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Write the linear differential equation as $\dot{\mathbf{x}}=M\mathbf{x}$, withIts characteristic polynomial is $(\lambda +4{)}^2+9$. Thus a choice of eigenvalues and eigenvectors isThe distinct eigenvalues give two linearly independent solutions, and the general complex solution isReal solutions have $a_2=\overline{a_1}$. Equivalently,They spiral counterclockwise towards the origin, which is a stable spiral.

In a forced linear differential equation, resonance in a differential equation occurs when exponential forcing excites a mode with the same exponent as its eigenvalue, producing a polynomial factor such as $t$ multiplying that exponential. Merely matching an exponent is not enough: its projection onto that mode must be nonzero. Here $t{e}^{{\lambda }_{j}t}$ still decays in absolute magnitude because the real part of ${\lambda }_j$ is negative; resonance refers to this extra factor, not necessarily to unbounded growth.

For each forcing vector, resolve it in the eigenvector basis:If $u_k$ is the coefficient of the $k$th eigenvector, its scalar equation is ${\dot{u}}_k={\lambda }_k{u}_k+{\sum }_j{c}_{kj}{e}^{{\lambda }_{j}t}$. Multiplying by $e^{-{\lambda }_{k}t}$ and integrating shows that the $j=k$ contribution is $c_{kk}t{e}^{{\lambda }_{k}t}$, whereas a $j\ne k$ contribution is $c_{kj}{e}^{{\lambda }_{j}t}/({\lambda }_j-{\lambda }_k)$. Therefore the exact conditions for no resonant response areUnder these conditions one particular solution iswhich contains no factor of $t$. For a real forcing, $p_2=\overline{p_1}$ and $q_2=\overline{q_1}$, and the two boxed conditions are conjugate to each other.

Let $Az=\lambda z$ with $z\ne 0$, allowing complex eigenvectors. A real symmetric matrix is Hermitian, so $z^{\ast }Az$ is real and $\lambda =z^{\ast }Az/(z^{\ast }z)$ is real. For eigenvectors $u,v$ of distinct real eigenvalues $\lambda ,\mu$, self-adjointness gives $\lambda u^{\ast }v=u^{\ast }Av=\mu u^{\ast }v$, hence $u^{\ast }v=0$. Thus the distinct eigenspaces are orthogonal.

Here $A=3I-J$, where $J$ is the all-ones matrix. Since $J(1,1,1{)}^T=3(1,1,1{)}^T$ and $J$ vanishes on the plane of coordinate sum zero,Every nonzero vector in the indicated eigenspace is an eigenvector. The repeated eigenvalue three has a two-dimensional eigenspace.

A complex symmetric nilpotent matrix provides the requested counterexample:Its characteristic polynomial is ${\lambda }^2$, so its only eigenvalue is zero. It is not diagonalizable: a diagonalizable matrix with only zero eigenvalues would itself be zero. Complex symmetry $N^T=N$ is weaker than the Hermitian condition $N^{\ast }=N$.

Expanding the characteristic polynomial along the first column givesFor $s\ne 1$, solving the two nullspace equations givesThese are all the eigenvalues and eigenvectors, with nonzero vectors understood. The three displayed spanning vectors form an eigenbasis, so $A$ is diagonalizable.

At $s=1$, the sole eigenvalue is three, with algebraic multiplicity three but the same two-dimensional $E_3$. Thus $A$ is diagonalizable exactly when $s\ne 1$. Indeed, at the exceptional parameter, $A=3I+N$ with $N\ne 0$ and $N^2=0$, so its Jordan normal form has one block of size two and one of size one. This illustrates how an eigenvalue collision can destroy diagonalizability without changing the two-dimensional eigenspace.

