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For a constant coefficient system, eigenvectors split the homogeneous equation into scalar exponential or power-law modes.

Linear operators $\alpha$ and $\beta$ on the same finite-dimensional vector space are conjugate when $\beta =s^{-1}\alpha s$ for some linear isomorphism $s$. Their matrices in any fixed basis are similar.

A unitary operator $U$ on a complex vector space with an inner product satisfies $U^{\ast }U=U{U}^{\ast }=I$ and therefore preserves inner products.

Mapping matter and halos from their initial positions adds the matter displacement contribution, giving $b=1+b_L$. Mass conservation requires the mass-weighted consistency relation $\int (dn/dM)(M/\bar{\rho })b(M)dM=1$.

A connected noncompact manifold is disconnected at infinity when the complement of some compact subset has at least two unbounded connected components. Equivalently, it has at least two ends.

The relevant general concept is an End; having several ends is a property of the space, not a synonym for one end.

The Jacobi identity is the compatibility condition on a Lie bracketTogether with an Alternating Lie bracket, it defines a Lie algebra. Unlike associativity, it relates three cyclic bracketings rather than equating two bracketings. Equivalently, using antisymmetry of a Lie bracket,Thus the linear map ${\mathrm{ad}}_x:y\mapsto [x,y]$ is a derivation of a Lie algebra: acting on a bracket is the sum of acting on each argument. This makes infinitesimal symmetries act consistently on each other's products. It also gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, the identity that makes the Adjoint representation of a Lie algebra a Lie algebra representation.

For the commutator in an associative algebra, the cancellation is directly visible:Each word occurs once with each sign. This proves the Jacobi identity without a restriction on the characteristic of a field.

For a computed check in the sl2 matrix commutator example,Here the last two terms are individually nonzero over characteristic zero; the identity records their cancellation. In the cross-product Lie algebra, the same cyclic cancellation is the vector triple product identity.

The historical connection is unusually concrete: Lie and Engel's volume I expands nested commutators of differential operators into twelve terms and cancels them cyclically, in Chapter 5, §26, equation (7). See Lie and Engel, §26, in Merker's translation, pp. 105–106.

The commutator of two elements of an associative algebra over a field isIt measures how much the order of multiplication matters: $[A,B]=0$ exactly when $AB=BA$. It is a bilinear map, it is alternating since $AA-AA=0$, and associativity makes it satisfy the Jacobi identity. Hence every associative algebra becomes a Lie algebra on the same vector space when equipped with this product. This holds in every characteristic of a field.

For square matrices, this is the matrix commutator, and it is defined even for singular matrices. The group commutator of invertible matrices is instead $AB{A}^{-1}{B}^{-1}$: it equals the identity element when the matrices commute, whereas $AB-BA$ equals the zero matrix. The Lie bracket from local group commutators explains how differentiation connects the two constructions.

A representation of $\mathfrak{g}$ on a vector space $V$ is a Lie algebra homomorphism $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$.

Every subset $B$ of a Lebesgue null set is a Lebesgue measurable set and is itself a null set. In fact the same conclusion follows directly from $m^{\ast }(B)=0$, without assuming measurability in advance. For any $A\subseteq \mathbb{R}$, monotonicity of Lebesgue outer measure gives $m^{\ast }(A\cap B)=0$ and $m^{\ast }(A\setminus B)\le m^{\ast }(A)$. ThereforeThe reverse inequality follows from countable subadditivity of Lebesgue outer measure, applied to the union $A=(A\cap B)\cup (A\setminus B)$. This is Carathéodory's criterion, proving measurability. If $B$ is a subset of an outer-null $N$, its outer measure is zero by the same monotonicity. Thus the restriction of outer measure to measurable sets is a complete measure.

If $g$ is locally Lebesgue integrable andthen $G^{\prime }(x)=g(x)$ at every Lebesgue point of $g$, and hence almost everywhere.

A point $x\in {\mathbb{R}}^n$ is a Lebesgue point of a locally Lebesgue integrable function $f$ whenThe Lebesgue differentiation theorem says that almost every point is a Lebesgue point.

Let nonzero ordinals $\xi ,\eta$ have leading exponents $\rho ,\sigma$. If $\eta$ is finite, then $\xi \eta$ has leading exponent $\rho$. If $\eta$ is infinite, so that $\sigma >0$, then $\xi \eta$ has leading exponent $\rho +\sigma$. This follows by multiplying the leading terms in Cantor normal form and using continuity of ordinal multiplication at limit ordinals.

