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Let a radius-$R$ circle have center $c\in {\mathbb{R}}^2$, and apply the real affine map $y=Mx+b$. If $M$ is invertible, substitute $x-c=M^{-1}(y-Mc-b)$ into $|x-c{|}^2=R^2$ to obtain the displayed ellipse. Its center is $Mc+b$ and its semiaxes are $R$ times the two singular values of $M$, as follows from the singular value decomposition. A rotation or reflection preserves both semiaxes and hence gives another circle. If $M$ has rank one, the image is a segment; rank zero gives one point. This real-plane description also handles maps that stretch real and imaginary coordinates differently and are not holomorphic complex maps.

A numerical algorithm is backward stable when its computed output is the exact output for input data perturbed by an amount comparable to the arithmetic's round-off error. The allowed input perturbations and their norm must be specified. A bound on backward error gives a forward error bound only after accounting for the problem's condition number. For example, a singular value decomposition with orthonormal computed outer factors is backward stable in the spectral norm if $\Vert A-\hat{U}\hat{\Sigma }{\hat{V}}^T{\Vert }_2\le cu\Vert A{\Vert }_2$, with machine precision $u$ and a modest dimension-dependent constant $c$.

A singular value decomposition of the charged-lepton Yukawa matrix uses independent unitary rotations of left- and right-handed leptons. Their gauge-covariant kinetic terms are unchanged because each rotation mixes identical Standard Model representations. After electroweak symmetry breaking, the nonnegative singular values give Dirac masses $m_i=y_{i}v/\sqrt{2}$ and diagonal Higgs boson couplings $m_i/v$. When the neutrinos are massless, rotating their left-handed fields with the charged partners preserves the diagonal weak charged current.

Split a thin singular value decomposition at $\ell$, writing $A=U_a{D}_a{V}_a^T+U_b{D}_b{V}_b^T$ and ${\Omega }_i=V_i^{T}X$. Suppose ${\Omega }_a$ has full row rank and the column space of $Q$ contains that of $AX$. With $P=Q{Q}^T$, the identity $(I-P)AX=0$ and the Moore-Penrose pseudoinverse giveSince $\Vert D_b{\Vert }_2={\sigma }_{\ell +1}$, submultiplicativity of the operator norm gives the displayed bound: $[-FI][-FI{]}^T=F{F}^T+I$ proves $\Vert [-FI]{\Vert }_2=\sqrt{1+\Vert F{\Vert }_2^2}$. This statement is deterministic. A Gaussian error estimate needs a separate probability bound on the two random factors.

For a finite matrix with singular values in nonincreasing order, truncating its singular value decomposition after $k$ terms minimizes the Frobenius norm error, whose square is the sum of the discarded squared singular values. It also minimizes the operator norm error, which equals the next singular value, or zero if none remain. See Mazeika, Section 4.3: Low rank approximation.

The problem of replacing a matrix by one of lower matrix rank is explicit in Eckart and Young's 1936 paper, ^[ref]. The publication predates the modern image compression example, showing how a general matrix question can support later applications.

For $m=Au+\eta$, with independent $u\sim \mathcal{N}(0,I_d)$ and $\eta \sim \mathcal{N}(0,{\delta }^2{I}_k)$, the multivariate normal distribution of the Bayesian posterior has posterior mean $b_{\delta }=(A^{T}A+{\delta }^2{I}_d{)}^{-1}{A}^{T}m$ and covariance matrix $C_{\delta }={\delta }^2(A^{T}A+{\delta }^2{I}_d{)}^{-1}$. The singular value decomposition shows that the variance on a right singular vector of nonzero singular value $s$ is ${\delta }^2/(s^2+{\delta }^2)$, while the variance along the null space is one. Thus, for fixed data, the mean tends to the minimum-norm least-squares solution and the covariance tends to the orthogonal projection onto the null space. Unobserved components retain their prior distribution even as the noise vanishes.

For the dihedral group of order $2m$ with odd $m$, let $D_k(r)=\mathrm{diag}({\zeta }^k,{\zeta }^{-k})$ and $D_k(s)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, where $\zeta =e^{2\pi i/m}$ and $1\le k\le (m-1)/2$. An invariant tensor defining a bilinear form satisfies $D_k(g{)}^{T}L{D}_k(g)=L$. The rotation forces $L_{11}=L_{22}=0$ and the reflection forces $L_{12}=L_{21}$. Thus the invariant tensor is a multiple of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, and a singlet Yukawa interaction gives two equal physical masses by singular value decomposition. Distinct inequivalent doublets admit no such bilinear form, by Schur lemma and their self-duality.

