Assume first an ultrarelativistic charged-current interaction with negligible electron mass in the centre-of-mass spin analysis. Choose the incident antineutrino along $+z$. Its right helicity gives spin $+z$; the interacting left-helicity electron moves along $-z$ and also has spin $+z$. The total initial projection is $+1$. The final antineutrino and selected electron likewise have total spin $+1$ along the outgoing antineutrino direction. The spin-one rotation therefore givesAn unpolarized electron target changes the normalization through the spin average, not this selected-current shape. For electron antineutrinos the charged-current diagram is the $s$-channel $W^{-}$ graph. A contact approximation further requires $m_e^2\ll s\ll m_W^2,m_Z^2$.

The full Standard Model also has neutral-current $Z$ exchange. In the contact limit it supplies a right-electron contribution as well as interference in the left-electron channel. With $g_L=1/2+{\sin }^2{\theta }_W$, $g_R={\sin }^2{\theta }_W$, the full high-energy shape is proportional to $g_L^2(1+\cos {\theta }^{\star }{)}^2/4+g_R^2$. Thus the pure squared angular form is an explicitly stated charged-current approximation, not an exact prediction for all Standard Model contributions. At momentum transfers comparable to the weak-boson masses, propagators must also be retained.^[ref]

Assuming that the two listed leptonic decay categories exhaust the leptonic modes, the inclusive hadronic branching fraction is $100-17.83-17.41=64.76\%$. ThusAt the simplest massless quark level, the W boson couples to $u\bar{d}$ and $u\bar{s}$ in three color charge states, so equal quark and lepton weak coupling constant values predictThis is a rough check of color multiplicity in a decay width and universal weak-current normalization. It is much less precise than the two-lepton ratio because hadronic quantum chromodynamics and electroweak interaction corrections are sizeable: the data require a net correction factor near $3.632/3.0024\simeq 1.21$. They must be included before extracting a quark-to-lepton coupling constant ratio. It would be incorrect to interpret the excess over three directly as a failure of weak universality. The displayed exclusive hadronic modes alone would underestimate the inclusive sum.

The Cabibbo-Kobayashi-Maskawa matrix is the unitary change of basis between down-type quark mass eigenstates and the combinations appearing in the weak charged current. In particular, $u\bar{d}$ and $u\bar{s}$ production carry $V_{ud}$ and $V_{us}$. Exclusive tau lepton decays into a charged pseudoscalar meson giveup to small radiative corrections. The pseudoscalar decay constants specify the vacuum-to-meson axial-current matrix elements, $⟨0|\bar{q}{\gamma }^{\mu }{\gamma }^{5}u|P(p)⟩=i{f}_P{p}^{\mu }$.

The observed branching fraction ratio is $0.70/10.83=0.06464$, and the kinematic factor is approximately $0.8622$. Hence the directly determined combination isWith an independent, representative lattice input $f_K/f_{\pi }\simeq 1.193$, this becomes $|V_{us}/V_{ud}|\simeq 0.2295$, close to the source's rounded expectation $0.224/0.975=0.2297$. Conversely that expectation and the hadronic input predict a branching fraction ratio near $0.0648$. Without the decay-constant input the table does not isolate a pure Cabibbo-Kobayashi-Maskawa matrix ratio. Setting $f_K=f_{\pi }$ would give approximately $0.0455$, demonstrating the importance of flavor symmetry breaking.^[ref]

Lepton universality means that the weak charged current couples equally to the three charged lepton families after accounting for masses and known radiative corrections. In the ratio of two tau lepton leptonic decay widths, the parent coupling constant, mean lifetime and $m_{\tau }^5$ cancel:The universal prediction is $B_{\mu }/B_e=0.97256/0.9999993=0.9725607$. The measured ratio is $17.41/17.83=0.97644$, givingif the quoted errors are treated as independent. This is consistent with unity at about $1.2$ standard deviations. An actual precision fit must retain the measured covariance and the small radiative corrections; the rounded table alone does not supply those correlations.

