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For a unitary matrix, $U{U}^{†}=1$, so $(U^{†}MU{)}^r=U^{†}{M}^{r}U$. Cyclicity of the trace then gives $\mathrm{tr}(U^{†}{M}^{r}U)=\mathrm{tr}(M^r)$ for both powers entering the action. Thus the action is invariant under unitary conjugation.

The spectral theorem for normal operators diagonalizes a finite-dimensional Hermitian matrix by a unitary matrix. Applying the invariance to that diagonal form givesOnly the eigenvalues enter, not the choice of eigenvectors.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac basis, with $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$. In this solution the sign of the chirality matrix is the one specified in the paper, ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Define $t_0=1$ and $t_i=-1$, with no summation in componentwise transformation formulas.

Complex conjugation changes the explicit $-i$ to $+i$. Three spatial gamma matrices each contribute another minus sign, soAlso $({\gamma }^5{)}^2=1$ and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ by the Clifford algebra.

The quantum time-reversal operator is antiunitary: it conjugates numerical coefficients, including the Dirac spinors and plane-wave exponentials in the mode expansion of a Dirac field. To fix the spin phase explicitly, put ${ϵ}_s=(-1{)}^{1/2-s}$ and useThis convention gives ${\hat{T}}^2=-1$ on a one-fermion state, since ${ϵ}_s{ϵ}_{-s}=-1$. In the transformed expansion, change variables from $(p,s)$ to $(p_T,-s)$. The Lorentz-invariant phase-space measure is unchanged, and $p_T\cdot x=-p\cdot x_T$. The coefficient of $b^s(p)e^{-ip\cdot x_T}$ is thereforeThe same calculation applies to the antiparticle coefficient. Thus, with this explicitly fixed spin convention,The spin phase in the Dirac time-reversal matrix matters here: a common change of the one-particle time-reversal phase replaces $B$ by $-B$; the phase-independent relation is $B\propto {\gamma }^{5}C$. It changes neither the defining conjugation property nor any bilinear result below. Normalize $B$ to be a unitary matrix. In the Dirac basis, ${\gamma }^0$ is real and commutes with $B$, so transforming the Dirac adjoint givesHere complex conjugation of the matrix defining the Dirac adjoint is essential. An explicit realization consistent with the sign of ${\gamma }^5$ used here is $C=i{\gamma }^2{\gamma }^0$ and $B={\gamma }^1{\gamma }^3$; it has $B^{-1}=-B$, so conjugation by $B$ and by $B^{-1}$ coincides. The usual changes of Dirac spinor basis by unitary matrices carry the adjoint and time-reversal matrix with them.

For the charge-conjugation matrix, the gamma matrix adjoint and transpose identities give ${\gamma }^{\mu T}=t_{\mu }{\gamma }^{\mu \ast }$. Consequently $B{\gamma }^{\mu T}{B}^{-1}={\gamma }^{\mu }$. Since $C={\gamma }^{5}B$ in the chosen phase convention,This proof uses a temporal Hermitian matrix and spatial skew-Hermitian matrices explicitly; the transpose identity is preserved when $C$ is transformed appropriately with the gamma matrices.

For the Fermi interaction, let $V_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }n$ and $A_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }{\gamma }^{5}n$. In the displayed Dirac basis, the inverse version of the conjugation relation also holds. Antiunitarity, $B^{-1}{\gamma }^{\mu \ast }B=t_{\mu }{\gamma }^{\mu }$ and $B^{-1}{\gamma }^{5\ast }B={\gamma }^5$ giveThe leptonic weak charged current has the same component signs. In the contraction of leptonic and hadronic currents, the two $t_{\mu }$ factors cancel. Its two independent operators thus keep their form while their coefficients become $g_V^{\ast }$ and $g_A^{\ast }$. The Hermitian-conjugate term transforms separately; its presence does not remove a relative complex phase between the vector and axial couplings.

