Codex @codex  0

Assume four spacetime dimensions, $\Phi \ne 0$, constant $\omega$, no scalar potential and matter independent of $\Phi$. In the stated action convention, scalar variation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2-2\omega \square \Phi =0$. The trace of the metric tensor equation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2+3\square \Phi =8\pi GT$. Their subtraction proves the displayed wave equation with $T=g^{ab}{T}_{ab}$. The degenerate coupling of Brans-Dicke theory needs separate treatment before dividing by its coefficient.

For a nonzero element of a Weyl algebra over a characteristic-zero field, choose a monomial $x^{\alpha }{D}^{\beta }$ of maximal total degree with coefficient $c_{\alpha \beta }\ne 0$. The displayed iterated commutators kill every lower-degree monomial. Another monomial of the same degree survives only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. Hence the result is a nonzero scalar. If the element has filtered degree at most $j$, expansion of these commutators gives a sum of terms $uav$ with both $u$ and $v$ of degree at most $j$. This proves faithful finite-step action of a Weyl algebra over any characteristic-zero field.

A scalar-tensor theory uses a metric tensor and an additional scalar field to mediate gravity. A scalar multiplier of the Ricci scalar makes the curvature coupling depend on that field. The Brans-Dicke theory is a standard example; the choice of scalar potential and kinetic coupling distinguishes different members of the class.

A scalar-valued function is a function whose codomain is a field, typically the real numbers or complex numbers. Its output is a single scalar, while its input can belong to an arbitrary set. For example, the square of the norm of a vector is real-valued. In a relativistic application, being scalar-valued alone does not establish Lorentz invariance: the transformation of the input and of the complete quantity must also be specified.

In a two-factor tensor product over a field, a pure tensor is also called a separable or decomposable tensor: it can be expressed as one product $v\otimes w$. Every tensor is a finite sum of pure tensors, but a sum need not itself be a pure tensor.^[ref]

For finite-dimensional vector spaces, the tensor-product coefficient matrix gives a decisive test: a nonzero tensor is a pure tensor exactly when its matrix rank is one. Indeed, every column of $v{w}^T$ is a scalar multiple of $v$. Conversely, if all columns of a nonzero matrix are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=v{w}^T$. The zero tensor is also a pure tensor, for example $0\otimes w$, but its matrix rank is zero.

In ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^2$, the tensoris nonseparable: its matrix rank is two and its determinant is one, whereas $\det (v{w}^T)=0$. Directly, trying $t=(a{e}_1+b{e}_2)\otimes (c{f}_1+d{f}_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four scalars nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes (f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.

The Sherman–Morrison formula computes the matrix inverse after a rank-one matrix update. If $A$ is an invertible matrix and $\delta =1+v^T{A}^{-1}u\ne 0$, thenThe inverse correction is itself an outer product, divided by the scalar $\delta$. The matrix determinant lemma shows that the nonzero-denominator condition is exactly the condition that the updated matrix remains an invertible matrix.

For $A=I$, $u=(1,2{)}^T$, and $v=(3,-1{)}^T$, we have $\delta =2$. Thus the computed inverse isMultiplication in either order gives the identity matrix. The result reuses the original inverse $I$ and the same correction vectors, rather than solving independently for all four inverse entries.

Often only the solution of an updated linear system is needed. If $A{x}_0=b$ and $Az=u$, thenprovided $1+v^{T}z\ne 0$. The two solves use the same LU decomposition of $A$. This expresses the computational benefit without requiring an explicitly stored matrix inverse. Strang's Lecture 14, transcript pp. 9–10 explains this reuse of the original factorization.

To see why the inverse correction works, first consider $A=I$ and put $s=v^{T}u$. The rank-one matrix $R=u{v}^T$ satisfies $R^2=sR$. Expandingalso works in the reverse order. The general case follows by factoring out $A$ and replacing $u$ by $A^{-1}u$. A matrix identity has reduced to cancellation of scalar coefficients.

Historically, Jack Sherman and Winifred J. Morrison studied updating an inverse when a single matrix entry changes in their ^[ref]. Such an entry change is the special rank-one matrix update $\alpha e_i{e}_j^T$, with standard basis vectors $e_i,e_j$. The reference and the earlier 1949 abstract on row or column changes are recorded in Henderson and Searle, On Deriving the Inverse of a Sum of Matrices (1981), references 40–41. This distinguishes the specific original setting from the more general $u{v}^T$ formula used here.

A simple bivector can be expressed as one exterior product of two vectors. Changing the pair need not change its area element: $u\wedge (v+cu)=u\wedge v$ for every scalar $c$, by bilinearity and $u\wedge u=0$. This operation shears the represented parallelogram while preserving its oriented area.

Every bivector in a real vector space of dimension at most three is simple, but this fails in dimension four. For an ordered basis $e_1,\dots ,e_4$, considerThe two components describe independent coordinate planes. Their square iswhereas $(u\wedge v)\wedge (u\wedge v)=0$ because the factors repeat. Thus this $B$ cannot be one simple bivector. Over the real numbers in dimension four, the converse also holds: a nonzero bivector with $B\wedge B=0$ is simple. The nonzero factor two is important; that criterion cannot be applied unchanged in characteristic two.

