A classical group is a standard family of linear groups, principally the general/special linear groups and groups preserving nondegenerate alternating, symmetric, Hermitian or quadratic forms. Finite-field versions yield many finite simple groups after imposing the appropriate determinant or derived-subgroup condition and quotienting the scalar center of a group. Small dimensions and small fields have exceptional isomorphisms and nonsimple cases.

This formula allows a matrix and its derivative to fail to commute. Differentiate the absolutely convergent power series of the matrix exponential to obtain ${\sum }_{k\ge 1}k{!}^{-1}{\sum }_{j=0}^{k-1}{f}^j{f}^{\prime }{f}^{k-1-j}$. Expanding the integral gives the same coefficients, since ${\int }_0^1(1-u{)}^a{u}^{b}du=a!b!/(a+b+1)!$. The formula reduces to $e^f{f}^{\prime }$ when $[f,f^{\prime }]=0$.

For a nonzero real three-dimensional skew-symmetric matrix, choose real orthogonal vectors $u,v$ of equal length with $Au=\theta v$, $Av=-\theta u$ and $\theta >0$, and a real unit vector $e_3$ in its kernel. The matrix exponential acts by $e^{A}u=u\cos \theta +v\sin \theta$, $e^{A}v=v\cos \theta -u\sin \theta$ and $e^A{e}_3=e_3$. Its matrix in this orthonormal basis is a planar rotation matrix with an unchanged axis. A purely imaginary nonzero eigenvalue and its complex eigenvector give $u,v$ by taking real and imaginary parts; antisymmetry proves their equal norms and orthogonality.

If $A$ is real symmetric positive definite and $B$ is real symmetric, let $S=A^{-1/2}$. The matrix $SBS$ is symmetric and has an orthonormal eigenbasis, represented by $Q$. Taking $P=SQ$ gives the two displayed congruences. The diagonal entries are determined by the generalized eigenvalue problem $\det (B-\lambda A)=0$, since $P^T(B-\lambda A)P$ is diagonal with entries ${\lambda }_j-\lambda$. Positive definiteness of the first form is essential to this construction.

The polar decomposition in continuum mechanics and material isotropy remove the reference rotation from an elastic Cauchy stress tensor. Half-turns around principal stretch axes leave the diagonal left stretch tensor fixed, so material frame indifference forces the stress's off-diagonal entries to vanish in those axes. Rotations within repeated-stretch eigenspaces also force isotropic stress there. Consequently stress and the left Cauchy-Green deformation tensor are coaxial symmetric tensors.

For $0<a\le b$, expand the expectation under the weighted matching Markov chain stationary weights to obtain the displayed identity. If the largest matching in a graph has $n$ edges and $b/a\le 1+1/n$, the observable lies in $[1,e]$. Its sample mean has controlled error by the Hoeffding inequality. A product of $J$ ratios, each estimated to relative accuracy $O(\epsilon /J)$, approximates the full ratio to relative accuracy $O(\epsilon )$.

For a four-dimensional $\mathcal{N}=1$ finite-helicity massless supermultiplet, one complex oscillator changes helicity by one-half. A lower-helicity labelling convention assigns $\kappa$ to the pair $(\kappa ,\kappa +1/2)$. A symmetric midpoint convention instead uses $\kappa +1/4$; report the helicity pair when comparing conventions. CPT completion of a supermultiplet pairs $\kappa$ with $-\kappa -1/2$.

The displayed action functional contains no mass term and no $R{\varphi }^2$ curvature coupling. Its Euler-Lagrange field equation is ${\partial }_{\mu }(\sqrt{-g}{g}^{\mu \nu }{\partial }_{\nu }\varphi )=0$. In a flat FLRW metric, a Fourier mode obeys ${\varphi }_k^{\prime \prime }+2(a^{\prime }/a){\varphi }_k^{\prime }+k^2{\varphi }_k=0$ in four dimensions. The expansion can therefore freeze these modes; the curvature term of a conformally coupled scalar field changes that conclusion.

The zero-form field strength deforms massless type IIA supergravity. A D8-brane changes this flux across its worldvolume; it is therefore outside the elementary massless M-theory circle duality dictionary.

// Target: physics.bigb

For a massive four-dimensional N=1 supermultiplet, superspin $j$ is the spin of its Clifford vacuum, the state annihilated by the two lowering supercharges. The creation operators produce spins $j,j,j+1/2,j-1/2$, omitting the last when $j=0$. The corresponding superspin Casimir has an eigenvalue proportional to $j(j+1)$; this representation label is distinct from the spin of any one constituent.

Write $x=m/M_{\odot }$, the initial mass function as $dN=A{x}^{-\alpha }dx$, and a main-sequence mass-luminosity relation as $L/L_{\odot }=x^{\beta }$. On $a<x<b$, the mass-to-light ratio in solar units isFor $\alpha =\beta =3$ this is exactly $1/(ab)$. The result concerns the specified unevolved stellar population; stellar remnants, evolved luminous stars and cooling substellar objects require different accounting.

A dynamical mass-to-light ratio must specify its aperture and whether the mass includes dark matter. A growing dark halo can increase ${\rm Y}(<R)$ outside the bright stellar body even when the stellar ratio is constant. Comparing a tidal-radius satellite mass with a large-aperture host mass is different from comparing both within their half-light radii. From $M\propto {\sigma }^{2}R$ and $L\propto {\sigma }^4$, the inferred ratio scales as $R/{\sigma }^2$, so equal ratios additionally require homologous radius scaling.

