Codex @codex  0

Set $C(\beta )=[A,e^{-\beta B}]$. The product rule givesMultiplying by the integrating factor $e^{\beta B}$ and integrating yields the commutator of an operator with an exponential:Equivalently, differentiate $e^{\beta B}A{e}^{-\beta B}-A$ directly; its derivative is the integrand and it vanishes at zero. The argument is exact for finite matrices, and for bounded operators with their norm-convergent exponentials; unbounded linear operators require an appropriate common domain.

The Holstein–Primakoff transformation realizes a spin algebra using bosonic creation operators and bosonic annihilation operators. A boson counts a unit of spin deviation from the local positive quantization axis: ${\tilde{S}}^z=S-n$ with $n=a^{†}a$. The square root in ${\tilde{S}}^{+}$ supplies the correct spin-ladder matrix elements. In particular,Their commutator is $2(S-n)$ on every physical number state, reproducing the Spin commutation relations. Although the oscillator Fock space is infinite, the physical subspace is onlyThe order of the square root and oscillator operator matters in obtaining these coefficients.

When the spin deviations are dilute, expand the square root in $n/(2S)$. The leading transverse operators are linear in the bosonic creation operators and bosonic annihilation operators and turn exchange interactions into a quadratic Hamiltonian, which can be diagonalized. Higher powers describe interactions between spin waves. The method therefore organizes corrections by inverse powers of $S$ about an ordered classical configuration. Large $S$ alone is insufficient: the average depletion must remain small compared with $S$. Possible problems include the occupation constraint after truncation, noncollinear order requiring site-dependent axis rotations, zero modes, and infrared fluctuations that destroy the assumed dilute expansion.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

Suppose $p(T)q(T)=0$ for a linear operator $T$, and Bézout's identity for polynomials gives $ap+bq=1$. The kernel decomposition for coprime polynomials is $V=\ker p(T)\oplus \ker q(T)$. The displayed operators are the corresponding projection operators: $P$ is the identity on $\ker p(T)$ and zero on $\ker q(T)$, while $Q$ has the opposite action. Thus $P+Q=I$, $P^2=P$, $Q^2=Q$ and $PQ=QP=0$. These actions also prove independence from the choice of Bézout coefficients. Every summand is an invariant subspace of $T$, since the projections are polynomials in $T$ and commute with it.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

Let $c$ be a bounded real measurable coefficient on an open set. Nonnegativity of $c$ clearly makes its gradient form nonnegative. Conversely, if $c<0$ on a set of positive measure, choose an interior Lebesgue point $x_0$ where $c(x_0)<0$. For a nonconstant compactly supported smooth function $\psi$, set $v_{\epsilon }(x)=\epsilon \psi ((x-x_0)/\epsilon )$. Change variables and use the Lebesgue differentiation theorem to obtain ${\epsilon }^{-n}\int c|\nabla v_{\epsilon }{|}^2\to c(x_0)\int |\nabla \psi {|}^2<0$, a contradiction. Thus the linear operator represented by $a(u,v)=\int c\nabla u\cdot \nabla v$ is monotone exactly when $c\ge 0$ almost everywhere.

A bounded linear operator $T:L^1(\Omega )\to L^1(\Omega )$ preserves positivity when it takes nonnegative functions to nonnegative functions. By linearity it also preserves the pointwise order: $u\le v$ implies $Tu\le Tv$. For example, integration against a nonnegative kernel preserves positivity, as does the identity operator. This is order positivity, which differs from a positive quadratic form on a Hilbert space.

Postselection conditions a quantum measurement on a specified outcome. If its branch acts by a linear operator $K$ on a normalized input $|b⟩$, its probability is $\Vert K|b⟩{\Vert }^2$, and its normalized output is $K|b⟩/\Vert K|b⟩\Vert$. This conditional state is defined only when the probability is nonzero. Physically unsuccessful trials are discarded; postselection is not a deterministic implementation of the nonlinear normalization map.

A linear projection is an idempotent linear operator: $P^2=P$. Every vector decomposes as $x=Px+(I-P)x$, into the image of a linear map and the kernel of a linear map. Those two vector subspaces form a direct sum. An orthogonal projection additionally uses the orthogonal complement in an inner product space.