Use a high-multiplicity hadronic mode with only one invisible neutrino, and reconstruct the total visible four-momentum $p_h$. In the tau lepton rest frame,Fit the endpoint of the hadronic invariant mass distribution, preferably together with its laboratory energy boundary at known beam energy. The two-dimensional boundary uses more information than the single largest observed mass. Detector smearing, unseen neutral products, backgrounds, beam radiation and hadronic spectral modeling must be included. Absence of a displaced endpoint generally sets an upper bound rather than a positive neutrino mass measurement. Strictly the flavour state contains several mass eigenstates; an unresolved endpoint constrains the corresponding weighted mass spectrum.

Write the leptonic partial decay width asHere $G_F$ is defined from muon decay, so the coupling denominator fixes the convention. Measure ${\Gamma }_{\ell }=B_{\ell }/{\tau }_{\tau }$. The ratio $B_{\mu }/B_e$ tests $|g_{\mu }/g_e{|}^2$ after the known phase space and radiative corrections. Comparing $B_e/{\tau }_{\tau }$ with the muon lifetime and the fifth power of the mass ratio tests $|g_{\tau }/g_{\mu }{|}^2$. Hadronic comparisons such as tau-to-pion versus pion-to-muon decay supply additional universality tests with their own hadronic and electromagnetic corrections.

In ${\tau }^{-}\to {\ell }^{-}{\bar{\nu }}_{\ell }{\nu }_{\tau }$, charge and missing four-momentum show an invisible neutral component beyond the visible charged lepton. The distributions are consistent with three-body V-A interaction decay; hadronic modes such as ${\tau }^{-}\to {\pi }^{-}{\nu }_{\tau }$ instead have two-body missing-mass kinematics. Production and decay accounting assigns $L_{\tau }({\tau }^{-})=L_{\tau }({\nu }_{\tau })=1$ and their antiparticles minus one. Searches for wrong-flavour and neutrinoless channels constrain violations. These observations establish a useful approximately conserved tau-family number in ordinary interactions, rather than proving an exact law from missing energy alone.

To establish that the invisible particle is distinct from ${\nu }_e$ and ${\nu }_{\mu }$, combine decay evidence with the absence of unsuppressed tau production by conventional electron- and muon-flavour beams, and direct tau neutrino charged current interactions. DONUT observed the short displaced tau lepton decay in nuclear emulsion after a neutrino interaction. ^[ref] Neutrino oscillation means individual family numbers are not exact conserved quantities during propagation, even though they are an excellent approximation for these local decay assignments.

Under the permitted singular value bounds, $m=4n$ givesFor the symmetric positive-definite matrix $H=G^{T}G$ in part (iii), its eigenvalues are the squared singular values of $G$, soApply the supplied conjugate gradient convergence bound to $H$. For $e_k=x_k-x_{\ast }$ and nonzero $e_0$, the function $(\sqrt{\kappa }-1)/(\sqrt{\kappa }+1)$ increases with $\kappa \ge 1$, givingThus the error in this weighted norm contracts at a guaranteed exponential factor of at most $1/2$ per iteration, up to the prefactor $2$. Here $\Vert e{\Vert }_H=\Vert Ge{\Vert }_2$. The Pythagorean identity in part (iii) also identifies $\Vert e_k{\Vert }_H^2$ with the excess squared linear least-squares problem objective, which consequently decreases by a bound proportional to $4^{-k}$. For the Euclidean norm, the same eigenvalue bounds give $\Vert e_k{\Vert }_2/\Vert e_0{\Vert }_2\le 62^{-k}$.

For relative accuracy $0<ϵ<1$ in the $H$-weighted norm, it is sufficient to takeCompute $c=G^{T}b$ once in $O(mn)$ operations. Each iteration applies $p\mapsto G^T(Gp)$ with two dense matrix multiplications and uses $O(n)$ further arithmetic operations; when $m=4n$ this costs $O(n^2)$. Hence the total cost isUsing finite termination of the conjugate gradient method in exact arithmetic refines this to $O(n^2\min \{n,\lceil {\log }_2(2/ϵ)\rceil \})$. Explicitly constructing $G^{T}G$ first would add $O(m{n}^2)=O(n^3)$ setup work and is unnecessary. These accuracy estimates concern relative error; an absolute tolerance replaces $2/ϵ$ by $2\Vert e_0{\Vert }_H/ϵ$ when $\Vert e_0{\Vert }_H>ϵ$. If $e_0=0$, the initial guess already solves the problem.