Use coordinates $\mathbf{r}$ measured from the center of mass $G$, so $\int \mathbf{r}dm=0$. The displacement from $P$ is $\mathbf{r}-\mathbf{X}$. Expand the defining inertia tensor:All terms linear in $\mathbf{r}$ integrate to zero. The remaining constant terms give the tensor parallel-axis theorem,For the uniform cube, let $M=8\rho a^3$. Symmetry makes the off-diagonal inertia tensor entries zero. Direct integration gives $\int x_i^{2}dm=M{a}^2/3$ for each coordinate, and therefore $I(G)=(2M{a}^2/3)I_3$. With $\mathbf{X}=a(1,1,1{)}^T$, the tensor parallel-axis theorem givesThe unit eigenvector $u=(1,1,1{)}^T/\sqrt{3}$ has eigenvalue $2M{a}^2/3$. Every vector in the plane $x+y+z=0$ is annihilated by $11^T$ and hence has eigenvalue $11M{a}^2/3$. One orthonormal basis of that eigenspace isThe principal moments are $2M{a}^2/3$ along the body diagonal and $11M{a}^2/3$ twice in its perpendicular plane. Any orthonormal pair in that plane supplies the remaining principal axes.

Let $m_{\alpha }(x)$ be the monic minimal polynomial. The Cayley-Hamilton theorem says ${\chi }_{\alpha }(\alpha )=0$. Divide ${\chi }_{\alpha }$ by $m_{\alpha }$; its remainder also annihilates $\alpha$, and a nonzero remainder of smaller degree would contradict minimality. Thus $m_{\alpha }$ divides ${\chi }_{\alpha }$.

Every characteristic root ${\lambda }_i$ is an eigenvalue, with a nonzero eigenvector $v_i$. Since $m_{\alpha }(\alpha )v_i=m_{\alpha }({\lambda }_i)v_i=0$, it must be a root of the minimal polynomial as well. This proves there are no possibilities beyondEvery such choice really occurs. For each $i$, take one Jordan block of size $r_i$ at ${\lambda }_i$, together with $n_i-r_i$ scalar blocks at that eigenvalue, and form their direct sum. The characteristic polynomial has the prescribed multiplicities. The nilpotent part of a block of size $r_i$ has its $r_i$th power zero and its $(r_i-1)$st power nonzero, so this component's minimal exponent is exactly $r_i$. A polynomial annihilates the direct sum precisely when it annihilates every component, establishing the displayed minimal polynomial without any extra restrictions.

Start with the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$. For the dimension count, complexify and rescale the timelike matrix so that the generators obey the Euclidean complex relation $\{{\gamma }_a,{\gamma }_b\}=2{\delta }_{ab}$. This changes neither the complex representation dimension nor its decomposition under the complexified even algebra. In $D=2k$, pair the generators intoDirect substitution gives $\{a_i,a_j\}=\{b_i,b_j\}=0$ and $\{a_i,b_j\}={\delta }_{ij}$. The commuting operators $n_j=b_j{a}_j$ satisfy $n_j^2=n_j$, so their occupation eigenvalues are zero or one. An irreducible module contains an all-zero occupation state $|0⟩$, obtained by lowering the occupied entries of a simultaneous eigenvector. The statesare nonzero and independent by their distinct joint occupation eigenvalues. They span an invariant module; irreducibility makes this the full module. Conversely this Fock construction supplies a representation on exactly these $2^k$ states. Products of occupation projectors and raising/lowering operators give its matrix units, so the generated algebra is the full matrix algebra. Therefore the minimal complex Dirac spinor has

This module is irreducible under the full Clifford algebra, but the Spinor representation of the Lorentz group is generated by the even bilinears ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\gamma }^{\mu },{\gamma }^{\nu }]$. The chirality matrix in the Euclideanized basis isIts two eigenspaces therefore remain invariant under Lorentz transformations. Each has half the Fock states, since any gamma matrix bijects them. The even algebra connects every pair of occupation states of the same parity and has no further invariant subspaces within either one. HenceThese are the irreducible Weyl spinors. Globally they are ordinary representations of $\mathrm{Spin}(D-1,1)$, the double cover; calling them representations of $SO(D-1,1)$ uses the conventional projective spinor terminology.

