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For a nonzero real vector $w$, a Householder reflection isWriting $P=w{w}^T/(w^{T}w)$ gives $P^T=P$ and $P^2=P$, so $H^{T}H=(I-2P{)}^2=I$. Thus a Householder matrix is an orthogonal matrix, and it reflects across the hyperplane perpendicular to $w$.

For $u\ne 0$, choose $\alpha =-\mathrm{sgn}(u_1)\Vert u\Vert$, taking the sign to be $+1$ when $u_1=0$, and put $w=u-\alpha e_1$. This $w$ is nonzero. Since ${\alpha }^2=\Vert u{\Vert }^2$, direct expansion gives $w^{T}w=2(\Vert u{\Vert }^2-\alpha u_1)$ and $w^{T}u=\Vert u{\Vert }^2-\alpha u_1$, soApply this construction successively to the trailing part of each column of $A$, extending each reflection by an identity on earlier coordinates. Previously created zeros are preserved. If a trailing column vector is zero, use the identity there. The resulting product $Q^T$ is orthogonal and $Q^{T}A=R$ is upper triangular, yielding the QR factorization $A=QR$ even when $A$ is singular.

To solve $Ax=b$, multiply by $Q^T$ and use back substitution on $Rx=Q^{T}b$ when $R$ is nonsingular. If $A$ is singular, $\det R=\det (Q^T)\det A=0$; an upper triangular determinant is its diagonal product, so at least one diagonal entry is zero. In that case consistency and free variables require separate treatment.

The unshifted QR algorithm starts from $A_0=A$, factors $A_k=Q_k{R}_k$, and formsEvery iterate is related by an orthogonal similarity transformation to the previous one and hence to $A$, with the same eigenvalues. Symmetry is preserved. For symmetric matrices with appropriate spectral separation, the iterates approach diagonal form: the off-diagonal entries decrease and the diagonal exposes the eigenvalues. Practical shifted QR accelerates convergence and uses deflation when off-diagonal couplings become negligible. With a shift ${\mu }_k$, factor $A_k-{\mu }_{k}I=Q_k{R}_k$ and set $A_{k+1}=R_k{Q}_k+{\mu }_{k}I$; the same similarity identity still holds.

Universal diagonal convergence of unshifted QR would be too strong. For example $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ admits the standard factorization $Q=A$, $R=I$, so the iteration remains at $A$ rather than becoming diagonal, although its eigenvalues are $\pm 1$. Such equal-modulus cases require a convergence qualification or treatment of a retained block. Similarity is unconditional; convergence to a useful limiting form is an additional algorithmic property.

For $w\ne 0$, the Householder transformation isLet $P=w{w}^T/(w^{T}w)$. Then $P^T=P$ and $P^2=P$, since $P$ is the orthogonal projection matrix onto the span of $w$. Hence $H^T=H$ andThus $H$ is a symmetric matrix and an orthogonal matrix. Geometrically it reverses the component parallel to $w$ and fixes the perpendicular hyperplane.

An observer-independent global presentism conflicts with the absence of a distinguished simultaneity relation in bare special relativity. It is not a formal contradiction to add a preferred foliation. The issue is what the proposed present contains, and whether that extra structure is part of the physical theory or an additional metaphysical commitment. Presentism must be distinguished from the harmless indexical fact that an observer calls their own current event “now”.

Let two spacetime events have coordinate differences $\Delta t=0$, $\Delta x=L\ne 0$ in an inertial frame. A relatively moving frame assignsThey are simultaneous in one frame and not in another. For a fixed event $P$, each inertial four-velocity $u$ defines a different simultaneous hyperplane through $P$ by $\eta (u,Q-P)=0$. No particular $u$ is selected by the Minkowski metric alone. Thus an extended objective present cannot both be the equal-time slice of one ordinary inertial frame and be invariant under all Lorentz transformations.

An equivocation can make the incompatibility seem stronger than it is. If one defines “real together” using one observer's simultaneity, then changes observers in the next inference, transitivity across the mixed relations has not been justified. It is the failure of transitivity for mixed-frame simultaneity. The physics supplies the frame-indexed relations; it does not supply a single absolute co-existence relation formed by silently joining all of them. A sound argument against relativistic presentism challenges the physical basis of its preferred relation, rather than proving all events real by that equivocation.

