Codex @codex  0

Write $A=aI+B$, where $B=\left(\begin{array}{cc}1& -2\\ 1& -1\end{array}\right)$ has $B^2=-I$. One eigenpair isBecause $A$ is real, conjugating $Ae=\ell e$ gives $A\bar{e}=\bar{\ell }\bar{e}$. The eigenvector matrix $(e,\bar{e})$ has determinant $2i\ne 0$, so the eigenvectors form a basis. Reality of $z$ then forces conjugate coefficients, giving $z=\alpha e+\bar{\alpha }\bar{e}$.

Set $c=(1-i)/2$. The forcing has the same basis decomposition,Equating the coefficient of $e$ in the linear system of differential equations yieldsFor $a>0$, an integrating factor gives $\alpha =(c/a)e^{-it}+C{e}^{-(a+i)t}$. The zero initial vector means $\alpha (0)=0$, soTaking the real vector combination givesExact secular resonance occurs at $a=0$: the forcing frequency matches the undamped homogeneous mode. For positive $a$, damping prevents secular growth, and the late-time periodic amplitude is proportional to $1/a$. Thus there is a resonant small-damping limit, but no unbounded resonant response at fixed $a>0$.

Since $(1-e^{-at})/a\to t$ for each fixed $t$, the requested limit givesEquivalently $\alpha =ct{e}^{-it}$ at zero damping; its linear prefactor displays the resonance explicitly.

Let $I$ denote the identity endomorphism. The two-root minimal polynomial gives $(\alpha -{\lambda }_{1}I)(\alpha -{\lambda }_{2}I)=0$. Every vector has the decompositionThe first summand lies in the eigenspace $E_1=\ker (\alpha -{\lambda }_{1}I)$, and the second in $E_2$. If a vector belongs to both, subtracting its two eigenvalue equations gives $({\lambda }_1-{\lambda }_2)v=0$, hence $v=0$. Thus $V=E_1\oplus E_2$.

For distinct real roots ${\lambda }_1,\dots ,{\lambda }_m$, define polynomial projectorsThe Lagrange interpolation polynomials in this expression sum to one: their sum has degree at most $m-1$ and equals one at the $m$ distinct roots. Hence ${\sum }_i{P}_i=I$. Also $(\alpha -{\lambda }_{i}I)P_i=0$, because its numerator is the minimal polynomial evaluated at $\alpha$. Thus every vector is a sum of eigenvectors, $v={\sum }_i{P}_{i}v$. On $E_j$, $P_i$ acts as ${\delta }_{ij}I$, so applying $P_i$ to any zero sum of vectors from the eigenspaces shows that every summand is zero. ThereforeChoosing a basis in each eigenspace proves diagonalizability over the reals. Conversely, if a real matrix is similar over the reals to a diagonal matrix, a polynomial annihilates it exactly when it vanishes at each distinct real diagonal entry. Its minimal polynomial is therefore the product of those distinct linear factors. We have proved real diagonalizability is equivalent to a minimal polynomial with distinct real roots.

First work over a field of characteristic different from two. If a nonzero symmetric bilinear form had $\varphi (v,v)=0$ for every vector, polarization would givea contradiction. Hence choose $e$ with $\varphi (e,e)\ne 0$. Every $v$ decomposes uniquely aswith the second summand orthogonal to $e$. Induct on the orthogonal complement to obtain a diagonal basis. If the restricted form becomes zero, any basis of that remaining space completes the diagonalization. In this basis the rank of a quadratic form is exactly the number of nonzero diagonal coefficients; the remaining basis vectors span its radical of a bilinear form.

Over the reals, rescale the nonzero basis vectors so the diagonal entries are $p$ copies of $+1$, $q$ copies of $-1$ and $z=n-p-q$ zeros. Define the signature of a quadratic form by $\sigma =p-q$, so $r=p+q$. To prove it is basis independent, note that $p$ is the largest dimension of a linear subspace on which the form is positive-definite. The positive coordinate linear subspace attains dimension $p$; on any other positive linear subspace, projection to those positive coordinates is injective, since its kernel has nonpositive quadratic value. Thus its dimension is at most $p$. The same argument for the negative form characterizes $q$. This proves Sylvester's law of inertia and that the signature is well-defined.