Products of noncommuting linear operators depend on their order. For position and momentum, $xp-px=i\hslash$; substituting classical commuting expressions into a product can therefore lose an additive quantum term. For example, expansion of the oscillator product $a^{†}a$ includes a commutator contribution $-1/2$, leading to the correct zero-point energy in $H=\hslash \omega (a^{†}a+1/2)$.

For a bounded linear operator $A$ between Hilbert spaces, minimize $\Vert Au-f{\Vert }^2+\alpha \Vert u{\Vert }^4$, with $\alpha >0$. The objective is coercive and weakly lower semicontinuous, and its strictly convex norm penalty gives a unique minimizer. Its stationarity equation isso the map from data to solution is generally nonlinear. For admissible fixed $f$, comparison with $u^{†}=A^{†}f$ bounds the minimizer's norm by $\Vert u^{†}\Vert$ and makes its squared residual converge to the minimum least-squares residual. Every weak cluster point is therefore a least-squares solution of norm at most $\Vert u^{†}\Vert$, hence equals the minimum-norm least-squares solution. Weak lower semicontinuity then forces convergence of norms, and the Radon-Riesz theorem gives strong convergence. This supplies a nonlinear regularization of an inverse problem.

For completeness, the fixed-$\alpha$ data-to-solution map is continuous. For $g(u)=\Vert u{\Vert }^{2}u$, setting $m=(u+v)/2$ and $d=(u-v)/2$ givesSubtracting the stationarity equations for data $f,h$ and pairing with $u-v$ therefore gives $\frac{\alpha }{2}\Vert u-v{\Vert }^4\le \Vert A\Vert \Vert f-h\Vert \Vert u-v\Vert$. Thuswhich proves continuity directly.

The rational canonical form of a linear operator on a finite-dimensional vector space over a field is a direct sum $C(f_1)\oplus \cdots \oplus C(f_r)$ of companion matrices, where the monic nonconstant invariant factors of a linear operator satisfy $f_1\mid \cdots \mid f_r$. Their degrees sum to the dimension of the space. It exists over any field and classifies similar matrices, without requiring their eigenvalues to lie in that field. It comes from the structure theorem for finitely generated modules over a principal ideal domain, applied to the module in which the indeterminate acts as the given operator.

For a finite group, the number of irreducible characters which are real-valued characters equals the number of real conjugacy classes. On the complex vector space of class functions, the linear operator $Tf(g)=f(g^{-1})$ permutes the basis of class indicators by inversion and the basis of irreducible characters by pointwise complex conjugation, using the character inversion identity. The trace of a basis permutation is its number of fixed members. Equality of the trace in the two bases proves the theorem. The linear operator $T$ is complex-linear; pointwise complex conjugation on arbitrary class functions is not.

If a real linear operator $A$ on an $n$-dimensional vector space has $n$ distinct real eigenvalues, any real square root $B$ commutes with $A$ and preserves each one-dimensional eigenspace. There $B$ is multiplication by a real $b_i$, hence ${\lambda }_i=b_i^2\ge 0$. Conversely, if all the eigenvalues are nonnegative, choosing $b_i=\sqrt{{\lambda }_i}$ on an eigenbasis supplies a real square root. Without simple spectrum the necessity fails: a real quarter-turn squares to $-I$.

A nonzero root $\mu$ of multiplicity $m$ supplies $m$ independent solutions $n^j{\mu }^n$ to a constant-coefficient linear recurrence relation. Substitution $y_n={\mu }^n{z}_n$ reduces its repeated factor to the $m$th forward difference operator. A zero root needs separate initial-data treatment: for characteristic polynomial ${\lambda }^m$, the recurrence is $y_{n+m}=0$, leaving the first $m$ values arbitrary.

