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A norm $p$ is nonnegative with $p(x)=0$ exactly when $x=0$, satisfies $p(ax)=|a|p(x)$, and obeys $p(x+y)\le p(x)+p(y)$. The Euclidean norm has positivity and absolute homogeneity immediately. For the triangle inequality, positivity of $\Vert x-ty{\Vert }_2^2$ for all real $t$ gives the Cauchy-Schwarz inequality $|x\cdot y|\le \Vert x{\Vert }_2\Vert y{\Vert }_2$ (with $y=0$ immediate). Hencewhich proves the triangle inequality after taking square roots.

For the prescribed set, use the Minkowski functionalSince $U$ is open and contains zero, some Euclidean ball of radius $r>0$ lies inside it. Since it is bounded, it lies inside a ball of radius $R$. These inclusions imply $\Vert x{\Vert }_2/R\le p_U(x)\le \Vert x{\Vert }_2/r$, so the functional is finite and positive for nonzero $x$; it is zero at zero. Symmetry gives $p_U(-x)=p_U(x)$, and substitution in the infimum gives $p_U(ax)=a{p}_U(x)$ for $a>0$, hence absolute homogeneity for all real scalars.

Convexity and $0\in U$ imply $sU\subseteq U$ for $0\le s\le 1$. Consequently, for $a>p_U(x)$ and $b>p_U(y)$, one has $x/a,y/b\in U$. Convexity then givesso $p_U(x+y)\le a+b$. Letting $a\downarrow p_U(x)$ and $b\downarrow p_U(y)$ proves the triangle inequality.

Finally, if $p_U(x)<1$, there is a $t<1$ with $x\in tU\subseteq U$. Conversely, if $x\in U$ and $x\ne 0$, openness permits $(1+\delta )x\in U$ for some $\delta >0$, so $p_U(x)\le 1/(1+\delta )<1$; zero is immediate. Thus $\boxed{U=\{x:p_U(x)<1\}}$, completing the norm from a bounded symmetric convex neighbourhood construction.

For $x\in \Omega$ and nonempty $A$, letThe Talagrand convex distance isEquivalently,For completeness, let $v_{\ast }$ be a closest point of the finite convex hull to zero. If $v_{\ast }\ne 0$, minimality gives $v_{\ast }\cdot (v-v_{\ast })\ge 0$ for every $v\in V$. Taking $\alpha =v_{\ast }/\Vert v_{\ast }\Vert$ proves that the supremum is at least $\Vert v_{\ast }\Vert$. Conversely each permitted $\alpha$ has ${\inf }_{v\in V}\alpha \cdot v\le \alpha \cdot v_{\ast }\le \Vert v_{\ast }\Vert$ by Cauchy-Schwarz. The weights are nonnegative because the hull has nonnegative coordinates. If $v_{\ast }=0$, both sides are zero. This proves the dual definition used below.

We prove Talagrand's convex distance inequality by induction on the number of factors. Write $\Omega ={\Omega }^{\prime }\times X_n$, with measure $\mu \otimes \nu$, and $F_A=d_T(\cdot ,A{)}^2$. For a last-coordinate value $z$, let $A_z=\{y^{\prime }:(y^{\prime },z)\in A\}$. Choose a section $B=A_b$ of largest probability $m=\mu (B)>0$, and put $a_z=\mu (A_z)/m\in [0,1]$. The total event probability is $p=\mathrm{Pr}(A)=m{\sum }_z\nu (z)a_z$.

If $A_z$ has positive probability, combine mismatch vectors from $A_z$ and from $B$. A weight $t$ on the second section contributes at most $t$ in the last coordinate. Convexity of the squared Euclidean norm gives, for every $0\le t\le 1$,The last-coordinate contribution is actually zero if $z=b$, so the displayed upper bound remains valid. Hölder's inequality and the induction hypothesis yieldFor a zero-probability section use $t=1$ directly, so no division by zero or exponent of an infinite quantity is needed.