With all intrinsic phases fixed to one and a real positive Fermi constant, invariance requires real coefficients in that convention. A common phase of $g_V$ and $g_A$ can instead be absorbed into a rephasing of the nucleon fields, and hence into their intrinsic time-reversal phases. The convention-independent condition, for $g_V\ne 0$, isThis is the relative weak phase condition for time reversal. If one coefficient vanishes, there is no relative phase to constrain; the remaining common phase can be removed. A nonreal ratio violates time-reversal symmetry even though the Lagrangian density includes its Hermitian conjugate.

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

Start with the assumed basis of eigenvectors, grouping them by eigenvalue. In each eigenspace, apply the Gram-Schmidt process, which replaces a linearly independent list by an orthonormal list with the same span. Every new vector stays within that eigenspace, since an eigenspace is a vector subspace.

By part (iv), different eigenspaces are orthogonal. The combined lists therefore form an orthonormal basis of the entire space, still consisting of eigenvectors. Let $U$ have these vectors as columns. Then $U$ is a unitary matrix, soThis gives the required unitary diagonalization of a normal matrix, including repeated eigenvalues.

For a unitary matrix, $A^{†}=A^{-1}$, so both defining products for a normal matrix are $I$. The matrix is invertible, hence no eigenvalue is zero. If $Ax=\lambda x$, then $A^{-1}x={\lambda }^{-1}x$, while part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Therefore $\overline{\lambda }={\lambda }^{-1}$, andThus all eigenvalues have unit modulus.

Introduce a parameter $t$ to keep track of total degree. The power series for the matrix exponential givesSet $S=X+Y$ and $C=\frac{1}{2}[X,Y]$, where the commutator is $[X,Y]=XY-YX$. Expanding a second matrix exponential givesSincethe two expressions agree through total degree two. More explicitly, applying the local matrix logarithm to the first expansion gives $\log (e^{tX}{e}^{tY})=t(X+Y)+\frac{1}{2}{t}^2[X,Y]+O(t^3)$. This proves the displayed order of the Baker--Campbell--Hausdorff formula. The Lie algebra is closed under the commutator, so its displayed exponent lies in the same Lie algebra. The expansion is a formal power series identity, or a convergent identity sufficiently near the identity matrix; this argument does not assert unrestricted global convergence.

The next two terms of the Baker--Campbell--Hausdorff formula are the two cubic commutators:Here $O(4)$ means total degree at least four in $X,Y$. If one continues by one more degree, the quartic contribution is $-\frac{1}{24}[Y,[X,[X,Y]]]$.

For the Pauli matrices, use the Pauli matrix multiplication lawPut $N=\mathbf{n}\cdot \mathbf{\tau }$. The antisymmetric term disappears when contracted with $n_a{n}_b$, so $N^2=|\mathbf{n}{|}^{2}I=I$. Consequently every even power of $N$ equals $I$, and every odd power equals $N$. Splitting the matrix exponential into even and odd terms of its power series yieldsThe Pauli matrices are Hermitian matrices, and the components of $\mathbf{n}$ are real, so $N^{†}=N$. For real $\alpha$, the resulting matrix $U$ satisfiesAlso $\mathrm{tr}N=0$ and $N^2=I$, so its two eigenvalues are $1,-1$. The eigenvalues of $U$ are therefore $e^{i\alpha },e^{-i\alpha }$, whose product is one. Thus $U$ is a unitary matrix with determinant one: it belongs to the SU(2) group.