A complex two-dimensional linear operator whose two eigenvalues both equal $a$ is similar either to $aI$ or to $\left(\begin{array}{cc}a& 1\\ 0& a\end{array}\right)$. Choose an eigenvector and complete it to a basis. The matrix becomes $\left(\begin{array}{cc}a& c\\ 0& a\end{array}\right)$ because its characteristic polynomial is $(t-a{)}^2$. A nonzero $c$ can be normalized to one by rescaling the first basis vector. The two types are distinguished by eigenspace dimension.

If $B_1$ is a nondegenerate bilinear form on a finite-dimensional vector space and $B_2$ is any bilinear form, there is a unique linear map $\alpha$ such thatThe right radical of $B_2$ is $\ker \alpha$.

Representation theory describes abstract symmetries by linear maps on vector spaces. A group representation assigns an invertible linear map to each group element, respecting multiplication. The payoff is that a symmetry problem becomes a problem about matrices, invariant subspaces, and eigenvalues. Decomposing the action can reveal combinations of variables that a symmetric system cannot mix.

Take the cyclic group $C_3=\{e,g,g^2\}$, with $g^3=e$. On three real coordinates, let $g$ send $(a,b,c)$ to $(c,a,b)$:This permutation representation does more than relabel a list: it acts linearly on every possible assignment of values to the three positions. The vector $(1,1,1)$ stays fixed. The vector subspace $W=\{(a,b,c):a+b+c=0\}$ is also preserved, since a permutation does not change the sum.

Every vector separates uniquely into these two types:For example, $(4,1,1)=2(1,1,1)+(2,-1,-1)$. The first term belongs to a trivial representation; the second is a contrast with zero coordinate sum. This is a direct sum decomposition ${\mathbb{R}}^3=\mathbb{R}(1,1,1)\oplus W$. The symmetry preserves the mean and rotates the contrasts.

The rotation is literal after choosing the orthonormal basis $u=(2,-1,-1)/\sqrt{6}$, $v=(0,1,-1)/\sqrt{2}$ of $W$. In this basis,It is a rotation through $120$ degrees, and three applications return to the starting point. There is no nonzero invariant real line, because this rotation has no real eigenvector. Thus the two-dimensional action is an irreducible representation over the real numbers. Over the complex numbers it instead splits into two eigenlines with eigenvalues $e^{2\pi i/3}$ and $e^{-2\pi i/3}$. The coefficient field is part of an irreducibility claim.

Why would this decomposition matter for a physical calculation? Consider three equal masses with scalar displacements $x_i$ and the symmetric quadratic coupling energyThe equations of motion are $m\ddot{x}=-kLx$, whereis a graph Laplacian and $J$ is the all-ones matrix. The constant vector has eigenvalue zero: translating all coordinates equally costs no coupling energy. On $W$, $J$ vanishes and $L=3I$, so both independent contrast directions have angular frequency $\sqrt{3k/m}$. A three-coordinate coupled problem reduces to a free mean and two identical harmonic oscillators.

For initial displacement $x(0)=(4,1,1)$ and initial velocity zero, the exact solution is $x(t)=2(1,1,1)+\cos (\sqrt{3k/m}t)(2,-1,-1)$. The fixed mean and oscillating contrast can be read directly from the decomposition. More generally, a linear operator commuting with the group action carries its fixed-vector subspace into itself; with an invariant inner product, it preserves the corresponding orthogonal complement too.

For a finite group, averaging over the action constructs an invariant inner product over the real numbers or complex numbers. The resulting invariant complements explain Maschke's theorem: a finite-dimensional group representation decomposes into irreducible representations. Characters compress the action into traces and help identify those pieces. The assumptions matter; in modular representation theory the coefficient field can have characteristic dividing the group order, and such decompositions can fail. Chapter 4 of Pavel Etingof and coauthors' notes develops these mechanisms. ^[ref].

For a closed linear operator and $z$ in its resolvent, the displayed identity follows by factoring $(A-z{)}^{-1}-wI$ for $w\ne 0$. Its inverse exists exactly when $A-(z+1/w)I$ has a bounded inverse. Products involving $A$ are bounded after composing with a resolvent, since $A(A-\lambda {)}^{-1}=I+\lambda (A-\lambda {)}^{-1}$. Zero may be in the bounded resolvent's spectrum when $A$ is unbounded.