The scalar estimate for Talagrand product induction isHere is its proof. For $a>0$ set $L=-\log a$. If $L\le 1/2$, take $t=2L$; the logarithm of the left expression is $L-L^2$. Put $h(L)=\log (2-e^{-L})-L+L^2$. Then $h(0)=h^{\prime }(0)=0$ andsince $(2u-1{)}^2-u=(4u-1)(u-1)\ge 0$ for $u=e^L\ge 1$. Thus $h\ge 0$. If $L\ge 1/2$, choose $t=1$. The already proved boundary case gives $e^{1/4}\le 2-e^{-1/2}\le 2-a$. The same direct choice covers $a=0$.

Average the resulting section bounds over $z$:where the last inequality is equivalent to $(m-p{)}^2\ge 0$. The induction starts on the zero-coordinate singleton, where every positive-probability event is the whole space and the moment is one. HenceWe assume $\mathrm{Pr}(A)>0$; a zero-probability event has no finite right-hand bound to prove. Markov's inequality also gives

For the subsequence application, let $L(x)$ be the longest increasing subsequence length for independent sequence coordinates. A witness of length $\ell$ uses only $\ell$ positions. If $L(x)=\ell \ge b$ and $y$ satisfies $L(y)\le a<b$, at least $\ell -a$ witness positions must change: all unchanged witness positions remain in increasing order. Put weights $1/\sqrt{\ell }$ on the witness and zero elsewhere. The dual distance definition giveswhere the latter expression increases with $\ell$ for $a\ge 0$. Therefore the increasing-subsequence certificate concentration bound isIf the lower event has probability zero the conclusion is immediate.

Let $m$ be an integer median, so $\mathrm{Pr}(L\le m),\mathrm{Pr}(L\ge m)\ge 1/2$. Taking the two thresholds in the bound gives, for $t>0$,For $t>m$ the lower event is empty. In a nonempty sequence $m\ge 1$. Integration of these tails gives an explicit mean-median estimate: use $m+t\le 2m$ for $t\le m$ and $m+t\le 2t$ for $t\ge m$, obtainingWriting $\mu =\mathbb{E}L$, the median property also gives $m\le 2\mu$. Consequently, for a universal constant $C$, all $u\ge 0$ satisfyThus the subsequence length is concentrated near its mean on the scale $\sqrt{\mu }$, rather than the generic coordinate-exposure scale $\sqrt{n}$. This argument applies directly to finite-valued independent sequences. Independent continuous samples follow by refining finite coordinate discretizations; the strict-order pattern stabilizes almost surely when ties have probability zero. A uniform random permutation is the rank sequence of independent continuous samples, so it has the same subsequence-length concentration. Arbitrarily dependent random sequences are not covered by the product-space assumption.

Suppose two distinct minimizers $x_1,x_2$ have the same minimum distance $d$ from $y$. Convexity puts their midpoint $z=(x_1+x_2)/2$ in $A$. Expanding the Euclidean norm givescontradicting minimality. Therefore the nearest point is unique. Together with the preceding existence proof, this defines the metric projection onto a nonempty closed convex set.

If the roots are parallel, their Euclidean norms are equal by reducedness. For nonparallel, nonorthogonal roots, the Cartan integers $a,b$ above are nonzero integers of the same sign and have product $1,2$ or $3$. AlsoThe absolute pairs $(|a|,|b|)$ are $(1,1)$, $(1,2),(2,1)$, or $(1,3),(3,1)$. HenceMore precisely, angles $\pi /3,2\pi /3$ require equal lengths; angles $\pi /4,3\pi /4$ require a long-to-short ratio $\sqrt{2}$; and angles $\pi /6,5\pi /6$ require a ratio $\sqrt{3}$. The exclusion of orthogonality matters: orthogonal irreducible components can be scaled independently.

For the sample, put $\bar{x}=n^{-1}{\sum }_i{x}_i$ and use the sample covariance matrixOrder its eigenvalues ${\hat{\lambda }}_j$ decreasingly and choose unit orthogonal eigenvectors ${\hat{\ell }}_j$. The sample principal components are the score columnsTheir sample means are zero and their sample covariance is diagonal, since ${\hat{\ell }}_j^{T}S{\hat{\ell }}_k={\hat{\lambda }}_j{1}_{\{j=k\}}$. The $j$th sample variance is ${\hat{\lambda }}_j$, and its explained variance of a principal component is ${\hat{\lambda }}_j/\mathrm{tr}(S)$. Using divisor $n$ instead changes the numerical variances by a common factor but leaves the component directions unchanged.

