$ and $|\kappa |<1$ on a half-line. The interior dispersion relation selects decaying spatial modes; the boundary equations must determine their coefficients without admitting a nonzero growing homogeneous mode. A uniform inverse estimate for the boundary system, expressed by the Uniform Kreiss--Lopatinskii condition, also rules out arbitrarily weak boundary control near limiting frequencies. Merely excluding one obvious unstable mode is weaker than proving a mesh-uniform estimate. Discrete summation by parts and suitable boundary penalties offer an energy-based alternative.

Finally, nonlinear scalar conservation laws can develop discontinuities, so linearized Fourier stability is only one part of the analysis. A monotone conservative scheme, such as the Engquist-Osher method under its CFL restriction, has an invariant state range and L1 contraction of a monotone conservative scheme, with a discrete entropy inequality selecting admissible shocks. These complement the Fourier and boundary estimates. For an evolution problem, the meaningful outcome is a stability bound in a specified norm, uniform over the chosen mesh family and compatible with the actual boundary and initial data.

An eigenvector of a matrix $A$ is a nonzero vector $v$ for which $Av=\lambda v$. The scalar $\lambda$ is its eigenvalue. For a fixed eigenvalue, the zero vector together with all corresponding eigenvectors forms the eigenspace $\ker (A-\lambda I)$. The requirement $v\ne 0$ prevents every scalar from being an eigenvalue merely because $A0=\lambda 0$.