For $D=2k+1$, adjoin ${\gamma }_{2k+1}=\pm {\gamma }_{\ast }$. It anticommutes with the old matrices and gives a Clifford module of the same dimension $2^k$. The new even generators $[{\gamma }_a,{\gamma }_{2k+1}]$ interchange the two old chirality eigenspaces, so the odd-dimensional Lorentz spinor is irreducible of complex dimension $2^k$, with no Weyl splitting. The product of all odd-dimensional gamma matrices is central and acts as a scalar; it cannot be a new anticommuting chirality operator. Its two signs distinguish the two complex odd-Clifford modules, whose restrictions to Spin group are equivalent. Majorana spinor reality conditions, when the signature permits them, are additional conditions and should not be confused with complex chirality.

For the reduction, take the massless zero modes on a flat seven-torus with no flux or projections. The eleven-dimensional fields are the metric $G_{MN}$, a three-form $A_{MNP}$ and a Majorana gravitino ${\Psi }_M$. Split $M=(\mu ,i)$ with $\mu =0,\dots ,3$ and $i=1,\dots ,7$. The metric supplies $G_{\mu \nu }$, seven vectors $G_{\mu i}$ and 28 scalar fields $G_{ij}=G_{ji}$. The three-form supplies a four-dimensional three-form $A_{\mu \nu \rho }$, seven two-forms $A_{\mu \nu i}$, $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ vectors $A_{\mu ij}$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ scalars $A_{ijk}$.

A massless two-form in four dimensions has one physical particle polarization and can be dualized to a scalar by $H_3=\ast d\sigma$; the seven two-forms thus supply seven more scalars. The four-dimensional three-form has no local particle polarization, since its four-form field strength is locally a constant times the volume form. The total bosonic content is consequentlyThe eleven-dimensional Majorana spinor has 32 real components, decomposing into eight four-dimensional Majorana spinors. Accordingly ${\Psi }_{\mu }$ supplies eight gravitini, while ${\Psi }_i$ supplies $7\times 8=56$ spin-$1/2$ fields after diagonalizing the gravitino/matter kinetic terms. The resulting theory has 32 real supercharges, hence is four-dimensional N=8 supergravity, withThe on-shell counts check the reduction: the metric has $11(11-3)/2=44$ polarizations and the three-form has $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, totaling 128. In four dimensions $2+28\cdot 2+70=128$ bosonic polarizations match $8\cdot 2+56\cdot 2=128$ fermionic ones. Higher Kaluza–Klein levels, fluxes or projections would be additional choices beyond this massless toroidal reduction.

The mixed-state expression in the PDF contains a mismatched bra in its second term. A probability mixture of the stated orthonormal eigenvectors must be interpreted as $\rho ={\sum }_k{p}_k|u_k⟩⟨u_k|$. Taken literally with a different bra, it need not be Hermitian or positive and is not generally a density operator.

Let $m={\sum }_k{p}_k{e}^{i{\varphi }_k}=\mathrm{Tr}(\rho U)$. A quantum channel and its measurement probabilities are linear in the density operator, so averaging the pure-state results givesTherefore the same frequency estimator yieldsMore generally, after the controlled operation and before the final Hadamard gate, the reduced control density operator isThis follows by a partial trace of the control blocks $\rho$, $\rho U^{†}$, $U\rho$ and $U\rho U^{†}$. It proves the mixed-state Hadamard-test quadratures for any auxiliary density operator, even one not diagonal in the eigenbasis. Conjugation by the final Hadamard gate maps $X$ to $Z$ and $Y$ to $-Y$, explaining the imaginary-part sign.

Prepare the auxiliary system in the maximally mixed state $\rho =I/N$. Then $\mathrm{Tr}(\rho U)=\mathrm{Tr}U/N$, so the two measurement bases giveThere is no need to know the eigenvectors of $U$. Choose a uniformly random member of any known orthonormal basis on each run; discarding the classical preparation label produces $I/N$. Alternatively, discard one half of a maximally entangled pair of $N$-level systems. This is trace estimation by a maximally mixed Hadamard test.