Standardizing means replacing a variable with positive sample variance by $(x_{ij}-{\bar{x}}_j)/s_j$. This gives principal component analysis on a correlation matrix, using $C=D^{-1/2}S{D}^{-1/2}$ with $D=\mathrm{diag}(S)$. It removes arbitrary differences of measurement units and prevents a large-variance coordinate from dominating solely through its scale. It also changes the question being optimized: small-variance and possibly noisy coordinates receive equal marginal weight. Principal component analysis based on the covariance matrix can be more appropriate when the variables have common meaningful units and absolute variability matters. Standardization is therefore a substantive choice, not an automatic improvement. A zero-variance variable must first be removed. In the word-rating example the variables share the same numerical range, but their across-word dispersions can differ; standardization gives each attribute equal initial variance.

For the standardized eight-variable analysis, total variance is $\mathrm{tr}(C)=8$. The first three fractions of explained variance of a principal component are approximately 59.6%, 19.1% and 10.1%, respectively. The first two account for about 78.8%, and the first three for 88.9%. The first direction is dominant but one direction alone loses substantial variation. Retaining two principal components is a plausible descriptive summary: their eigenvalues exceed $1$, whereas the third is below $1$. That cutoff is a heuristic, and the third principal component may still have interpretable content.

The reported coefficient rows are not unit-length vectors. The printed rounded rows have squared lengths $4.7514$, $1.5222$ and $0.8063$, consistent with their eigenvalues to the precision of the coefficient table. They are consistent with the scaled principal component loading convention $r_j=\sqrt{{\lambda }_j}{\ell }_j$. Such rows remain eigenvectors, but the score direction with unit length is ${\ell }_j=r_j/\sqrt{{\lambda }_j}$, or more accurately the row divided by its actual Euclidean norm when using rounded entries. Under this convention $r_{ij}$ is the correlation of standardized attribute $i$ with the variance-one score for component $j$:This explains both their sizes and why they must not be treated as the unit vectors from parts (i) and (ii).

The first principal component has positive coefficients on every attribute, strongest on friendliness, goodness, niceness, bravery and strength. With the favorable endpoints coded in the positive direction, it describes a broad favorable or impressive evaluation, with a potency contribution. The second contrasts movement and speed, with some size and strength, against the first three evaluative attributes; it is chiefly an activity dimension. The third is dominated by size, with smaller positive strength and negative movement/speed coefficients, distinguishing size or potency from activity. These are interpretations of directions of variation, not evidence of causal latent traits. Reversing a principal component's overall sign changes neither its meaning as an axis nor its explained variance; endpoint coding fixes only how its positive side is described.

Take the norm to be the Euclidean norm. The solution is $\mathbf{q}(t)=e^{tA}\mathbf{q}(0)$, so optimal growth is the operator norm $G(t)=\Vert e^{tA}{\Vert }_2$, equivalently the largest singular value of the matrix exponential. Let $s(A)$ be the spectral abscissa and let $w(A)={\lambda }_{\max }((A+A^{\ast })/2)$ be the numerical abscissa. The Euclidean semigroup growth bounds areFor the lower bound, the spectral radius of $e^{tA}$ is $e^{ts(A)}$ and is bounded by its operator norm. For the upper bound, differentiating $\Vert \mathbf{q}{\Vert }_2^2$ gives $2\mathrm{Re}({\mathbf{q}}^{\ast }A\mathbf{q})\le 2w(A)\Vert \mathbf{q}{\Vert }_2^2$, and integration gives the result. A non-normal matrix can have $w(A)>s(A)$, allowing transient amplification even when all its eigenvalues have negative real parts.

For the triangular matrix, $s(M)=-1/R$. The eigenvalues of its Hermitian part are $-3/(2R)\pm \frac{1}{2}\sqrt{1+R^{-2}}$, soThe upper exponent becomes positive when $R>\sqrt{8}$, even though the long-time spectrum is stable.