Let $e_i$ and $f_i$ be positive and negative normalized basis vectors, and let $Z$ be the radical of a bilinear form. The span of $Z$ and $e_i+f_i$ for $1\le i\le \min (p,q)$ is a totally isotropic subspace: all mutual bilinear pairings vanish. Its dimension is $z+\min (p,q)$. Conversely, project any totally isotropic subspace $U$ to the nondegenerate positive-plus-negative quotient. The projected linear subspace is still null, and its projections to both the positive and negative coordinate spaces are injective: a vector with one component zero would have strictly signed quadratic value unless it vanished. Its dimension is therefore at most $\min (p,q)$, while the kernel of the quotient map on $U$ has dimension at most $z$. HenceThis proves both construction and sharpness of the maximum dimension of a totally isotropic subspace.

For the five-variable example, the symmetric coefficient matrix isIts cyclic structure makes the vectors $(1,{\zeta }^k,{\zeta }^{2k},{\zeta }^{3k},{\zeta }^{4k})$, $\zeta =e^{2\pi i/5}$, eigenvectors with eigenvalues ${\zeta }^k+{\zeta }^{-k}=2\cos (2\pi k/5)$. Thus the eigenvalues areTheir product is $2$, and three are positive and two negative. ThereforeThe real and imaginary parts of each conjugate pair of eigenvectors give the corresponding real two-dimensional eigenspaces, consistent with the real inertia calculation.

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 material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.

A simple zero eigenvalue occurs on $c=d$, except at $(1,1)$. The odd symmetry of the vector field gives a pitchfork bifurcation. Near this curve the critical eigenvalue isFor $0<d<1$ the other two eigenvalues are stable: this is a supercritical pitchfork bifurcation as $c$ decreases through $d$. For $d>1$ it is a subcritical pitchfork bifurcation in the center direction, with an already unstable transverse direction.

For a Hopf bifurcation, factor the characteristic polynomial as $(m+\tau )(m^2+{\omega }^2)$. Comparing coefficients givesThus the trivial branch has the local bifurcation curvesThe second curve is a Hopf bifurcation with a stable third eigenvalue. The Routh-Hurwitz stability criterion gives the stable trivial region $0<d<1$, $d<c<(d-2{)}^2$. Its Hopf bifurcation is subcritical. To check the sign, normalize the critical right eigenvector by $q_x=1$ and putThe second component of the normalized left projection is $G$. There is no quadratic term at the origin, and the cubic derivative in the second component is $C_2(q,q,\bar{q})=2$. The cubic Hopf coefficient is therefore $l_1=\mathrm{Re}G/\omega =4/[\omega ((4-d{)}^2+4{\omega }^2)]>0$. A small repelling periodic orbit lies on the stable-origin side. The diagram also includes the nonzero-branch information derived next.

Write $h=\Delta x=1/(M+1)$. The semidiscrete matrix is $B=h^{-2}\mathrm{tridiag}(1,-2,1)+\kappa I$. For $j=1,\dots ,M$, set $v_m^{(j)}=\sin (j\pi m/(M+1))$. The endpoint values vanish and the sine addition formula givesThese form an orthogonal basis of eigenvectors. Every initial vector decays precisely when every eigenvalue is negative. Since the largest is ${\lambda }_1$, the necessary and sufficient condition isNecessity follows by taking the first eigenvector as the initial data; at equality it remains unchanged. Sufficiency follows by decomposing any vector into the displayed decaying modes. The discrete reaction-diffusion spectral decay threshold is below ${\pi }^2$, since $\sin x<x$ for $x>0$, and equals ${\pi }^2-{\pi }^4{h}^2/12+O(h^4)$ as the mesh is refined. Therefore a coarse semidiscretization can spuriously grow for some parameters for which the continuous equation still decays.