Multiplying the two exact Pauli matrix exponentials, and using ${\tau }_1{\tau }_2=i{\tau }_3$, givesIn particular, the ${\tau }_3$ term has a minus sign. Its quadratic Taylor expansion isTaking $X=i\alpha {\tau }_1$, $Y=i\beta {\tau }_2$, the Pauli matrix commutator identity gives $[X,Y]=-2i\alpha \beta {\tau }_3$. The quadratic exponent in the Baker--Campbell--Hausdorff formula is therefore $Z=i\alpha {\tau }_1+i\beta {\tau }_2-i\alpha \beta {\tau }_3$. In $e^Z=I+Z+\frac{1}{2}{Z}^2+O(3)$, only the linear part of $Z$ contributes to $Z^2$ through degree two. The anticommutator ${\tau }_1{\tau }_2+{\tau }_2{\tau }_1=0$ gives $Z^2=-({\alpha }^2+{\beta }^2)I+O(3)$, reproducing the exact product's quadratic expansion.

A complex square matrix $U$ is a unitary matrix when $U^{†}U=I$, equivalently $U^{-1}=U^{†}$. Here $†$ denotes the adjoint matrix, the conjugate transpose. Such a matrix preserves the standard complex inner product and its Euclidean norm.

A real symmetric matrix $A$ is a Hermitian matrix, so $A^{†}=A$. For arbitrary complex $\mathbf{x}$,The imaginary cross terms cancel because $A$ is Hermitian. Reversing the signs gives the same identity:In particular, $(A+iI)\mathbf{x}=0$ implies $|\mathbf{x}{|}^2=0$, so $A+iI$ has zero kernel of a linear map. It is a square matrix, and the rank-nullity theorem therefore makes it invertible. The same argument proves invertibility of $A-iI$.

The proposed matrixis consequently well-defined. For any $\mathbf{y}$, put $\mathbf{x}=(A+iI{)}^{-1}\mathbf{y}$. The two norm identities show directly thatTo verify the unitary matrix identity explicitly, take adjoints:The two shifts commute, because both are polynomials in $A$. HenceThusThe imaginary shifts are invertible, and their equal norm changes cancel in the quotient. This is the Cayley transform of a Hermitian matrix in the $(A-iI)(A+iI{)}^{-1}$ convention.

Use $c>0$, the necessary parameter range for the bounds involving $1/c$. The construction comes from the singular value decomposition of matrix Fourier blocks. For each inequivalent unitary irreducible representation $\rho$ of dimension $d_{\rho }$, consider the Hilbert space ${\mathcal{H}}_{\rho }=M_{n\times d_{\rho }}(\mathbb{C})$ with the Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{tr}(A^{\ast }B)$ and the operatorIts operator norm is at most one, since $\rho (x)$ is a unitary matrix and $\Vert f(x){\Vert }_{\mathrm{op}}\le 1$. Thus all its singular values ${\lambda }_{\rho ,j}$ lie in $[0,1]$.

Two identities control their weighted moments:Here the expectation in $S_4$ is uniform on the $|G{|}^3$ solutions of the constraint, and the trace on ${\mathcal{H}}_{\rho }$ used to compute each moment is the ordinary operator trace. We justify the identities explicitly to fix their normalizations and the noncommutative order.

The adjoint operator is $T_{\rho }^{\ast }(A)={\mathbb{E}}_{y}f(y{)}^{\ast }A\rho (y)$, soFor rectangular matrices, the operator $A\mapsto LAR$ has trace $\mathrm{tr}(L)\mathrm{tr}(R)$, as is seen on the matrix-unit basis. Squaring the previous operator therefore givesThe required representation theory consists of unitarization of a finite-group representation, the Schur orthogonality relations, and the regular representation decomposition. The last gives the character expansion of the identity deltaSumming the two operator traces with weights $d_{\rho }$ selects $x=y$ in the first, and $w=x{y}^{-1}z$ in the second. Multiplication by $|G|$ converts each independent-variable expectation to its conditional expectation. This proves $S_2,S_4$. In particular $S_4$ is real and nonnegative, despite the apparently complex summands in its original formula. Since $S_4\le S_2\le n$, the nonvacuous hypothesis has $0<c\le 1$.