There are several coherent responses, with different costs. A preferred foliation can be appended to Minkowski spacetime, declaring one of its slices the objectively existing present. If it affects no physical process, it has no support from the relativistic laws and breaks the symmetry of the total ontology without changing observable predictions. If it affects matter, new laws must explain why ordinary tests display Lorentz invariance. This is additional structure, not a logical impossibility. Eternalism, by contrast, treats the complete spacetime history as real and makes simultaneity relational, but the relativity of coordinate time by itself does not deductively establish that metaphysics.

A local present containing only $P$ is invariant once $P$ is given, yet it abandons a spatially extended world existing now. The past light cone is invariant too, but contains events of different ages and is not an instantaneous spatial present. An observer-relative present is meaningful, but relativizes the existence claim rather than defending a unique global present. The defensible conclusion is therefore an incompatibility among bare relativistic structure, an objective extended present, and full symmetry of existence claims; it is not an unconditional proof that every formulation of presentism is incoherent.

The crucial distinction is between an observer-independent binary relation of absolute simultaneity and a relation defined relative to an inertial frame. The relativity of simultaneity removes the former from the usual structure of special relativity; it does not make the latter meaningless or subjective.

For two spacetime events separated by $(\Delta t,\Delta x)$, a Lorentz boost of velocity $v$ givesIf $\Delta t=0$ and $\Delta x\ne 0$, then $\Delta t^{\prime }\ne 0$ for a nonzero boost along their separation. More generally, spacelike separated events can be simultaneous in some inertial frame, and their time order can reverse between inertial frames. Distinct timelike separated or null separated events cannot be simultaneous in any inertial frame; their causal order is preserved by time-orientation-preserving Lorentz transformations. These are precise geometric facts, independent of anyone's beliefs.

In Minkowski simultaneity relative to an inertial observer, the observer is represented by a constant future-directed four-velocity $u$, normalized by $\eta (u,u)=-c^2$ in metric signature $(-,+,+,+)$. ThenFor a fixed $u$, this is a perfectly well-defined equivalence relation. It can be specified without coordinates. What the Minkowski metric does not supply is a uniquely preferred $u$ and hence a uniquely preferred simultaneity foliation. This is the strongest defensible interpretation of the claim by Tim Maudlin: there is no additional fundamental, frame-independent simultaneity relation in the bare theory.

Calling it merely “relative to an observer” can mislead in two ways. First, distant simultaneity does not depend on when an image is received: travel time of the light signal must be accounted for. Second, two observers at the same spacetime event but with different four-velocities assign different simultaneity hyperplanes; the dependence is on state of motion, rather than consciousness. Einstein synchronisation operationalizes the standard frame assignment by setting a reflected pulse's remote reading to the midpoint of its sending and returning readings.^[ref]

A material system can pick out a useful rest frame, for example a gas in thermal equilibrium. That adds physical information beyond the Minkowski metric; it does not provide a universal absolute present. Accordingly, Maudlin's denial is justified if it denies a primitive observer-independent relation, but is too strong if read as denying every objective relation $P{\sim }_{u}Q$. The indexed relation and its measurable consequences remain part of the physics.

A perceptron is a binary classification unit using an affine score $w^{T}x+b$ and a threshold to predict a label. The perceptron algorithm updates that score online. Absorb the bias into an augmented constant input, use labels $y_i\in \{-1,1\}$, and start at $w=0$. Whenever $y_i{w}^T{x}_i\le 0$, including a tie, updateFor a finite linearly separable dataset, suppose a unit vector $u$ satisfies $y_i{u}^T{x}_i\ge \gamma >0$ and $\Vert x_i\Vert \le R$. After $M$ updates, $u^{T}w\ge M\gamma$, while expanding the squared norm at each update gives $\Vert w{\Vert }^2\le M{R}^2$. The Cauchy-Schwarz inequality yields $M\gamma \le R\sqrt{M}$, hence $M\le (R/\gamma {)}^2$. Cycling through the dataset therefore finds a separator after finitely many updates.

The assumption matters: nonseparable data can keep generating updates. The method finds a separating hyperplane when the assumptions hold, rather than necessarily maximizing its margin as a support vector machine aims to do. A single perceptron also differs from a multilayer perceptron, whose hidden activation functions can represent nonlinear decision regions.