An inverse problem reconstructs an unknown object $x$ from observed data $y$ and a specified model $A(x)=y$. The associated forward problem predicts the observations from $x$. In image deblurring, $x$ contains sharp pixel values and a matrix $A$ mixes neighbouring pixels; in tomography, the observations are integrals through the object. The appeal is that indirect measurements can reveal otherwise inaccessible structure. The difficulty is that the forward problem may erase precisely the distinctions we want to recover.

Consider this completely computable two-pixel image deblurring model:Each measured pixel retains $55\%$ of its own value and receives $45\%$ of its neighbour's. Adding measurement noise $e=(0.01,-0.01{)}^T$ gives $y^{\delta }=(0.56,0.44{)}^T$, whose direct reconstruction is $A^{-1}{y}^{\delta }=(1.1,-0.1{)}^T$. The Euclidean norm of the reconstruction error is $0.1\sqrt{2}$, ten times the data error $0.01\sqrt{2}$. A negative second pixel appears even though both observed pixels are positive.

Despite this wrong reconstruction, its residual of an inverse problem is exactly zero: the predicted data equal the noisy observations. Thus matching the data perfectly does not certify reconstruction accuracy. The relevant question is how large a change in the unknown could hide within the measurement uncertainty. A condition number quantifies that sensitivity for an invertible matrix; the null space identifies changes that no improvement in precision can reveal.

The exact singular value decomposition explains this amplification. With the orthonormal basisthe mean mode survives unchanged while the difference mode shrinks tenfold. Direct inversion multiplies difference-mode measurement noise by ten. The condition number is ${\kappa }_2(A)={\sigma }_{\max }/{\sigma }_{\min }=10$. The beauty of the singular value decomposition is this separation into independent channels: it identifies which information survives and which needs protection by regularization of an inverse problem.

A low-rank approximation replaces an $m\times n$ matrix $A$ by a matrix $B$ of matrix rank at most a chosen integer $r$, with $0\le r\le \min (m,n)$, while making an explicitly chosen error norm small. The point is to describe many entries using a few shared patterns: an image compression may retain broad brightness patterns while discarding finer ones. The approximation is useful when the saved storage or computation justifies the lost detail. A small matrix rank alone says nothing about the error.

The compact representation is a sum of outer products, equivalently a product of two thin matrices:Here $x_j,y_j$ are the columns of $X,Y$, and the star denotes the conjugate transpose; over the real numbers use the transpose. The rank bound for a matrix product gives $\mathrm{rank}B\le r$. Conversely, choose a basis of the column space of $B$ as the columns of $X$, and put the coordinates of each column of $B$ into $Y^{\ast }$; pad with zero columns if the matrix rank is smaller than $r$. Each nonzero outer product is a rank-one matrix: it repeats a single column pattern at different strengths across the columns.

Keeping the two factors stores $r(m+n)$ scalars instead of $mn$. This saves space only when $r(m+n)<mn$. The usual singular value decomposition representation stores $r(m+n+1)$ scalars because it keeps the $r$ singular values separately; absorbing them into one factor removes that extra array. These are representation sizes at equal scalar precision, before file overhead. See Uschmajew and Vandereycken, §9.2: thin factors and their storage, and Illinois CS 357, Low-rank approximation: storing singular vectors and values.

The factors also give $Bz=X(Y^{\ast }z)$ without forming the full matrix $B$. Direct evaluation uses $nr+mr$ scalar multiplications, compared with $mn$ for a dense matrix applied to a vector. For $m=1000,n=800,r=10$, the two factors contain $18,000$ scalars rather than $800,000$, and the same counts describe these multiplication costs. Constructing the factors is a separate cost. Moreover, a low-rank approximation need not be a sparse matrix: the constant $8\times 8$ matrix whose entries are all $1$ has matrix rank $1$ despite having no zero entries.