The circuit directly estimates the normalized trace. To demand fixed absolute accuracy $\epsilon$ in the unnormalized trace requires normalized accuracy $\epsilon /N$; straightforward independent frequency sampling consequently uses order $N^2/{\epsilon }^2$ runs at fixed confidence. Thus the construction should not be confused with a dimension-independent guarantee for arbitrary absolute trace precision.

Write $C_U=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes U$ for the controlled unitary gate. For a normalized eigenvector $|u⟩$, the first Hadamard gate and quantum phase kickback giveThe final Hadamard gate therefore produces the unentangled outputThe auxiliary system is unchanged, while its eigenphase becomes accessible through interference of the control qubit. The special unitary group assumption is not needed for this calculation; unitarity and the normalized eigenvector suffice.

The primary vorticity Jacobian determinant vanishes because ${\nabla }^2\psi =-s\psi$. The quadratic temperature Jacobian determinant isThe mean temperature correction therefore obeys $\dot{c}=-4{\pi }^{2}c-\alpha \pi ab/2$. Its Jacobian determinant with the primary streamfunction isProjecting on the retained primary modes gives, with $D=s+Q{\pi }^2/s$,For a nonzero steady roll, $b=(s^2+Q{\pi }^2)a/(R\alpha )$ and $c=-\alpha ab/(8\pi )$. Substitution in the second equation gives, within this truncation,The positive denominator ensures branches only on the supercritical side for every $Q\ge 0$. To verify dynamical stability as well as existence, the critical right eigenvector is $(1,\alpha /s)$; normalizing the left eigenvector gives linear eigenvalue derivativeQuadratic slaving gives $c=-{\alpha }^2{a}^2/(8\pi s)$, and projection of its feedback gives the cubic amplitude equationIts nonzero branches are stable in the center direction, while all other eigenvalues remain negative at a simple onset. Reflection about the mid-width sends $a\to -a$, so the bifurcation is a supercritical pitchfork bifurcation for all $Q$, in the generic case of a single critical roll mode.

The cubic saturation of quasistatic magnetoconvection result is identical in Fourier truncation and perturbation theory at the orders needed here. At quadratic order the only forced mode is precisely the retained horizontally uniform $\sin 2\pi z$ temperature mode. At cubic order its feedback projects on the primary mode with the same coefficient. The generated third vertical harmonic is orthogonal to that primary projection and contributes only to higher-order corrections. Thus both methods give the same linear threshold and cubic solvability condition; the truncated finite-amplitude branch is not an exact solution of the full PDE at all amplitudes.

A spatial finite difference method and time discretization for a linear evolution problem give a recurrence $U^{n+1}=B_{h,k}{U}^n$. To obtain numerical stability on $0\le nk\le T$, one needs a bound $\Vert B_{h,k}^n\Vert \le C_T$ uniform in the permitted grid sizes and time steps. More generally a growth estimate $\Vert B_{h,k}^n\Vert \le C{e}^{cnk}$ is sufficient. The finite-time stability versus power boundedness distinction matters: growth per step of $1+O(k)$ can be compatible with the latter finite-time estimate, whereas unrestricted geometric growth at a fixed Courant number is not.

The eigenvalue method diagonalizes $B_{h,k}$ into modes. If $B_{h,k}$ is a normal matrix, the spectral theorem gives $\Vert B_{h,k}^n{\Vert }_2={\max }_j|{\mu }_j{|}^n$. Thus $|{\mu }_j|\le 1$ gives contraction, and $|{\mu }_j|\le e^{ck}$ gives the preceding growth estimate. A diagonalizable matrix $B=V\Lambda V^{-1}$ has $\Vert B^n{\Vert }_2\le \Vert V{\Vert }_2\Vert V^{-1}{\Vert }_2{\max }_j|{\mu }_j{|}^n$, so its condition number must remain bounded uniformly as the mesh changes. A stable-looking list of eigenvalues alone is insufficient.