The two equations are $q_1^{\prime }=-q_1/R$ and $q_2^{\prime }=q_1-2q_2/R$. Solve the first, then use an integrating factor in the second. Writing $s=t/R$, $a=e^{-s}$, $b=e^{-2s}$ and $c=R(a-b)$ givesThe maximizing initial state is the unit eigenvector of $B^{\ast }B$ belonging to its largest eigenvalue. HereFor $t>0$, defineThe optimal initial state for triangular stable shear is thereforeMultiplication by any nonzero complex scalar gives the same growth ratio; at $t=0$, every nonzero initial state is optimal.

For $t\gg R$, $d\to \infty$ and $r\to 0$. Thus the optimal initial state tends to $(1,0{)}^T$, and $G(t)\sim e^{-t/R}\sqrt{1+R^2}$. The initial first component excites the slower eigenmode and its amplified second component.

For $t\ll R$, the exact expression gives $d=(R^{-1}+t/2)[1+O(t/R)]$. Accordingly the short-relative-time optimal state for triangular shear usesThis retains both the unequal damping rates and the shear. If $R\gg 1$, it reduces to $r\sim (\sqrt{t^2+4}-t)/2$, the maximizing direction for the shear matrix $\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. If time also tends to zero at fixed $R$, it instead tends to $r=\sqrt{1+R^{-2}}-R^{-1}$, the largest-eigenvalue direction of the Hermitian part of $M$. Merely setting $t/R$ small does not justify setting $t$ small; the exact formula resolves both possibilities.

The gradient of $x_3^2-x_1^2-x_2^2$ is $2(-x_1,-x_2,x_3)$. On the upper cone, its Euclidean norm is $2\sqrt{2}{x}_3$. Thus the upward unit normal to an upper circular cone isPut $x_h=(x_1,x_2)$ and $p_h=(p_1,p_2)$. The hypotheses imply $|x_h|=x_3$ and $|p_h|\le p_3$. The Cauchy-Schwarz inequality therefore givesThis is also a direct instance of self-duality of a second-order cone: both $p$ and the chosen unit normal belong to the closed Lorentz cone.

For $y\ne 0$, the squared Euclidean norm of the orthogonal projection residual is nonnegative:This proves the Cauchy-Schwarz inequality $|x\cdot y|\le |x||y|$. Equality holds exactly when $x$ is a scalar multiple of $y$. If $y=0$, equality is automatic. Equivalently, equality means the two vectors are linearly dependent, including the cases with a zero vector.

For a point $C$ equidistant from the coordinate vertices,must be independent of $i$. Thus all $C_i$ are equal, and their sum is one. The centroid is thereforeFor every $x$ in the base probability simplex, the Cauchy-Schwarz inequality gives $1=(1,\dots ,1)\cdot x\le \sqrt{n}|x|$. Equality forces the same coordinates as $C$. Thus $C$ is the unique closest point of a probability simplex to the origin.

The angle $\alpha$ between $OC$ and any edge from the origin to $e_i$ satisfiesConsequently

False. Consider the real matrixIt is block upper triangular, so its characteristic polynomial is the same as that of $A_0$ and it has the required eigenvalues. Its second column, however, has squared Euclidean norm $1+c^2+c^2=2$, whereas every column of an orthogonal matrix has norm one. ThusUnit-modulus eigenvalues alone do not imply orthogonality; they do not control the geometry of the eigenvectors.

A full QR decomposition of a real $m$ by $n$ matrix with $m\ge n$ writeswhere $Q$ is an orthogonal $m$ by $m$ matrix and $R$ is upper triangular $n$ by $n$. In the full-column-rank case $R$ is invertible. Equivalently the reduced factorization is $A=Q_{1}R$ with $Q_1^T{Q}_1=I_n$. Householder transformations can construct the full factorization.