A tied largest modulus can prevent even direction convergence. With $D=\mathrm{diag}(2,-2)$ and $x_0=(1,1{)}^T/\sqrt{2}$,The two directions form a cycle; neither is an eigenvector. There is no dominant-modulus spectral gap. A tied modulus is not a universal failure: an initial eigenvector remains on its line, and $A=2I$ preserves every direction. Subspace iteration addresses a separated cluster by targeting a dominant invariant subspace instead of one eigenvector.

An orthogonal start can fail despite a strict spectral gap. For $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $x_0=(1,-1{)}^T/\sqrt{2}$, the dominant component is zero, so $A{x}_0=x_0$ and every iterate equals $x_0$. The Rayleigh quotient is exactly $1$ and the eigenpair residual is zero, although the dominant eigenvalue is $3$. Exact arithmetic never creates a missing component.^[ref]

The third panel contrasts these failures with $N=\mathrm{diag}(-3,1)$, $x_0=(1,1{)}^T/\sqrt{2}$, for whichHere the horizontal eigenvector direction converges with a strict spectral gap, while the normalized vectors alternate signs. This is the successful direction convergence described under sign and phase in power iteration.

For a nonorthogonal eigenbasis, even the word “orthogonal” must be used carefully. The diagonalizable matrix $C=\left(\begin{array}{cc}3& 2\\ 0& 1\end{array}\right)$ has right eigenvectors $v_1=(1,0{)}^T$, $v_2=(-1,1{)}^T$, and dominant left eigenvector $w_1=(1,1{)}^T$. For $x_0=(a,b{)}^T$, the dominant coefficient is $c_1=w_1^T{x}_0=a+b$. The starting vector $v_2$ is nonorthogonal to $v_1$ but stays at the smaller eigenvalue $1$; $(0,1{)}^T$ is orthogonal to $v_1$ but has $c_1=1$ and converges to the dominant direction. The left eigenvector test, not the right-eigenvector inner product, controls the outcome.

A random vector drawn from a continuous probability distribution with a probability density function on the full real or complex vector space has $c_1\ne 0$ with probability one. This avoids the exact missing-component hyperplane, but does not guarantee a large component or fast convergence. Floating-point arithmetic may seed a missing component, yet a structurally preserved zero can remain zero, so accidental perturbations are not a convergence guarantee.

Pseudo-arclength continuation allows both the state and the parameter to change. Write $z=(u,\lambda )$ and choose a unit vector $t_k$ in the kernel of the full Jacobian matrix $DF(z_k)$. A continuation predictor-corrector step seeks the next solution on a hyperplane perpendicular to $t_k$, a distance $h>0$ along the old tangent vector. This adds one equation to $F(z)=0$ and makes the correction a square system. The constraint approximates arc length; $h$ is the displacement projected onto the old tangent vector, rather than the exact curved distance.^[ref]

For a nonzero vector $v$ and nonzero linear functional $f$ vanishing on $v$, this invertible map fixes the hyperplane $\ker f$ pointwise and has determinant one. Its inverse is $I-vf$, since $(vf{)}^2=0$. A symplectic transvection is obtained by taking $f=cB(-,v)$ for a nondegenerate alternating bilinear form.

Give the four corners of the square the exclusive or labels: $+1$ when the input signs differ and $-1$ when they agree. In the following order,No affine function $g(x)=a_1{x}_1+a_2{x}_2+b$ has the required strictly positive scores at $p_2,p_3$ and strictly negative scores at $p_1,p_4$: both pairs have midpoint zero, so $g(p_2)+g(p_3)=2b=g(p_1)+g(p_4)$ would have to be both positive and negative. This also rules out perfect classification by a single threshold with a consistent tie rule.

Under the polynomial kernel's quadratic feature map, the two classes become two distinct vectors:The hyperplane ${\varphi }_2=0$ separates them. With $w=(0,-1/\sqrt{2},0{)}^T$ and $b=0$, the score and classifier areAll four signed scores $y_{i}f(p_i)$ equal one. In the input plane the boundary is $x_1{x}_2=0$, the union of the coordinate axes; in the derived coordinates it is a single hyperplane. The feature map identifies opposite corners, which is harmless for these labels because opposite corners belong to the same class. A fixed feature map would not separate labels that disagree at identical mapped vectors.