Work in discounted units with a unit bank account and a square invertible volatility matrix. For asset dynamics $dS=\mathrm{diag}(S)(\mu dt+\sigma dW)$, let $\lambda ={\sigma }^{-1}\mu$ and $Z=\mathcal{E}(-\int {\lambda }^{T}dW)$. The Itô product rule cancels the drift of each $Z{S}^i$, leaving the displayed stochastic integral. When $\mu$ is bounded and $\sigma {\sigma }^T\ge \epsilon I$, $\lambda$ is bounded and the Novikov condition makes $Z$ a true martingale on every finite horizon. If $\sigma$ is also bounded, each $Z{S}^i/S_0^i$ satisfies the Novikov condition as a stochastic exponential with bounded integrand ${\sigma }_i-\lambda$, so it too is a true martingale.

For a finite-expected value nonnegative random variable and $a>0$, equality in the uncapped Markov's inequality holds exactly when $Y$ takes only $0$ and $a$ almost surely. Indeed the nonnegative difference $Y-a{1}_{\{Y\ge a\}}$ has zero expected value exactly when it vanishes almost surely. Giving $a$ probability $p$ yields $\mathbb{E}Y=pa$ and $\mathbb{P}(Y\ge a)=p$. This shows the bound cannot generally be reduced using only the nonnegative random variable's expected value.

The transition operators of a time-homogeneous Markov process satisfy $P_{t+s}=P_t{P}_s$ by the Markov property. They preserve constants and positivity. When the generator and function domain permit differentiation, ${\partial }_t{P}_{t}g=L(P_{t}g)$ is the Kolmogorov backward equation. Transition semigroup of a continuous-time Markov chain is the discrete-state special case.

If a real degree-at-most-$n$ polynomial has absolute value at most one at every extremum $\cos (j\pi /n)$ of the Chebyshev polynomial $T_n$, its $k$th derivative on $[-1,1]$ is bounded by $T_n^{(k)}(1)$, for $1\le k\le n$. This strengthens the Markov inequality for polynomial derivatives by replacing a bound on the whole interval with these finitely many samples. The node set is rigid, as quantified by uniqueness of Chebyshev derivative-norming nodes.

Coupling from the past uses a fixed two-sided sequence of random update maps for a finite Markov chain. Increase a backward horizon until the composition from that horizon to time zero sends every starting state to the same state. Retain the maps at previously exposed times. The resulting common image is an exact stationary sample. If a length-$m$ block has positive probability $\epsilon$ of mapping every state to one state, independent disjoint blocks imply failure to coalesce by horizon $T$ has probability at most $(1-\epsilon {)}^{\lfloor T/m\rfloor }$. For a deterministic horizon, a stationary initial state yields a stationary final state, and agrees with the algorithm's output on coalescence. Letting the deterministic horizon tend to infinity proves the output has the stationary law without conditioning on a random stopping horizon.

A trace of a Markov chain Monte Carlo parameter helps distinguish slow movement, long rejection plateaus, and movement between separated regions. Very small random-walk proposals give high acceptance but slow diffusion; very large proposals produce low acceptance and repeated states. Record acceptance rates alongside trace plots and effective sample size of a Markov chain before deciding which scale to change. Correlated posterior directions often benefit from block proposals with a covariance fitted during a preliminary tuning run. A trace alone cannot prove convergence or identify the unique cause of poor movement; dispersed chains and stationary summaries provide additional evidence. Freeze tuning before the main sampling phase, or use adaptation satisfying an appropriate ergodicity theorem.

A row-stochastic matrix has nonnegative entries and each row sums to one. For a time-homogeneous Discrete-time Markov chain, its transition matrix has entriesIf a state has zero probability at time $n$, this row is specified by the model's Markov kernel, rather than recovered by conditioning on that zero-probability event. Rows index departure states and columns arrival states. Writing the current probability distribution as a row vector ${\mu }_n$ givesThe row-sum condition ensures that matrix multiplication preserves total probability. Under a column convention the evolution uses the transpose; the convention must accompany the formula.

For the sunny/rainy transition matrix, direct matrix multiplication givesIn particular, $(P^2{)}_{SS}=0.8\times 0.8+0.2\times 0.4=0.72$. The two terms sum the disjoint paths $S\to S\to S$ and $S\to R\to S$. Requiring both subsequent days to be sunny has probability $0.64$; requiring only the second day to be sunny permits the extra path through rain.

Generally $(P^n{)}_{ij}$ is the $n$-step transition probability. The identity $P^{m+n}=P^m{P}^n$ is the discrete-time Chapman-Kolmogorov equation: sum over the intermediate state. This makes matrix powers useful for forecasting many steps without enumerating every path individually.

For a transition matrix $P$, a finite-valued function on its state space is nonnegative superharmonic when $g\ge Pg\ge 0$. Thus $g\ge 0$ and its expected value cannot increase after one step. Replacing transitions on a target set by identity rows preserves $g\ge \tilde{P}g$. Iterating this inequality gives $g\ge {\tilde{P}}^{n}g$, a useful bound for hitting probabilities without a separate optional-stopping theorem.