Let $Q^{T}b=\left(\genfrac{}{}{0ex}{}{c_1}{c_2}\right)$ with $c_1\in {\mathbb{R}}^n$. Since orthogonal transformations preserve the Euclidean norm,Therefore the linear least-squares problem has the full-rank solutionBack substitution solves this triangular system without forming $A^{T}A$. If the matrix is rank deficient, use a rank-revealing pivoted factorization and minimize over its independent columns; the minimizer need not be unique. The original PDF minimizes over $x\in {\mathbb{R}}^n$; the converted TeX's minimization over $b$ is a transcription error.

Uniqueness in law means that, for a fixed initial value, every two weak stochastic solutions of the stochastic differential equation have the same distribution as entire solution paths, even if their filtered probability spaces and driving Brownian motions differ. For an equation on $(0,\infty )$, initially consider the stopped path and lifetime at the boundary.

We first justify that the boundary cannot intervene when $\gamma \ge 2$. For any solution of the displayed Bessel equation, let ${\tau }_{\epsilon ,L}$ be its first exit from $(\epsilon ,L)$, where $0<\epsilon <r<L$. Applying Itô to $R^2$ givesThus the annulus exit has finite expectation and is almost surely finite. The one-dimensional generator is $\frac{1}{2}{\partial }_{rr}+\frac{\gamma -1}{2r}{\partial }_r$. Its harmonic scale functions are $\log r$ for $\gamma =2$ and $r^{2-\gamma }$ for $\gamma >2$, as direct differentiation verifies. They are bounded on the stopped annulus, so optional sampling givesBoth tend to zero as $\epsilon \downarrow 0$. A continuous path hitting zero before ${\tau }_L$ would hit every inner level first, so that event has probability zero. Exhausting the upper levels proves that zero is not hit at any finite time. Nor can an upper explosion occur: localization of $R^2$ gives $\mathbb{P}({\tau }_L\le t)\le (r^2+\gamma t)/L^2\to 0$. These are the Hitting-zero classification for a Bessel process arguments needed for the dimension-at-least-two case.

Now suppose $\gamma$ is an integer and take a $\gamma$-dimensional Brownian motion $U$ from any deterministic vector $u$ with $|u|=r$. For $\rho =|U|>0$, differentiation of the Euclidean norm givesDefine the radial noise $\beta ={\sum }_i\int U_i/\rho d{W}_i$, filling the integrand by a fixed unit vector on the zero set, as in part (b). It has bracket $t$ and is a Brownian motion. the Itô formula, initially before a possible zero, givesThe boundary argument just given makes this a global positive weak stochastic solution. The assumed uniqueness in law therefore identifies every solution's path law with that of $|U|$. Orthogonal invariance of Brownian motion makes this radial law independent of the selected point $u$ on the sphere. ConsequentlyThis proves equality of process laws, not merely equality of one-time marginal distributions.

Differentiability at $p$ means there is a linear map $A=Df{|}_p$ such thatThis is the Fréchet derivative definition; the same linear approximation must work in every direction, not merely along coordinate axes.

The chain rule says that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, then $g\circ f$ is differentiable at $p$ andFor its proof, put $A=Df{|}_p$, $B=Dg{|}_{f(p)}$, and $k=f(p+h)-f(p)=Ah+r(h)$. Since a linear map between these finite-dimensional normed spaces is bounded, $\Vert k\Vert =O(\Vert h\Vert )$. Differentiability of $g$ gives $g(f(p)+k)=g(f(p))+Bk+s(k)$ with $s(k)=o(\Vert k\Vert )$. ThereforeIf $k=0$, take $s(0)=0$; otherwise $\Vert s(k)\Vert /\Vert h\Vert =(\Vert s(k)\Vert /\Vert k\Vert )(\Vert k\Vert /\Vert h\Vert )\to 0$. This proves the required full linear approximation.

For the Euclidean norm, at every $x\ne 0$ the ordinary one-coordinate differentiation givesAt the origin its difference quotient in direction $e_i$ is $|t|/t$, whose right and left limits are $1$ and $-1$. Thus none of its coordinate partial derivatives exists at the origin.