As in part 2(b)(v), the Gaussian assumption alone does not make the printed singular value interval a certainty. For example, the event that every entry of $G$ has absolute value below a sufficiently small positive number has positive probability and forces $\Vert G{\Vert }_F<\sqrt{n}$, contradicting the asserted lower endpoint. The displayed rate is therefore conditional on the singular-value bounds authorized by the paper. The cost and convergence derivation are exact under that condition; an unconditional probabilistic statement would need an additional failure-probability estimate.

The least-squares normal equations areThe full column rank of $G$ gives $z^{T}Hz=\Vert Gz{\Vert }_2^2>0$ for nonzero $z$, so $H$ is a symmetric positive-definite matrix. Apply the Conjugate gradient method to this linear system. To verify that its solution is the desired one, let $x_{\ast }$ solve the displayed equations. Then $G^T(G{x}_{\ast }-b)=0$, and for any $x$ the Pythagorean identity givesThus $x_{\ast }$ is the unique minimizer of the linear least-squares problem.

One can apply $H$ without forming the Gram matrix: compute $Hp=G^T(Gp)$. More explicitly, initialize $x_0$, $r_0=c-H{x}_0$, $p_0=r_0$; until the linear-system residual vanishes, use the Conjugate gradient method recurrenceEach dense application of $H$ this way costs $O(mn)$ operations. The condition number satisfies ${\kappa }_2(H)={\kappa }_2(G{)}^2$, where ${\kappa }_2(G)={\sigma }_{\max }(G)/{\sigma }_{\min }(G)$; this is the conditioning cost of the least-squares normal equations.

For any nonzero vector $y\in {\mathbb{R}}^k$, orthonormality gives $\Vert Q_{k}y{\Vert }_2=\Vert y{\Vert }_2>0$. Since $A$ is a symmetric positive-definite matrix,The preceding part already gives symmetry, so $T_k$ is a symmetric positive-definite matrix. In fact the spectral theorem gives the useful boundswhich place its eigenvalues between the smallest and largest eigenvalues of $A$.

Multiplying the Arnoldi decomposition by $Q_k^T$ shows that the leading square block isIt is an upper Hessenberg matrix by the Arnoldi iteration. Because $A=A^T$, it is also a symmetric matrix. Entries below the first subdiagonal vanish by the Hessenberg structure, and their reflected entries above the first superdiagonal vanish by symmetry. ConsequentlyThe extra last row has only its final entry possibly nonzero. With the usual choice of positive normalization factors, the individual column relations are the three-term Lanczos algorithm recurrencewhere the first term is omitted for $j=1$. This reduction is an exact-arithmetic statement; round-off error can affect the orthogonality in an implementation.

The claim is true, withIt is a symmetric positive-definite matrix: for every nonzero vector $z$, the fact that $A$ is an invertible matrix gives $z^{T}Mz=\Vert Az{\Vert }_2^2>0$. Since $A{x}_{\ast }=b$, every candidate $x$ satisfiesThe generalized minimal residual method therefore minimizes precisely this weighted norm of the error over the linear span of $Q_k$. The same matrix $M$ works for every $k$ and does not depend on $b$.

For the small QR decomposition, each of the $k$ Givens rotations eliminates one subdiagonal entry of the upper Hessenberg matrix. At column $j$, apply the rotation to two rows in columns $j,\dots ,k$, costing $O(k-j+1)$ arithmetic operations. Summing gives $O(k^2)$, and applying the rotations to the right-hand side costs $O(k)$. Backward substitution in a triangular system costs $O(k^2)$.