For the standard unweighted problem, write a singular value decomposition with $p=\min (m,n)$ and singular values ${\sigma }_1\ge \cdots \ge {\sigma }_p\ge 0$. Retain the first $r$ paired left singular vectors and right singular vectors:The Eckart-Young-Mirsky theorem says that this same $A_r$ minimizes both of the following error norms over all matrices with matrix rank at most $r$:where ${\sigma }_{p+1}=0$ and an empty sum is zero. Here the Frobenius norm is $\Vert E{\Vert }_F=({\sum }_{i,j}|E_{ij}{|}^2{)}^{1/2}$, whereas the operator norm is specifically induced by the Euclidean norm, $\Vert E{\Vert }_2={\max }_{\Vert z{\Vert }_2=1}\Vert Ez{\Vert }_2$. The first aggregates errors in all entries; the second bounds the worst output error for a unit vector input. These formulas apply to real and complex matrices. ^[ref]

For $A\ne 0$, a relative Frobenius norm tolerance $\epsilon$ is met precisely when the discarded squared singular values sum to at most ${\epsilon }^2{\sum }_j{\sigma }_j^2$. Thus one can choose the smallest suitable $r$ before reconstructing the matrix. Many individually small discarded singular values can still have a large combined error. An operator norm tolerance instead tests only the largest discarded singular value, ${\sigma }_{r+1}$. The two questions can therefore select different ranks.

The following stipulated $8\times 8$ matrix makes the tradeoff visible. Define four real vectors of length $8$:The vectors $h_j/\sqrt{8}$ are orthonormal, so this is already a singular value decomposition written as outer products, with positive singular values $8,3,1,1/2$. All entries lie between $7/16$ and $25/16$ and can be displayed as brightnesses. The first component is uniform brightness, the second distinguishes broad blocks, and the last two supply finer alternating patterns. This is a designed arithmetic example, not measured image data.

Here $A_1=h_0{h}_0^T$ has every entry equal to $1$, while $A_2=h_0{h}_0^T+3h_1{h}_1^T/8$ has entries $11/8$ within the two diagonal blocks and $5/8$ in the other two blocks. For instance, $A_{11}=25/16$ becomes $(A_2{)}_{11}=11/8$: its error is $3/16$. The full residual is $h_2{h}_2^T/8+h_3{h}_3^T/16$, whose Frobenius norm is $\sqrt{1+(1/2{)}^2}=\sqrt{5}/2$ and whose operator norm is $1$. Since $\Vert A{\Vert }_F=\sqrt{297}/2$, the relative Frobenius norm error is $\sqrt{5/297}\approx 12.97\%$. Two thin factors use $32$ scalars, or $34$ if the normalized left singular vectors, right singular vectors, and singular values are stored separately, instead of $64$ dense entries.

Retaining three components reduces the Frobenius norm error to $1/2$ while using $48$ scalars in two factors. Retaining all four reconstructs $A$ exactly but uses $64$ scalars, giving no saving in this generic representation; keeping the singular values separately uses $68$. Our particular sign patterns could be encoded more economically, but that would exploit an additional structure beyond the general low-rank approximation. The striking feature is that numerical error and storage can both be read from the decomposition, rather than judged only by how smooth the picture looks.

The constraint is deliberately matrix rank at most $r$. The truncation has $\mathrm{rank}{A}_r=\min (r,\mathrm{rank}A)$. If $\mathrm{rank}A<r$, the best approximation is $A$ itself, with zero error. Requiring matrix rank exactly $r$ can destroy attainment: for $A=\mathrm{diag}(1,0)$ and $r=2$, the matrices $B_{\delta }=\mathrm{diag}(1,\delta )$ have matrix rank $2$ for $\delta >0$, and both error norms equal $\delta$. Their error has infimum zero, but no rank-two matrix attains it. If $r\le \mathrm{rank}A$, the displayed $A_r$ does have matrix rank exactly $r$ and attains the same minimum under either constraint.

Optimality also does not always mean uniqueness. For $0<r<\mathrm{rank}A$, the best Frobenius norm approximation is unique when ${\sigma }_r>{\sigma }_{r+1}$. ^[ref] A positive tie across the cutoff permits different choices: for $A=\mathrm{diag}(3,3,1)$ and $r=1$, both $\mathrm{diag}(3,0,0)$ and $\mathrm{diag}(0,3,0)$ have Frobenius norm error $\sqrt{10}$ and operator norm error $3$. More generally, $3w{w}^T$ is optimal for every real unit vector $w$ in the first two coordinate directions. Rotating which tied direction is kept changes the approximation. Repeated singular values entirely inside the retained group do not change its sum of outer products. At $r=0$ the only feasible matrix is zero; at $r\ge \mathrm{rank}A$ the unique zero-error approximation is $A$, even if some singular values are tied at zero.