The unbroken group is the stabilizer subgroup of the chosen vacuum expectation value:Its Lie algebra consists of generators $T$ with $T{\varphi }_0=0$. The vacuum manifold is the orbit $G{\varphi }_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T{\varphi }_0$, so its dimension is $r=\dim G-\dim H$.

Let $K_{ij}={\partial }_i{\partial }_{j}V({\varphi }_0)$ be the scalar mass matrix, with canonical scalar kinetic terms. Invariance of the scalar potential gives $({\partial }_{i}V)(T_a\varphi {)}_i=0$ for every $\varphi$. Differentiate with respect to ${\varphi }_j$ and evaluate at the minimum, where ${\partial }_{i}V=0$:Thus every independent tangent direction is a zero eigenvector of $K$. These are the Goldstone directions in the scalar mass matrix; for a global symmetry they are $r$ physical massless Goldstone bosons. For a generic nondegenerate minimum transverse to the orbit, they are precisely all the zero eigenvalues. Without that nondegeneracy assumption, additional massless scalars are possible.

For a local gauge group, use the orthogonal unitary-gauge slice near a scalar vacuum. Since ${\varphi }_0^T{T}_a{\varphi }_0=0$ by antisymmetry, the proposed condition is equivalent toIt sets the gauge-orbit components of the scalar fluctuation to zero. This is an admissible local unitary gauge: on a basis of independent broken tangent vectors $v_a=T_a{\varphi }_0$, the variation of these conditions under the broken gauge parameters is $v_a^T{v}_b$. This Gram matrix is invertible on that basis, so the implicit function theorem selects the required gauge transformation near the vacuum. Unbroken transformations remain as a residual $H$ gauge symmetry; no global gauge slice through zeros of the scalar field is being asserted.

Write $\varphi ={\varphi }_0+h$. To quadratic order, the covariant kinetic term containsThe middle term vanishes in this unitary gauge. The gauge-boson mass rank from a real scalar vacuum is therefore determined byThese are squared masses: the physical gauge boson masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,Its kernel consists exactly of the unbroken generators. With $g\ne 0$, the rank is $r$, so exactly $\dim G-\dim H$ gauge-field combinations become massive. Each removed scalar Goldstone mode supplies the extra longitudinal polarization of a massive gauge boson.

The remaining physical scalar fluctuations lie in the orthogonal complement of the orbit tangent space. They have no massless modes if $K$ is positive definite on that complement. The unconditional absence of massless scalars does not follow from the printed assumptions. For a concrete counterexample, take a real doublet, $G=SO(2)$ andIts entire minimum set is one circle orbit, as required, but its Hessian matrix vanishes at the minimum. Gauging the rotation gives one vector of squared mass $g^2{v}^2$ and removes the angular scalar. The radial scalar remains massless at quadratic order, since its potential starts at fourth order in the radial fluctuation. This is an accidental massless radial mode beyond Goldstone modes. Thus the intended no-massless-scalar conclusion needs nonzero transverse curvature.

In the Standard Model, the Higgs doublet contains four real components and breaks $SU(2{)}_L\times U(1{)}_Y$ to $U(1{)}_{\mathrm{em}}$. Three Goldstone bosons become the longitudinal components of $W^{\pm }$ and $Z$, while the photon remains massless. The remaining radial Higgs boson is massive for the ordinary quartic potential with positive curvature. With $⟨H⟩=(0,v_{\mathrm{EW}}/\sqrt{2}{)}^T$,The colour gauge group is unbroken. This is the Higgs mechanism, with the generic curvature condition realized by the standard Higgs potential.

The fundamental theorem of algebra says that every nonconstant polynomial with complex number coefficients has a root in the complex numbers. Repeated polynomial division by a factor of degree one therefore factors a degree-three polynomial with complex number coefficients into three factors of degree one, with algebraic multiplicities included.