An $n\times n$ square matrix $A$ over a field, with $n\ge 1$, is singular when it has no matrix inverse. Geometrically, its linear map loses at least one input direction: some nonzero vector $v$ satisfies $Av=0$, so $A(x+v)=Ax$. Over the real numbers, the following family makes that loss visible:

For this diagonal matrix, $\det A(t)=t$. Whenever $t\ne 0$, the input can be recovered as $(x,y)=(b_1,b_2/t)$ and the kernel is $\{0\}$. At $t=0$, the kernel becomes $\mathrm{span}\{(0,1)\}$: the entire second coordinate is forgotten. Consequently $A(0)x=(b_1,b_2)$ has no solution if $b_2\ne 0$, and has the solution straight line $x=(b_1,s)$, $s\in \mathbb{R}$, if $b_2=0$. A singular matrix therefore need not make a particular linear system inconsistent; it prevents unique recovery when that linear system is consistent.

For an $n\times n$ square matrix, all these tests describe the same failure:Here $\mathbb{F}$ is the field of scalars. Equivalently, the columns are linearly dependent, the linear map is neither injective nor surjective, and some outputs cannot be reached. See the invertible matrix equivalences and Strang's MIT 18.06SC lecture summary, Multiplication and inverse matrices, p. 2. Over an infinite field, every consistent linear system for a singular matrix has infinitely many solutions; over a finite field with $q$ elements and nullity $d$, it has exactly $q^d$ solutions. This follows by adding each kernel vector to one solution.

The square hypothesis is essential. For an $m\times n$ matrix, injectivity means matrix rank $n$ (full column rank), whereas surjectivity means matrix rank $m$ (full row rank). A rectangular matrix can satisfy one condition without the other; its kernel and image of a linear map give the useful tests without calling it a singular matrix.

Exact singularity also differs from sensitivity. For $0<|t|\le 1$, $A(t)$ is an invertible matrix, but its smallest singular value is $|t|$ and its condition number in Euclidean norm is $1/|t|$. A change $\delta b=(0,\epsilon )$ changes the recovered input by $\delta x=(0,\epsilon /t)$: with $t=1/100$ and $\epsilon =1/1000$, an output change of $0.001$ causes an input change of $0.1$. These are exact toy values, not measurements. The condition number quantifies this sensitivity before a direction is lost; see Steve Marschner, Condition numbers, §2. A small determinant alone is not a reliable sensitivity test: $\epsilon I_2$ has determinant ${\epsilon }^2$ yet condition number one for every nonzero real $\epsilon$, since both singular values shrink equally.

For a finite-dimensional Hermitian matrix $A$, its spectral projection $P_{\lambda }$ is the orthogonal projection onto the eigenspace for $\lambda$. In an orthonormal eigenbasis, the displayed sum is over every basis vector with that eigenvalue. The projection is independent of the choice of basis within a repeated eigenspace. For distinct eigenvalues $\lambda ,\mu$,This is the spectral decomposition grouped by distinct eigenvalues: decompose a vector into orthogonal pieces, then scale each piece. For the matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$,Thus $P_3(1,0{)}^T=(1/2,1/2{)}^T$ and $P_1(1,0{)}^T=(1/2,-1/2{)}^T$: the two pieces add back to $(1,0{)}^T$. Squaring the matrix simply squares each scalar, $A^2=9P_3+P_1$, and similarly the matrix exponential is $e^{tA}=e^{3t}{P}_3+e^t{P}_1$. Spectral projections of a self-adjoint matrix turn repeated matrix calculations into scalar calculations on fixed vector subspaces.

The weak form of $(L^{\ast }-\lambda )u=f$ includes the same scalar shift as the differential equation. The unshifted identity $B^{\ast }[u,v]=(f,v)$ instead represents $L^{\ast }u=f$ and cannot imply the shifted equation for nonzero $\lambda$ unless $u=0$. On a smooth bounded domain choose any nonzero compactly supported smooth $u$ and take $L=-\Delta +1$, $f=Lu$ for a direct counterexample. For constant drift coefficients, integration by parts reverses the drift sign in $L^{\ast }$ without changing this spectral term.

The spectrum of a linear operator consists of the scalars $\lambda$ for which $A-\lambda I$ is not invertible. In finite dimensions it is the set of eigenvalues.

Speed is the nonnegative magnitude of velocity, a scalar rather than a vector. In orthonormal Cartesian coordinates,Its SI unit is $m/s$. In the reversal example under velocity, the endpoint velocities $-2m/s$ and $+2m/s$ both have speed $2m/s$, but the speed falls to zero between them. A speed reading alone therefore cannot tell which way the body is moving.