Select every singular value satisfying ${\lambda }_{\rho ,j}\ge \sqrt{c/2}$. Index the selected pairs by $p=1,\dots ,r$, set ${\rho }_p=\rho$, $n_p=d_{\rho }$, and choose unit right singular vectors $A_p$ and unit left singular vectors $B_p$ with $T_{{\rho }_p}{A}_p={\lambda }_p{B}_p$. DefineBoth are $n\times n_p$ matrices, proving (i). Repeated group representations in this list are literally the same chosen representative, rather than different equivalent realizations.

Use the printed gamma matrix conventions throughout, in particular ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. The transpose identity for ${\gamma }^0$ gives $C{\gamma }^0=-{\gamma }^{0}C$, and its Hermitian conjugate gives the corresponding identity for $C^{†}$. Thus the Dirac adjoint of the charge-conjugated field isReordering the anticommuting components of the fermion bilinear supplies a minus sign:Invariance for arbitrary fields therefore requires $C^{†}C=I$, so the charge-conjugation matrix is a unitary matrix. This is the required normalization constraint; it is not a bosonic reordering calculation.

For a spinor matrix $\Gamma$, the same calculation and the given $C^{-1}=C^T$ giveWith $V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ and $A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, the supplied transpose identities implyConsequently the vector current is C-odd and the axial current is C-even: $V^{\mu }\mapsto -V^{\mu }$ and $A^{\mu }\mapsto A^{\mu }$.

For parity, put $\Lambda =\mathrm{diag}(1,-1,-1,-1)$. The gamma matrix identities ${\gamma }^0{\gamma }^{\mu }{\gamma }^0={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$ and ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$ giveThus $V^0$ is P-even and $\mathbf{V}$ P-odd, whereas $A^0$ is P-odd and $\mathbf{A}$ P-even. The axial current transforms as a pseudovector.

The Yukawa matrices for up-type and down-type quarks are diagonalized by separate unitary changes of basis. Let $u_L^{\mathrm{weak}}=U_u{u}_L^{\mathrm{mass}}$ and $d_L^{\mathrm{weak}}=U_d{d}_L^{\mathrm{mass}}$. The generation-diagonal weak charged current in the weak basis becomesThus the CKM matrix records the mismatch between the two left-handed mass bases, not an additional non-unitary interaction.

An $N\times N$ unitary matrix has $N^2$ real parameters. The $2N$ quark-field phases can remove $2N-1$ phases from $V$: the common phase changes neither current. This leaves $(N-1{)}^2$ physical real parameters, comprising $N(N-1)/2$ mixing angles and $(N-1)(N-2)/2$ CP-violating phases. Hence three generations give four parameters: three angles and one phase; four generations give nine: six angles and three phases. This count assumes the usual nondegenerate masses, so that extra mass degeneracies do not enlarge the allowed basis transformations.

A Dirac-to-Weyl gamma-matrix similarity transform with the required signs isIt is a unitary matrix, with $U^{-1}=U^{†}=2^{-1/2}\left(\begin{array}{cc}{I}_2& I_2\\ -I_2& I_2\end{array}\right)$. Multiplication givesThe matrix $U$ commutes with every ${\tilde{\gamma }}^i$: their common block structure uses the same antisymmetric $2\times 2$ block matrix, and the scalar blocks of $U$ commute with each Pauli matrix. Hence $U{\tilde{\gamma }}^i{U}^{-1}={\tilde{\gamma }}^i$, exactly the spatial matrices specified in the Weyl representation of the gamma matrices. A similarity transformation preserves every Clifford algebra relation, so no new consistency check is needed for the transformed representation.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

At the only algebraic normal value $Re=-10$, the generator is a skew-Hermitian matrix, so its matrix exponential is a unitary matrix. Both eigenvalues have zero real part; henceThere is no distinguished initial amplitude ratio and no perturbation-energy growth. The relations $x_2(0)=i{r}_{\pm }{x}_1(0)$ select the individual orthogonal eigenmodes, but arbitrary nonzero combinations are equally optimal because all singular values are one. If $Re$ is restricted to its physical positive domain, this part has no applicable normal case at all. This resolves the inconsistency in the original printed parameters rather than assuming an unjustified optimal direction.