Computing $\Vert b{\Vert }_2$ costs $O(n)$ and the final dense matrix multiplication $Q_k{y}_k$ costs $O(nk)$. Thus the total, with the Arnoldi decomposition already supplied, isHere $1\le k<n$, since $Q_{k+1}$ has orthonormal columns in ${\mathbb{R}}^n$. In the standard dense arithmetic model, forming the general matrix–vector product gives the sharp $\Theta (nk)$ cost. The pair above records that two-parameter bound; looser bounds such as $O(n^2)$ are also possible because $k<n$. Treating ${\tilde{H}}_k$ as an unstructured dense matrix would unnecessarily replace its $O(k^2)$ factorization by an $O(k^3)$ computation.

Write $x=Q_{k}y$ and $\beta =\Vert b{\Vert }_2$. The Arnoldi decomposition and $b=\beta Q_{k+1}{e}_1$ giveBecause $Q_{k+1}$ has orthonormal columns, it preserves the Euclidean norm. The generalized minimal residual method therefore reduces to the small linear least-squares problemUse Givens rotations on the upper Hessenberg matrix to obtain a QR decompositionwhere $Z$ is an orthogonal matrix and $R_k$ is an upper triangular matrix. Then solve $R_k{y}_k=g_{1:k}$ by backward substitution in a triangular system and form $Q_k{y}_k$. The resulting minimum linear-system residual has Euclidean norm $|g_{k+1}|$.

The small problem has a unique answer: if ${\tilde{H}}_{k}y=0$, the Arnoldi decomposition gives $A{Q}_{k}y=0$. The invertible matrix $A$ implies $Q_{k}y=0$, and the orthonormality of $Q_k$ then gives $y=0$. Thus ${\tilde{H}}_k$ has full column rank and $R_k$ is nonsingular.

Write $\ell =r/2$; the requested expression requires the positive even integer $r\ge 2$. Split the thin singular value decomposition after $\ell$ columns:Here $\Vert D_b{\Vert }_2={\sigma }_{\ell +1}=\Vert A-A_{\ell }{\Vert }_2$. Each column of a Gaussian random matrix is an orthogonally invariant random vector: its probability density function is proportional to $\exp (-\Vert z{\Vert }_2^2/2)$, and an orthogonal matrix preserves both $\Vert z{\Vert }_2$ and volume. Thus $V^{T}X$, where $V=[V_a,V_b]$, again has independent and identically distributed standard normal entries. In particular both displayed blocks have the Gaussian distributions needed below.

First prove a deterministic error bound for a randomized range finder. Put $P=Q{Q}^T$ and suppose ${\Omega }_a$ has full row rank. Its Moore-Penrose pseudoinverse is ${\Omega }_a^{†}={\Omega }_a^T({\Omega }_a{\Omega }_a^T{)}^{-1}$, so ${\Omega }_a{\Omega }_a^{†}=I_{\ell }$. The QR decomposition implies $(I_m-P)AX=0$, henceMultiplication by ${\Omega }_a^{†}$ expresses the missed leading columns in terms of the trailing columns. Substitution into $(I_m-P)A$ gives the exact identityFor any matrix $F$, the spectral norm of $[-FI]$ is $\sqrt{1+\Vert F{\Vert }_2^2}$, because its product with its transpose is $F{F}^T+I$. Using submultiplicativity of the operator norm, the orthonormality of $U_b$, and the contraction property of the orthogonal projection, we obtainThis estimate is deterministic and also covers rank-deficient $A$; if ${\sigma }_{\ell +1}=0$, its right-hand side vanishes.

Now use the paper's permitted singular value estimates. Apply the lower estimate to the tall matrix ${\Omega }_a^T$ of size $r\times \ell$, rather than to its wide transpose. It givesThe upper estimate for ${\Omega }_b$, transposing if necessary, givesThus, with $K=2/(1-1/\sqrt{2})$,Since $n/r>1$, the deterministic estimate yieldsThis proves the requested conclusion under the explicitly authorized estimates.

The probabilistic qualification is necessary. A Gaussian random matrix does not always have all its singular values between the two endpoints printed in the hint. Already a $1\times 1$ example would require $|Z|\le 2$ for every standard normal random variable $Z$, which fails with positive probability. The proof above holds on the event where the two block estimates hold. In a probabilistic formulation one supplies estimates for that event, usually allowing a positive margin in both endpoints; no failure probability is specified by the paper.