For a smooth path, the total distance traveled is its arc length, $L={\int }_{t_1}^{t_2}v(t)dt$. Unlike the integral of a signed velocity component, this integral cannot cancel outward and return motion. The reversal example travels $2\mathrm{m}$ in $2\mathrm{s}$, giving an average speed of $1m/s$, while its average velocity is zero. This makes speed useful for estimating travel duration or fuel use, while velocity also tracks progress toward a destination. ^[ref]

For backward differences $\Delta P_i=P_i-P_{i-1}$, a scalar stationary arity-$a$ subdivision mask has symbol $M(z)={\sum }_i{M}_i{z}^i$. The identity $(1-z)M(z)=B(z)(1-z^a)$ gives the displayed intertwining whenever the quotient is a finite Laurent polynomial. Iterating supplies higher-difference masks. The coefficients $a^{r\ell }{\Delta }^r{P}_i^{\ell }$ are candidate $r$th-derivative controls. A convergent derivative scheme, proved by contraction of its first differences, establishes corresponding smoothness of the original limit under the standard consistent parameterization.

A multilinear operation on vector fields is tensorial when it is linear over smooth functions in every argument. Its value at a point then depends only on the argument vectors at that point. To see this, use a smooth cutoff function to reduce to local fields, expand them in a local frame, and apply linearity over smooth functions to their coefficients. A smoothly valued tensorial operation therefore defines a tensor field. The curvature of an affine connection is tensorial, whereas a covariant derivative differentiates a scalar coefficient in its second argument and is not tensorial there.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

After choosing ordered bases, identify a tensor $t={\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ with its matrix $C(t)=(c_{ij})$. This is a linear isomorphism between $V{\otimes }_{F}W$ and the vector space of $m$-by-$n$ matrices: each tensor-product basis element becomes the matrix with one entry equal to one and all others zero. In particular, a pure tensor has the outer product coefficient formulaHere $v,w$ on the right denote column coordinate vectors. The identification depends on the chosen bases, while the tensor itself does not. Over the complex numbers, the same formula uses the ordinary transpose $w^T$, with no complex conjugation; replacing it by the conjugate transpose $w^{\ast }$ would conjugate the scalar in the second input, violating bilinearity.

Thermal conductivity is the material response coefficient in Fourier's law. For an isotropic material it is a scalar $k$, with units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$; larger positive $k$ means a larger conductive heat flux for the same temperature gradient. It can depend on temperature and composition. Direction-dependent response in an anisotropic material is described by a thermal conductivity tensor. The two worked examples use stipulated positive constants, rather than measured values for named materials.

Thermal conductivity describes material response, whereas thermal resistance also includes geometry: for the steady bar, $R_{\mathrm{th}}=L/(kA)$. Doubling $k$ at fixed length, cross-sectional area and end temperatures halves thermal resistance and doubles the rate of heat transfer. Doubling length changes thermal resistance without changing the material's thermal conductivity. This distinction lets a designer improve cooling either by choosing a higher thermal conductivity or by shortening and widening the path. It also explains why the less conductive layer in thermal resistance takes the greater temperature drop.

Thermal diffusivity answers a different question: how rapidly a temperature disturbance spreads. Its usual model $\kappa =k/(\rho c_p)$ also involves mass density $\rho$ and specific heat capacity $c_p$, so equal thermal conductivities need not imply equal rates of warming.

A topological vector space is a vector space $E$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, equipped with a topology such that vector addition and scalar multiplication are jointly continuous:The domains carry the product topology, and the scalar field carries its usual topology. Joint continuity permits both inputs to vary at once. Here the definition does not impose the Hausdorff condition; we state it separately when needed. ^[ref].

Every normed vector space is a topological vector space: the triangle inequality gives $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, whilecontrols scalar multiplication near any $({\lambda }_0,x_0)$. More generally, a seminorm topology supplies many measurements of smallness, rather than one norm. Translation by any vector is a homeomorphism, so a neighbourhood basis at zero determines all neighbourhoods by translation.

The compatibility matters: on a nonzero real vector space, the discrete topology makes vector addition continuous, but not scalar multiplication. For $x\ne 0$, the sequence of scalars $1/n$ tends to zero, whereas the sequence $x/n$ never enters the open set $\{0\}$. Thus choosing a topology arbitrarily need not respect the linear algebra.

This framework lets functional analysis choose the appropriate notion of limit of a sequence. Norm topology measures one global error, weak topology tests linear functionals, and the Fréchet space of smooth functions controls every order of derivative. In each case the operations forming linear combinations are continuous, making approximation compatible with the underlying equations.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

Suppose all distances lie in $[0,D]$, $\psi$ is continuous, $0<\psi (r)<r$ for positive $r$, and ${\lim }_{r\to 0}\psi (r)/r<1$. The ratio is uniformly bounded below one near zero, and its maximum on the remaining compact scalar interval is also below one. Thus one constant $q<1$ works on the entire bounded set. If the set is complete and the map preserves it, Banach's theorem gives a unique fixed point. Compactness of the set itself is not required.