Arrange the interior values into a vector $U$. Represent the five-point Dirichlet Laplacian as a Kronecker sum $L_h=h^{-2}(D\otimes I+I\otimes D)$, where $D$ is the tridiagonal matrix with diagonal $-2$ and adjacent entries one. This Kronecker sum is real symmetric; the real sampled potential has a real diagonal matrix $D_V$. ThusThe generator $-i{H}_h$ is a skew-Hermitian matrix. ConsequentlyThe matrix exponential is a unitary matrix, either by differentiating its product with its adjoint or by unitary diagonalization of a normal matrix. For the two-dimensional discrete L2 norm $\Vert U{\Vert }_h^2=h^2{\sum }_{m,n}|U_{m,n}{|}^2$, this givesApplying the same equation to a difference proves stability of a numerical method with a mesh-independent constant one. This is norm conservation of a semidiscrete Schrödinger equation. It concerns continuous time after spatial discretization; an arbitrary subsequent time integrator need not preserve this stability.

The printed coordinates do not discretize the stated square. For $[-1,1{]}^2$ with $M$ interior points in each direction, use $h=2/(M+1)$, $x_m=-1+mh$, $y_n=-1+nh$, and sample $V$ there. The printed $h=1/(M+1)$ and unshifted coordinates instead describe a grid on $[0,1{]}^2$. The matrix proof is valid for either geometry with its corresponding boundary values, so this transcription-independent statement flaw does not alter the stability conclusion.

A matrix is Hermitian when $A^{†}=A$, and unitary when $U^{†}U=U{U}^{†}=I$.

If $A\mathbf{u}=\lambda \mathbf{u}$ with $\mathbf{u}\ne 0$, thenso every eigenvalue is real. If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, thenDistinct eigenvalues therefore have orthogonal eigenvectors.

The normalized eigenvectors form an orthonormal basis, so the matrix $U=({\mathbf{u}}_1\cdots {\mathbf{u}}_n)$ satisfies $U^{†}U=I$ and is unitary. Withthe equations $A{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ say $AU=UD$, hence $A=UD{U}^{†}$. This is the finite-dimensional unitary diagonalization of a normal matrix.

A unitary conjugate of an arbitrary diagonal matrix need not be Hermitian: $U=(1)$ and $D=(i)$ give $UD{U}^{†}=(i)$. It is Hermitian exactly when all diagonal entries of $D$ are real.

For the displayed matrix, one valid decomposition isThe columns are orthonormal eigenvectors, so direct multiplication gives $UD{U}^{†}=A$.

A complex square matrix $A$ is a normal matrix if it commutes with its conjugate transpose:By unitary diagonalization of a normal matrix, there are a unitary matrix $U$ and a diagonal matrix $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ such thatThe matrix 2-norm is invariant under multiplication by unitary matrices, soFor a diagonal matrix,with equality on a coordinate vector belonging to an eigenvalue of greatest modulus. Hencethe matrix 2-norm of a normal matrix.

Biunitary transformations diagonalize the up- and down-type Yukawa matrices:The neutral Higgs couplings are then diagonal, but the charged weak current becomesA $3\times 3$ unitary matrix has nine real parameters. Rephasing the six quark mass eigenfields removes five phases because their common phase is baryon number. The Cabibbo-Kobayashi-Maskawa matrix therefore has four physical parameters: three mixing angles and one CP-violating phase.

Let $H_h$ be the periodic centered second-difference matrix minus the real diagonal matrix containing $V(x_m)$. It is a Hermitian matrix, so the semidiscrete system iswith a skew-Hermitian matrix generator. ConsequentlyIts exact propagator $e^{-it{H}_h}$ is a unitary matrix, and therefore the semidiscretization is stable in the discrete $2$-norm, uniformly for all times and mesh sizes.