Nor can the requested inequality with one universal $C$ hold almost surely for every $A$. For example, take $n=m=3$, $r=2$, $A=\mathrm{diag}(L,1,1)$ with $L>1$, and a sampling matrixAt $X_0$, the sampled column space misses the first coordinate, so $\Vert A-\hat{A}{\Vert }_2=L$, whereas $\Vert A-A_1{\Vert }_2=1$. Choose $L>C\sqrt{3/2}$. Because $A{X}_0$ has full column rank, its orthogonal projection depends continuously on $X$ near $X_0$ through $AX(X^T{A}^{T}AX{)}^{-1}{X}^T{A}^T$. The strict violation persists in an open neighborhood of $X_0$. The joint Gaussian probability density function is positive there, so that neighborhood has positive probability. This identifies the missing qualification without weakening the conditional calculation.

Because $r<n$, choose a unit vector $v$ in the null space of $X^T$, so that $v^{T}X=0$. Let $J\in {\mathbb{R}}^{m\times n}$ have orthonormal columns, for example the first $n$ columns of $I_m$, and choose a real number $\alpha >1$. SetIts singular value along $v$ is $\alpha$, and its other singular values are $1$. Thus ${\sigma }_1(A)=\alpha >1={\sigma }_2(A)$. Meanwhile $AX=JX$. If $X$ has full column rank, the column space of the thin QR decomposition is exactly $\mathrm{range}(JX)$, which is orthogonal to $Jv$. Hence $PJv=0$ for $P=Q{Q}^T$, andConversely, an orthogonal projection is a contraction in the Euclidean norm, so $\Vert (I_m-P)A{\Vert }_2\le \Vert A{\Vert }_2=\alpha$. ThereforeIf $X$ is rank deficient, complete the QR decomposition within $J(v^{\perp })$: this space has dimension $n-1\ge r$ and contains all columns of $JX$, so the required orthonormal completion exists there. The same proof applies. For rank-deficient data the choice of unused columns of $Q$ matters; the construction specifies a valid choice. If $Q$ instead contains only a basis of $\mathrm{range}(AX)$, the conclusion holds without any completion. A fixed sampling matrix can thus miss the unique largest singular value completely.

The thin QR decomposition gives $AX=QR$. Its columns already belong to the column space of $Q$, so the orthogonal projection $Q{Q}^T$ leaves them fixed:This identity uses only $AX=QR$ and $Q^{T}Q=I_r$; it remains valid if $R$ is singular, provided such a factorization with orthonormal $Q$ is used.

The orthogonal projection $P=Q{Q}^T$ givesThe first summand has every column in the column space of $Q$, and the second has every column in its orthogonal complement. They therefore have zero Frobenius inner product. The Pythagorean identity, together with $Q^{T}Q=I_r$, gives the stronger exact formulaConsequentlywith equality exactly when $M=Q^{T}A$. Thus solving the linear least-squares problem for each column chooses the best coefficients in the prescribed column space.

Put $P=Q{Q}^T$. Since $Q$ has $r$ columns, the rank bound for a matrix product gives $\mathrm{rank}(PA)\le r$. The Eckart-Young-Mirsky theorem and part (a) then giveOne can also see the lower bound directly. The restriction of $PA$ to the $(r+1)$-dimensional linear span of $v_1,\dots ,v_{r+1}$ has a nonzero null space, because its matrix rank is at most $r$. For a unit vector $z$ in that null space, $\Vert (A-PA)z{\Vert }_2=\Vert Az{\Vert }_2\ge {\sigma }_{r+1}$ by the singular value decomposition.

Choose the first $r$ left singular vectors:Their orthonormality gives $Q_{\ast }^T{Q}_{\ast }=I_r$. The orthogonal projection $Q_{\ast }{Q}_{\ast }^T$ fixes $u_j$ for $j\le r$ and annihilates $u_j$ for $j>r$. Applying it to the singular value decomposition therefore gives