A strict gap does not guarantee uniqueness for the operator norm. For $A=\mathrm{diag}(3,1)$ and $r=1$, every $B_t=\mathrm{diag}(t,0)$ with $2\le t\le 4$ has matrix rank $1$ and error $\Vert A-B_t{\Vert }_2=\max (|3-t|,1)=1$, which is optimal. Only $t=3$ minimizes the Frobenius norm, since $\Vert A-B_t{\Vert }_F=\sqrt{(3-t{)}^2+1}$. An unchanged worst direction can hide extra error in another direction under the operator norm.

Finally, these guarantees concern the stated norms and the sole matrix rank constraint. They do not establish that a discarded pattern is noise or irrelevant to a particular task: the finer patterns in our constructed example were put there intentionally. Adding entrywise weights or restrictions on the factors changes the problem, so the Eckart-Young-Mirsky theorem above does not automatically solve it. For historical context, the original 1936 matrix problem is documented in the existing Eckart-Young-Mirsky theorem section.

A matrix decomposition expresses a matrix as a matrix product of factors with simpler structure. An LU decomposition uses triangular factors, a QR decomposition uses an orthogonal matrix and a triangular factor, and a singular value decomposition separates input directions, stretching factors and output directions. The factorization is an identity; the chosen structure determines what it helps compute.

For an original two-by-two example,The outer factors are a lower triangular matrix and an upper triangular matrix, each with unit diagonal as in a unit triangular matrix; the middle factor is a diagonal matrix. Their matrix product is $M$, and the final diagonal entry $5/2$ is the Schur complement of the first entry $2$. This is the scalar version of the factorization under block elimination.

To solve $Mz=(3,4{)}^T$, solve with the three factors in their displayed order from left to right: first the lower triangular factor, then the diagonal factor, then the upper triangular factor. The intermediate vectors are $(3,5/2{)}^T$ and $(3/2,1{)}^T$, giving the concise answer $z=(1,1{)}^T$. A new right-hand side uses the same factors. This is why Reusing LU factors for multiple right-hand sides is useful: factorization records work that otherwise would be repeated, while forward substitution and back substitution apply it cheaply.^[ref]

Different decompositions reveal different features. A Cholesky decomposition uses positive definiteness of a symmetric matrix, while a singular value decomposition also applies to rectangular matrices and exposes matrix rank. The existence and numerical behaviour of the factors depend on their own hypotheses; for example, an unpivoted LU decomposition can fail when a chosen pivot element is zero even though the matrix is an invertible matrix.

The Moore-Penrose inverse extends the useful part of a matrix inverse to rectangular and singular matrices. A linear system $Ax=b$ may have no exact solution, or may have many. The pseudoinverse resolves both ambiguities: $x_{\ast }=A^{+}b$ is the minimum-norm least-squares solution. First minimize the Euclidean norm $\Vert Ax-b{\Vert }_2$; among all vectors attaining that minimum, choose the one with smallest Euclidean norm $\Vert x{\Vert }_2$. It does not pretend that inconsistent equations can all be satisfied.

For $A\in {\mathbb{C}}^{m\times n}$, its Moore-Penrose inverse is the unique $n\times m$ matrix satisfying the four Penrose equations. The singular value decomposition gives a direct construction:Here $U,V$ are unitary matrices, the star is the conjugate transpose, and ${\Sigma }^{+}$ is the $n\times m$ rectangular diagonal matrix obtained by replacing each nonzero singular value ${\sigma }_j$ by $1/{\sigma }_j$ and leaving zeros at zero. Over the real numbers, use orthogonal matrices and transposes. The pseudoinverse reverses the stretches that can be reversed; it discards output components the original linear map cannot produce and does not invent an input component in its kernel. ^[ref]

The two products have especially clear meanings:They are orthogonal projection matrices: $A{A}^{+}$ selects the producible part of an output, while $A^{+}A$ selects the detectable part of an input. They are generally not identity matrices. For an invertible matrix, both are identities and $A^{+}=A^{-1}$.