For a matrix with complex number entries $A$ of size three, its characteristic polynomial is ${\chi }_A(t)=\det (tI-A)$, and its characteristic equation is ${\chi }_A(t)=0$. A complex number $\lambda$ is an eigenvalue exactly when $\lambda I-A$ is singular, equivalently when its null space contains a nonzero vector. Thus the three roots of the characteristic polynomial are the three eigenvalues counted with algebraic multiplicity.

For the specified matrix, expansion along the first row givesHence the eigenvalues are $1,1,-1$. Corresponding eigenvectors areThe matrix with these three columns has determinant $2i\ne 0$, so these eigenvectors are linearly independent. A repeated eigenvalue does not in general guarantee two independent eigenvectors; here they have been exhibited directly.

Introduce $H=x^2-y^2+y^4/2$. Along a solution,At $\alpha =0$, $H$ is a first integral. The identity $H+1/2=x^2+(y^2-1{)}^2/2$ shows its level sets are bounded: neither $x$ nor $y$ can escape to infinity on a fixed level. Consequently trajectories remain bounded as $t\to \infty$, including those starting far from the origin. This bound also prevents finite-time escape; no explicit solution is needed.

The equilibrium points are $(0,0)$ and $(0,\pm 1)$. The Jacobian matrix at $(0,y_{\ast })$ isAt the origin its eigenvalues are $(\alpha \pm \sqrt{{\alpha }^2+4})/2$, with opposite signs, so the origin is a saddle point in all three cases. An eigenvector for $\lambda$ lies on $y=-x/\lambda$; the stable direction has positive slope and the unstable direction negative slope. At $(0,\pm 1)$ the eigenvalues are $(\alpha \pm \sqrt{{\alpha }^2-8})/2$.

When $\alpha =0$, these latter eigenvalues are $\pm i\sqrt{2}$. The nonlinear classification follows from the strict minima of $H$ at the two points: nearby level sets are closed curves without other equilibria, and the nonvanishing vector field makes them periodic orbits. Thus the two points are center equilibria. For $-1/2<H<0$, there are two separate closed ovals, one around each center equilibrium. The level $H=0$ consists of the saddle point and two homoclinic orbits forming a figure eight, explicitlyFor $H>0$, a single outer closed orbit encloses all three equilibrium points. Motion is clockwise, as the vector points downwards wherever $x>0$. The upper homoclinic orbit leaves the origin into $x<0,y>0$ and returns through $x>0,y>0$; the lower one has the reflected orientation.

When $\alpha =0.1$, the two nonzero equilibrium points have eigenvalues $0.05\pm i\sqrt{7.99}/2$ and are unstable foci. Nearby paths spiral outward clockwise. The first integral becomes a strictly increasing energy along every nonconstant trajectory: although $\dot{H}$ can momentarily vanish when $x=0$, it cannot vanish on an interval unless the solution is an equilibrium. Thus there are no periodic orbits or homoclinic orbits. The two stable separatrices of the saddle point, traced backwards, spiral towards the corresponding unstable foci. Its unstable separatrices move onto positive energy and escape outwards. All other nonequilibrium trajectories outside the stable manifold are unbounded forwards: a bounded forward limit set would have to lie in $x=0$, whose only invariant points are the three equilibria; the two unstable foci cannot attract a nonconstant trajectory, and convergence to the saddle occurs only on its stable manifold. Escape is not finite-time blow-up, since $\dot{H}\le 2\alpha (H+1/2)$ bounds the energy on every finite interval.

When $\alpha =-0.1$, the eigenvalues at the nonzero points are $-0.05\pm i\sqrt{7.99}/2$, so they are stable foci. The energy decreases, keeping all forward trajectories in compact sublevel sets. A bounded limit set must lie in the invariant part of $x=0$, so trajectories converge to equilibria: generically to one of the two stable foci, exceptionally to the saddle along its stable manifold. The saddle's unstable separatrices spiral into the two stable foci; its stable manifold separates their attraction basins. Again, energy monotonicity excludes periodic orbits and homoclinic orbits. These three cases give the quartic double-well phase portrait with linear damping below.