In the truncated tensor algebra, let ${\mathfrak{g}}^N(V)$ be the linear span of all iterated Lie brackets of elements of $V$ of length at most $N$, with bracket $[a,b]=a\otimes b-b\otimes a$. Brackets of length greater than $N$ vanish. The free step-N nilpotent Lie group isThe exponential and logarithm are the finite seriesIts multiplication is the tensor multiplication above, its identity is $e=(1,0,\dots ,0)$, and its inverse isThe truncated Baker--Campbell--Hausdorff formula shows that this multiplication stays in the exponential of the Lie algebra. This is the connected simply connected Lie group of the free step-$N$ nilpotent Lie algebra, rather than a discrete free nilpotent group.

Give the first layer $V$ its usual Euclidean norm. An absolutely continuous curve $\gamma$ in $G^N(V)$ is horizontal iffor an absolutely continuous control $x$ in $V$. In the truncated tensor algebra this equation is ${\dot{\gamma }}_t={\gamma }_t\otimes {\dot{x}}_t$, truncated at degree $N$. Its horizontal length is $\int |{\dot{x}}_t|dt$. The Carnot-Carathéodory distance isThe first-layer Lie brackets span the full Lie algebra, so horizontal curves reach every endpoint. Thus this infimum is finite; it is a left-invariant distance inducing the usual manifold topology. Dilations ${\delta }_{\lambda }(a{)}^{(k)}={\lambda }^k{a}^{(k)}$ give $d_{CC}({\delta }_{\lambda }g,{\delta }_{\lambda }h)=\lambda d_{CC}(g,h)$ for $\lambda >0$. Equivalently $d_{CC}(e,g)$ is the least length of a bounded variation control whose step-$N$ path signature is $g$.

For $p\ge 1$ set $N=\lfloor p\rfloor$. A weak geometric p-rough path is a continuous path $\mathbf{X}:[0,T]\to G^N(V)$, usually based at $e$, withIts increments ${\mathbf{X}}_{s,t}={\mathbf{X}}_s^{-1}\otimes {\mathbf{X}}_t$ obey ${\mathbf{X}}_{s,u}={\mathbf{X}}_{s,t}\otimes {\mathbf{X}}_{t,u}$. Equivalently their degree-$k$ tensor components have finite $p/k$-variation and satisfy the path signature algebraic constraints. This is a group-valued definition; being a limit of smooth path signatures in the same $p$-variation topology is the stronger definition of a geometric p-rough path.

For a real skew-symmetric matrix, $P^T=-P$. A scalar quadratic form satisfies $y^{T}Py=(y^{T}Py{)}^T=y^T{P}^{T}y=-y^{T}Py$, so it is zero. ThereforeThe finite-dimensional linear system has the global solution $e^{tP}{y}_0$. Equivalently, $(e^{tP}{)}^T{e}^{tP}=e^{-tP}{e}^{tP}=I$, so its exact flow is an orthogonal matrix and preserves the Euclidean norm.

The Runge-Kutta method has update $y_{n+1}=y_n+h{\sum }_i{b}_i{k}_i$. Expanding its squared Euclidean norm gives the exact identityThe stated per-step relation makes this difference zero. Induction from the initial vector then provesFor a positive step size the relation is also necessary for exact per-step norm conservation. This algebraic identity applies whenever the stage equations define a step; it does not silently assume solvability of arbitrary implicit stages.

Two norms $\Vert \cdot {\Vert }_a,\Vert \cdot {\Vert }_b$ are Lipschitz equivalent norms when constants $c,C>0$ satisfy $c\Vert x{\Vert }_a\le \Vert x{\Vert }_b\le C\Vert x{\Vert }_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a Cauchy sequence for one norm exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one norm is complete, a Cauchy sequence for the other converges using the first norm and hence also the second. Interchanging the norms proves completeness is preserved in both directions.

Write $x={\sum }_i{x}_i{e}_i$ and set $C={\sum }_i\Vert e_i\Vert$. If $\Vert x{\Vert }_2=1$, then $|x_i|\le 1$, and the triangle inequality givesScaling a nonzero $x$ gives $\Vert x\Vert \le C\Vert x{\Vert }_2$ for every $x$. The reverse triangle inequality now impliesso $x\mapsto \Vert x\Vert$ is continuous, indeed Lipschitz, in the Euclidean norm.