This gives a useful order of operations for least-squares solutions: project the requested output onto the column space, then invert the restriction of $A$ from $(\ker A{)}^{\perp }$ to its column space. Adding a kernel vector leaves the residual unchanged but increases the squared Euclidean norm by the Pythagorean theorem in an inner-product space, because $A^{+}b$ is orthogonal to that kernel. The pseudoinverse therefore provides one canonical answer without hiding either lost information or irreconcilable data.

The name records a history with several contributors. E. H. Moore gave an earlier generalization in 1920; M. H. Pearl's 1966 paper records that work, Arne Bjerhammar's 1951 rediscovery for maximal-rank rectangular matrices, and Roger Penrose's independent 1955 treatment. ^[ref] Penrose's original paper proves existence and uniqueness from the four Penrose equations for arbitrary complex matrices, including rectangular ones. ^[ref]

Minimizing $\Vert y-X\theta {\Vert }^2$ gives the displayed equations by differentiation. Equivalently, the regression residual $r=y-X\hat{\theta }$ satisfies $X^{T}r=0$: it is orthogonal to every column of the design matrix. If $X$ has full column rank, the minimizer is unique and the fitted values are the orthogonal projection of $y$ onto its column space. A rank-deficient design matrix can give multiple coefficient vectors, while the projected fitted vector is still unique.

For an intercept and one predictor, put $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2$ and $S_{xy}={\sum }_i(x_i-\bar{x})(y_i-\bar{y})$. The equations give $\hat{\beta }=S_{xy}/S_{xx}$ and $\hat{\alpha }=\bar{y}-\hat{\beta }\bar{x}$, provided $S_{xx}>0$. In the five-point example in linear regression, $\bar{x}=2$, $\bar{y}=2.8$, $S_{xx}=10$ and $S_{xy}=10$, giving $\hat{\beta }=1$ and $\hat{\alpha }=0.8$.

The normal equations describe the mathematical optimum, but explicitly forming and inverting $X^{T}X$ can amplify numerical problems. A QR decomposition computes the same least-squares solution through orthogonal coordinates; a singular value decomposition also exposes nearly dependent columns. This connects statistical fitting to numerical linear algebra and the choice of a stable algorithm. ^[ref]

The operator Schmidt rank of a bipartite linear operator is its shortest expansion as a sum of product linear operators. Expanding in local orthonormal operator bases gives a coefficient matrix; its matrix rank is the operator Schmidt rank, by a singular value decomposition. It is unchanged by invertible local operator-basis changes. A rank-one projector onto a bipartite pure state of Schmidt rank $d$ has operator Schmidt rank $d^2$: its expansion contains all $d^2$ independent products of local matrix units.

Suppose three measurements constrain two unknowns:The linear system asks for $x_1=1$, $x_2=2$, and $x_1+x_2=6$, which are incompatible. A least-squares solution makes a balanced compromise. The matrix has full column rank, soThe fitted output and residual areThus the two direct measurements are each overshot by one, and the measured sum is undershot by one. The residual has Euclidean norm $\sqrt{3}$ and is orthogonal to the column space, the plane $z=y_1+y_2$ in output coordinates $(y_1,y_2,z)$.

For any perturbation $h\in {\mathbb{R}}^2$, the orthogonality of $r$ to $Ah$ givesSince the kernel of $A$ is trivial, every nonzero $h$ increases the residual. The least-squares solution is unique in this example. The formula using $A^{T}A$ is convenient for these small exact calculations; an actual numerical computation can use a QR decomposition or singular value decomposition to avoid the extra conditioning cost of the least-squares normal equations. For full column rank, the squared singular values of $A$ are the eigenvalues of $A^{T}A$, so its condition number in the Euclidean norm is ${\kappa }_2(A^{T}A)={\kappa }_2(A{)}^2$. ^[ref]

The left- and right-chiral components of a charged Standard Model fermion transform in different electroweak group representations. A bare Dirac mass term $-m{\bar{f}}_L{f}_R+h.c.$ is therefore not invariant under the full electroweak gauge group. The Higgs doublet supplies the missing transformation properties. With $\tilde{H}=i{\sigma }_2{H}^{\ast }$, the gauge-invariant Yukawa interactions arewhere each $Y_f$ is a flavor-space Yukawa matrix. For example, hypercharges in the down-quark term add to $-1/6+1/2-1/3=0$, while those in the up-quark term add to $-1/6-1/2+2/3=0$. Thus the Higgs and conjugate Higgs are required in different terms.