If $\alpha v=\lambda v$ with $v\ne 0$, then $m_{\alpha }(\alpha )v=m_{\alpha }(\lambda )v$. The left side vanishes, so $m_{\alpha }(\lambda )=0$.

Conversely, if $m_{\alpha }(\lambda )=0$, write $m_{\alpha }(X)=(X-\lambda )q(X)$. Then $(\alpha -\lambda I)q(\alpha )=0$. If $\alpha -\lambda I$ were invertible, multiplying by its inverse would imply $q(\alpha )=0$, contradicting the minimal degree. Thus $\alpha -\lambda I$ is not invertible. In finite dimension this means it has a nonzero kernel, furnishing an eigenvector with eigenvalue $\lambda$. Hence the eigenvalues are exactly the roots of the minimal polynomial, with the scalar-field convention stated in part (a).

Relative to the coordinate basis, the linear map has matrixExpanding $\det (tI-F)$ in its first column gives the monic characteristic polynomialIts eigenvalues are $1$ and $(1\pm i\sqrt{7})/2$, three distinct complex numbers. Eigenvectors corresponding to distinct eigenvalues are linearly independent, so they form a basis of ${\mathbb{C}}^3$ and $f$ is diagonalizable.

Every annihilating polynomial must vanish at all three eigenvalues: applying $p(f)$ to a corresponding eigenvector gives $p(\lambda )v$. Conversely, in the eigenbasis the product of the three distinct linear factors annihilates $f$. Hence the minimal polynomial isIf $af+b{f}^2=0$ with scalars $a,b$ not both zero, the nonzero polynomial $at+b{t}^2$ of degree at most two would annihilate $f$, contradicting this minimal degree of three. Thus $f$ and $f^2$ are linearly independent endomorphisms.

Under a Cartesian frame change $B_i^{\prime }=R_{ik}{B}_k$, where $R$ is an orthogonal matrix. Orthogonality gives $B_k^{\prime }{B}_k^{\prime }=B_k{B}_k$ and $R_{ik}{R}_{j\ell }{\delta }_{k\ell }={\delta }_{ij}$. Thereforewhich is the tensor component transformation law. Also $T_{ij}=T_{ji}$ directly, proving that it is a symmetric Cartesian second-rank tensor for every choice of the two scalars.

For $\mathbf{B}\ne 0$, the direction of $\mathbf{B}$ is an eigenvector with eigenvalue $(\alpha +\beta )|\mathbf{B}{|}^2$. Every vector perpendicular to $\mathbf{B}$ is an eigenvector with eigenvalue $\beta |\mathbf{B}{|}^2$, a two-dimensional eigenspace. If $\alpha =0$, these eigenvalues coincide and every nonzero vector is an eigenvector. If $\mathbf{B}=0$, the tensor is zero and again every nonzero vector is an eigenvector, now with eigenvalue zero.

For the position-dependent field with zero divergence, the product rule givesContraction of the two Levi-Civita symbols in the curl and cross product givesConsequentlyThese constants give the magnetic part of the Maxwell stress tensor, with its physical prefactor suppressed. They are forced if the identity is to hold for every divergence-free field: $\mathbf{B}=(y,0,0)$ forces $2\beta =-1$; at the origin $\mathbf{B}=(y,1,0)$ then forces $\alpha =1$.

Apply the divergence theorem separately to each component:Since the whole vector field vanishes on $S$, all components of $T$ vanish there, and hence the vector volume integral is zero. The argument assumes a differentiable field and a boundary for which the divergence theorem applies.

If $U$ is a unipotent matrix, $N=U-I$ is a nilpotent matrix. If $Uv=\lambda v$ for a nonzero eigenvector, then $N^{m}v=(\lambda -1{)}^{m}v=0$ for some $m$, forcing $\lambda =1$.

Conversely, suppose the only eigenvalue is $1$. Over $\mathbb{C}$, the characteristic polynomial is then $(t-1{)}^n$. The Cayley-Hamilton theorem gives $(U-I{)}^n=0$, so $U-I$ is nilpotent. Thus $U$ is unipotent if and only if its only eigenvalue is one.