Suppose there were no positive lower bound on the Euclidean unit sphere. We could choose $x_j$ with $\Vert x_j{\Vert }_2=1$ but $\Vert x_j\Vert <1/j$. The Bolzano-Weierstrass theorem supplies a subsequence converging in the Euclidean norm to $x$. Continuity of both norms gives $\Vert x{\Vert }_2=1$ and $\Vert x\Vert =0$, contradicting definiteness. Therefore some $m>0$ satisfies $\Vert x\Vert \ge m$ on the unit sphere. Scaling again proves the finite-dimensional equivalence of norms:

Use geometrized units $G=c=1$. For a stationary asymptotically flat spacetime, normalize the timelike Killing vector field $\xi$ to have squared norm $-1$ at infinity. The Komar mass isHere the oriented binormal is chosen so that an ordinary positive-mass static solution gives a positive integral. Equivalently, on a static slice with lapse $N=\sqrt{-{\xi }^2}$ and outward spatial unit normal $s^i$,This second expression fixes the orientation unambiguously and is convenient for the Majumdar–Papapetrou solution.

The spatial metric is ${\gamma }_{ij}=H^2{\delta }_{ij}$ and $N=H^{-1}$. On a large coordinate sphere, $s^i=H^{-1}{\hat{r}}^i$ and $dA=H^2{r}^{2}d\Omega$. ThusThe original PDF has the Euclidean norm in each source denominator, which is missing in the TeX aid. With that norm, the harmonic function has asymptoticsThereforeIf $M_i$ are retained as length parameters while $G$ is restored, the physical mass is ${\sum }_i{M}_i/G$. This is the charge at infinity. A Komar integral is not generally surface-independent here, since Einstein-Maxwell theory has the electromagnetic stress-energy tensor between the surfaces; in particular it should not be confused with a sum of vacuum horizon Komar charges.

Write $y\in {\mathbb{R}}^n$ for the observed response and $x_j$ for column $j$ of the working design matrix $X$. The usual LARS convention gives every covariate unit Euclidean norm. If an unpenalized intercept is fitted separately, first centre $y$ and the nonconstant covariates, and restore the intercept afterward. For an intercept-free model no centring is required. The algebra below also applies to an unscaled full-rank design; only the literal equal-angle interpretation uses unit-length columns. The penalty is defined on the regression coefficients of whichever working design is used, since rescaling covariates changes the units of an L1 penalty.

Initialize ${\hat{\beta }}^0=0$, ${\hat{\mu }}^0=X{\hat{\beta }}^0=0$ and residual $r^0=y$. At iteration $k$, letThese $c_j^k$ are residual inner products, often called residual correlations. If $C^k=0$, all normal equations are satisfied and the path has reached ordinary least squares. Otherwise the active set ${\mathcal{A}}^k$ consists of the selected columns tied at the maximal absolute correlation. Initially select the maximizing column, or all such columns in a tie. Ordinary least angle regression retains previously selected columns and adds each new one when it ties the active correlations. Put $s_j^k=\mathrm{sign}(c_j^k)$ for active $j$.

To define the direction, form the signed active design $X_{\mathcal{A}}^s=(s_j^k{x}_j{)}_{j\in {\mathcal{A}}^k}$ and its Gram matrix $G_{\mathcal{A}}=(X_{\mathcal{A}}^s{)}^T{X}_{\mathcal{A}}^s$. Full column rank makes this matrix positive-definite. Let $1_{\mathcal{A}}$ denote the vector of ones of length $|{\mathcal{A}}^k|$, and setThen $\Vert u^k{\Vert }^2=1$ and $(X_{\mathcal{A}}^s{)}^T{u}^k={\alpha }^k{1}_{\mathcal{A}}$. This is the equiangular direction in least angle regression. Define $a_j^k=x_j^T{u}^k$ for every covariate, and define the regression coefficient direction by $d_j^k=s_j^k{w}_j^k$ on ${\mathcal{A}}^k$, zero otherwise. It satisfies $X{d}^k=u^k$.