After electroweak symmetry breaking, choose unitary gauge with $H=(0,(v+h)/\sqrt{2}{)}^T$. The fermion mass matrices becomeA singular value decomposition $U_{fL}^{†}{M}_f{U}_{fR}=\mathrm{diag}(m_{f,1},m_{f,2},m_{f,3})$ defines nonnegative physical tree-level masses. The mismatch $U_{uL}^{†}{U}_{dL}$ is the Cabibbo-Kobayashi-Maskawa matrix in the weak charged current. The corresponding physical Higgs boson couples diagonally as $-(m_f/v)h\bar{f}f$ in the minimal one-doublet theory. This relates its coupling strength to each mass, but the measured Yukawa couplings remain independent input parameters: the minimal theory does not predict their hierarchy.

The minimal renormalizable Standard Model contains no right-handed neutrino, so it has neither a neutrino Dirac Yukawa term nor a gauge-invariant renormalizable left-handed Majorana mass term. Its neutrinos are massless. Observed neutrino oscillations require going beyond that minimal field content or operator set: adding right-handed singlets permits Dirac masses, while the dimension-five Weinberg operator gives a Majorana neutrino mass of order $v^2/\Lambda$ after symmetry breaking. These alternatives should be distinguished from the charged-fermion Yukawa mechanism.

Use the basis of solution for the two doublets, treating the barred left-handed fields as the components to which the printed representation assignment applies. Write $x=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $y=({\psi }_R^1,{\psi }_R^2{)}^T$. The doublet contribution to the Yukawa interaction is $H{x}^{T}Ly$, so invariant tensors obeyThe first equation forces $L_{11}=L_{22}=0$, since their factors are ${\alpha }^{\pm 2}\ne 1$. The second sets $L_{12}=L_{21}$. Thus the doublet invariant tensor is $x_1{y}_2+x_2{y}_1$. The product of either doublet with $R_1$ has no trivial representation, whereas $R_1\otimes R_1=R_0$. ThereforeIf instead the unbarred left-handed doublet is chosen to transform by $D_1$, its barred components transform by $D_1^{\ast }$; in that equivalent basis the upper block is $y{I}_2$. Exchanging the two left components relates these conventions. The singular value decomposition gives the same physical masses in either basis.

Let $v_H=⟨H⟩$ be the vacuum expectation value in the convention $M=v_H\lambda$. For canonically normalized fermion kinetic terms, physical charged lepton masses are the nonnegative values in the singular value decomposition of $M$, equivalently the square roots of the eigenvalues of $M^{†}M$.

In case (i),The opposite algebraic eigenvalues $y$ and $-y$ of the doublet block do not represent different physical masses. The unavoidable twofold degeneracy conflicts with the three distinct nonzero measured charged lepton masses, for any assignment of the three families.

In case (ii),One can fit the tau lepton mass by choosing $|v_H{y}_3|$, but the electron and muon remain massless. Thus neither exact dihedral group assignment with this single singlet complex scalar field reproduces the observed spectrum. Additional symmetry-breaking Yukawa interactions or other sources of mass are required; a basis change cannot remove the degeneracy or change the matrix rank.

Use logarithms to base two and set $0{\log }_{2}0=0$. For a density matrix with eigenvalues $({\lambda }_j)$, the Von Neumann entropy isThus it is the Shannon entropy of the spectrum. In dimension $d$, each term is nonnegative, and entropy vanishes exactly when one eigenvalue is one, namely for a pure state. The Jensen inequality applied to the strictly concave function $-x{\log }_{2}x$ gives $S(\rho )\le {\log }_{2}d$, with equality precisely at the maximally mixed state $I/d$. Unitary conjugation preserves eigenvalues, so $S(U\rho U^{†})=S(\rho )$. Product eigenvalues giveThe finite-dimensional entropy is continuous because the eigenvalues depend continuously on the matrix and $-x\log x$ is continuous also at zero.

Define the partial trace over ${\mathcal{K}}_2$ by choosing an orthonormal basis $(e_b)$ and settingEquivalently it is uniquely characterized by $\mathrm{Tr}({\rho }_{1}A)=\mathrm{Tr}(\rho (A\otimes I))$ for every operator $A$ on ${\mathcal{K}}_1$. This identity proves basis independence. Define ${\rho }_2$ similarly, with $\mathrm{Tr}({\rho }_{2}B)=\mathrm{Tr}(\rho (I\otimes B))$. Hermiticity is preserved by the displayed sum. For each $u\in {\mathcal{K}}_1$,Taking the trace in a product orthonormal basis gives $\mathrm{Tr}{\rho }_1=\mathrm{Tr}\rho =1$, and the same proof works for ${\rho }_2$. Thus both reductions are density matrices. This is the partial trace positivity from product vectors argument.