Every nonzero ordinal $\alpha$ has a greatest exponent $\beta$ such that ${\omega }^{\beta }\le \alpha$. It is the leading exponent in the Cantor normal form of $\alpha$, andfor a positive finite integer $n$ and $\gamma <{\omega }^{\beta }$.

A distributive lattice is a lattice in which meet distributes over join and, equivalently, join distributes over meet.

The Clapeyron equation connects the slope of a phase coexistence curve to the latent heat and the change in volume during a first-order phase transition. A steep boundary can indicate a large latent heat, a small change in volume, or both. This lets a measured boundary constrain caloric properties and predicts how changing pressure shifts a melting temperature.

For a pure substance in bulk phase coexistence, label the initial and final thermodynamic phases $\alpha$ and $\beta$. Both have the same absolute temperature $T$, pressure $p$, and chemical potential. With $s$, $h$, and $v$ denoting entropy, enthalpy, and volume on one common basis, define $\Delta s=s_{\beta }-s_{\alpha }$, $\Delta h=h_{\beta }-h_{\alpha }$, and $\Delta v=v_{\beta }-v_{\alpha }$. ThenThe derivative follows the phase coexistence curve; it is not a partial derivative within one thermodynamic phase. Use molar volume $v_m$ with molar latent heat $L_m$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or specific volume $v$ with latent heat $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Both give $dp/dT$ in $\mathrm{Pa}{\mathrm{K}}^{-1}$. For molar mass $M$, $L_m=ML$ and $\Delta v_m=M\Delta v$, so the predicted slope is independent of the chosen basis. This formulation assumes a simple compressible system, fixed composition, and negligible interface contributions; a curved interface can instead require the Young–Laplace equation.

For positive latent heat in the stated conversion direction, the sign of the slope is the sign of $\Delta v$. Converting liquid to gas normally increases volume and gives a positive slope. The negative melting-point slope of water illustrates the opposite sign. At a thermodynamic critical point, the coexisting gas and liquid merge: both $L$ and $\Delta v$ vanish, so their ratio needs a limiting analysis rather than substitution of zero into the Clapeyron equation.^[ref]

A short derivation uses the molar Gibbs free energy $g_m$, equal to the chemical potential for a pure substance. Equality persists along the boundary, so $d{g}_{m,\alpha }=d{g}_{m,\beta }$ with $d{g}_m=-s_{m}dT+v_{m}dp$. Therefore $-\Delta s_{m}dT+\Delta v_{m}dp=0$. Equality of $g_m=h_m-T{s}_m$ also gives $\Delta h_m=T\Delta s_m$. Eliminating $\Delta s_m$ yields the Clapeyron equation.

The propagator $e^{i\gamma Bt{S}_z/\hslash }$ multiplies each magnetic component $|s,m⟩$ by $e^{im\gamma Bt}$.

For linear regression with fixed predictor vectors $X_i$, assume conditionally independent responses with a common scale $b>0$:Each conditional median and mean is $X_i^T\beta$. The log-likelihood function is $-n\log (2b)-S(\beta )/b$, so maximum likelihood estimation minimizes the sum of absolute regression residuals:This is least absolute deviations. With only an intercept, it reduces to the Laplace location and scale maximum likelihood problem, so a sample median is a fitted intercept. Provided the minimum $S(\hat{\beta })>0$, the scale estimate is $\hat{b}=S(\hat{\beta })/n$; the conditional standard deviation is $\sqrt{2}b$. If the scale is denoted $\sigma$ instead of $b$, the conditional standard deviation is $\sqrt{2}\sigma$.

Minimizing a sum of absolute losses does not automatically give a unique coefficient vector. Even an intercept-only fit to an even sample size can have an interval of solutions, and redundant predictor columns can introduce further nonuniqueness. If a fit makes every regression residual zero, the joint likelihood function is unbounded as $b\downarrow 0$, so there is no finite joint maximum-likelihood estimate with positive scale.

The linear growth of absolute loss limits sensitivity to an extreme response at fixed predictors compared with squared-error loss. It does not protect against every outlier: an extreme predictor can have large regression leverage. The numerical example above changes a response in an intercept-only problem, and makes no general claim of immunity to such predictor changes.

A finite contour based at an integrable branch point, or a closed contour around an integer-order pole, gives a solution analytic at $z=0$. It is therefore a constant multiple of the unique Frobenius power-series solution there.

Let $f_r(m)$ be the lower-shadow size prescribed by the binomial representation in the Kruskal-Katona theorem. ThenInduction using Pascal's identity proves the inequality and drives the standard ground-set induction for the theorem.