For a bipartite pure state $\rho =|\psi ⟩⟨\psi |$, expand $\psi$ in a product orthonormal basis with coefficient matrix $C$. A singular value decomposition gives the Schmidt decompositionwith orthonormal families on both sides. Taking the two partial traces makes all cross terms vanish:The spectra may have different numbers of zeros but have identical nonzero eigenvalues. Consequently the reduced entropies coincide:This common value is the entanglement entropy. In infinite-dimensional trace-class systems the same partial-trace and Schmidt arguments apply, and the equality remains valid with the value $+\infty$ allowed; the bound ${\log }_{2}d$ is specifically finite-dimensional.

For completeness, the basic mixing and correlation properties also admit short proofs. Define quantum relative entropy by $D(\tau \Vert \sigma )=\mathrm{Tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ when supports are included, and by $+\infty$ otherwise. Here is its nonnegativity proof without merely citing an operator inequality. Diagonalize $\tau ,\sigma$ with eigenvalues $r_i,s_j$, and let $q_{ij}$ be the squared overlaps of their eigenbases. Both row and column sums of $q$ are one. The scalar inequality $r\ln (r/s)\ge r-s$, with zero cases taken by limits, givesEquality forces $r_i=s_j$ on every nonzero overlap, hence $\tau =\sigma$. For $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, expanding the trace definition yieldswhich proves Concavity of Von Neumann entropy.

The complementary entropy bounds for a quantum mixture areTo prove the upper bound, write the spectral pure-state decompositions of the ${\rho }_i$ as one ensemble with weights $w_{ij}=p_i{\lambda }_{ij}$. Form the column matrix with columns $\sqrt{w_{ij}}|{\psi }_{ij}⟩$. Its two Gram operators have the same nonzero spectrum: one is $\overline{\rho }$, and the other, $G$, has diagonal $w$. Relative entropy against $\mathrm{diag}w$ gives $0\le D(G\Vert \mathrm{diag}w)=H(w)-S(G)$. Since $H(w)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$ and $S(G)=S(\overline{\rho })$, the bound follows. Zero-weight columns are omitted.

Finally the support of $\rho$ is contained in $\mathrm{supp}{\rho }_1\otimes \mathrm{supp}{\rho }_2$. Indeed, a vector in a marginal kernel gives a sum of nonnegative product-vector expectations equal to zero; positivity forces the corresponding product vectors into the kernel of $\rho$. The operator identity $\log ({\rho }_1\otimes {\rho }_2)=\log {\rho }_1\otimes I+I\otimes \log {\rho }_2$ on this support and the partial-trace identities giveThis proves Subadditivity of Von Neumann entropy, with equality exactly for a product state. Purify ${\rho }_{12}$ by adjoining a third system, using its spectral decomposition. Equality of pure-state marginal entropies gives $S({\rho }_{13})=S({\rho }_2)$ and $S({\rho }_3)=S({\rho }_{12})$. Subadditivity on systems $1,3$ then gives $S({\rho }_2)\le S({\rho }_1)+S({\rho }_{12})$; swapping $1,2$ gives the Araki–Lieb inequality $\boxed{|S({\rho }_1)-S({\rho }_2)|\le S({\rho }_{12})}$. These finite-dimensional proofs establish the usual basic entropy bounds, rather than attributing all mixed-state behavior to pure-state Schmidt decompositions.

Principal component analysis replaces correlated measurements by orthogonal linear coordinates ordered by their variance. For centered observations collected in an $n\times p$ matrix $Z$, write the sample covariance matrix as $S=Z^{T}Z/(n-1)$. A unit loading vector $a$ gives score vector $Za$ with sample variance $a^{T}Sa$. The Rayleigh quotient is maximized by a unit eigenvector of $S$ for its largest eigenvalue.

The subsequent principal components maximize the same quadratic form subject to orthogonality to the earlier loading vectors. By the spectral theorem, if $S=Q\Lambda Q^T$ with descending eigenvalues, the scores $ZQ$ have diagonal covariance $\Lambda$. Their fractions of explained variance are ${\lambda }_j/{\sum }_i{\lambda }_i$. Repeated eigenvalues identify an eigenspace rather than a unique axis.

Keeping the first $q$ principal components gives the rank-$q$ reconstruction $\hat{Z}=Z{Q}_q{Q}_q^T$. Its squared reconstruction error isFor any rank-$q$ orthogonal projection $P$, retained variance is $\mathrm{tr}(SP)={\sum }_j{\lambda }_j{q}_j^{T}P{q}_j$. The weights $q_j^{T}P{q}_j$ lie in $[0,1]$ and sum to $q$, so this is at most the sum of the largest $q$ eigenvalues. Projecting onto any candidate reconstruction subspace is its best least-squares reconstruction; this proves the minimum-error property and agrees with the singular value decomposition. Thus PCA finds the linear subspace preserving the greatest variance, equivalently minimizing squared reconstruction error. The upper-left sketch shows the first axis aligned with the elongated cloud; the second is perpendicular. A scree plot and substantive interpretability help choose $q$.

These methods depend on measurement scale. A variable measured in large numerical units can dominate covariance-based principal component analysis; standardizing variables gives principal component analysis on a correlation matrix. Scores summarize patterns, but the method neither establishes causation nor guarantees that the largest-variance directions best predict a separate response.

Use $Q=T_3+Y$ and $T_a={\tau }_a/2$. For each generation $r=e,\mu ,\tau$, the leptons areBoth are color singlets. The conjugate antilepton fields transform in the conjugate representations and have opposite hypercharge; they are already included by the conjugate fields in the Lagrangian. The two components of $L_r$ have electric charges $0,-1$, and $R_r$ has charge $-1$. There is no right-handed neutrino representation in this field content. These assignments use the electroweak representation and hypercharge table in the same hypercharge normalization as the scalar doublet.

For a doublet of hypercharge $Y$, defineThe $SU(2)$ term is absent for a singlet. Under the local transformation $\psi \mapsto e^{iY\omega }U\psi$, the gauge fields transform so that $D_{\mu }\psi \mapsto e^{iY\omega }U{D}_{\mu }\psi$. Consequently the required gauge-covariant kinetic terms areThe fields of definite chirality are defined with $P_{L,R}=(1\mp {\gamma }_5)/2$. Since $\bar{L}$ has hypercharge $+1/2$, the Yukawa interactionis invariant: its hypercharges add to zero, and $\bar{L}\varphi$ contracts the two doublet indices. A bare charged-lepton mass violates these assignments. No renormalizable neutrino Yukawa term is available without a right-handed neutrino. A gauge-invariant scalar potential depending on ${\varphi }^{†}\varphi$ can select the nonzero vacuum, without altering these kinetic and Yukawa constructions.

There is an explicit local unitary gauge transformation whenever $r_{\varphi }=(|{\varphi }_1{|}^2+|{\varphi }_2{|}^2{)}^{1/2}\ne 0$:Thus, with the real field $\rho =\sqrt{2}{r}_{\varphi }-v$,This proves the polar coordinates for the electroweak Higgs doublet constructively. The gauge choice is local about the nonzero vacuum; it need not be nonsingular at zeros of the doublet.

The three identical-generation kinetic terms permit unitary flavor changes. A singular value decomposition gives $U_L^{†}{Y}_{\ell }{U}_R=\mathrm{diag}(y_e,y_{\mu },y_{\tau })$ with $y_r\ge 0$. Rotate both components of each left doublet together, and the right singlets independently. In this basis,The left/right cross terms combine into this Dirac mass. Since the neutrinos have no mass matrix, their simultaneous left rotation preserves a diagonal charged current. This is charged-lepton Yukawa matrix diagonalization; the neutrinos remain massless in the stated renormalizable model.

The scalar kinetic term givesIntroduce $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The combinations $W^{+}=(A_1-i{A}_2)/\sqrt{2}$, $W^{-}=(W^{+}{)}^{\ast }$, $Z=c_W{A}_3-s_{W}B$ and $A_{\gamma }=s_W{A}_3+c_{W}B$ therefore diagonalize the quadratic terms:The factor $1/2$ applies to the real neutral field, not to the complex charged field. The massless photon corresponds to the unbroken generator $T_3+Y$, which annihilates the vacuum. This gives the electroweak doublet gauge-boson mass matrix and its diagonal spectrum.