Indeed,Both eigenvalues are one, but the Jordan block gives growth proportional to $n\sim T/k$, violating uniform finite-time numerical stability. A nonnormal Fourier amplification matrix can similarly exhibit large amplification even with its eigenvalues inside the unit disk. Eigenvector conditioning and the full powers of the amplification operator are essential.

For constant-coefficient schemes on a periodic or infinite grid, a Fourier transform diagonalizes scalar translation-invariant stencils. This yields von Neumann stability analysis using the Fourier amplification symbol $G(\theta )$. The discrete Parseval identity turns a bound on $G(\theta {)}^n$ into a discrete L2 norm estimate. For systems, the symbol is a matrix; a uniform power bound for matrix Fourier symbols is required, rather than a bound on its spectral radius alone.

For the heat equation, a central finite difference in space and the explicit Euler method in time giveThe scalar condition $|G|\le 1$ for every frequency is exactly $0\le r\le 1/2$. With zero Dirichlet boundary conditions, the spatial matrix is symmetric and diagonalized by sine modes, leading to the same uniform sufficient restriction. By contrast, the Backward Euler diffusion scheme has $G(\theta )=[1+4r{\sin }^2(\theta /2){]}^{-1}$ and is stable for every $r\ge 0$.

For the advection equation $u_t+a{u}_x=0$, $a>0$, centered space with the explicit Euler method gives $G=1-i\nu \sin \theta$, $\nu =ak/h$, and $|G{|}^2=1+{\nu }^2{\sin }^2\theta$. At fixed positive Courant number, repeated powers are unbounded as the mesh is refined. The upwind finite difference scheme instead haswhich gives contraction precisely for $0\le \nu \le 1$. This comparison shows why the spatial stencil and time method must be analyzed together.

Boundary treatment can destroy the normality or mode decomposition available on a periodic grid, so a finite-interval scheme needs its own uniform operator estimate. Multilevel time schemes likewise require the root condition for a multistep method and uniform bounds for the associated amplification matrices; simple scalar root tests do not establish these bounds for arbitrary systems. Once consistency of a numerical method, a well-posed linear evolution problem and the required numerical stability estimate hold, the Lax equivalence theorem gives numerical convergence. Eigenvalues provide a powerful route to that estimate when the accompanying eigenvector and boundary analysis is justified.

For a uniform material and an affine static base deformation, the deformation gradient, incremental elastic moduli and reference mass density are constant. With zero incremental body force, the equation reduces to $c_{\alpha i\beta j}{v}_{j,\beta \alpha }={\rho }_0{\ddot{v}}_i$. Let $\nu$ be a unit material propagation direction, $m$ a constant polarization and $s=t-{\nu }_{\alpha }{\xi }_{\alpha }/c$. Direct differentiation of the plane wave givesThus every twice differentiable profile $f$ satisfies the incremental equation ifThis is the acoustic tensor eigenvalue problem. The acoustic tensor is a real symmetric matrix: interchange the index pairs in the incremental elastic moduli and then interchange the dummy material indices to obtain $Q_{ji}=Q_{ij}$. Its eigenvectors give the displacement polarizations. A positive eigenvalue gives a real nonzero speed $c$; arbitrary-profile propagation imposes precisely this algebraic condition. Boundary conditions determine which combinations of these local plane waves are allowed in a finite body.

Put $G=g({\rho }_p-{\rho }_0)/{\rho }_0$ and $b=G\varphi >0$, the reduced gravity. Use the Boussinesq approximation, hydrostatic pressure, uniform vertical velocity and concentration profiles, negligible fluid entrainment and drag, and a flat bed. Integrating volume conservation, momentum conservation and the particle budget over the layer givesThe hydrostatic pressure resultant is ${\rho }_{0}b{h}^2/2$; the last right-hand side is the deposition flux. With $D={\partial }_t+u{\partial }_x$, these areThe quasilinear coefficient matrix in variables $(h,u,b)$ isIts eigenvalues are $u,u\pm c$, where $c=\sqrt{bh}$. Therefore the characteristic curves and their compatibility ordinary differential equations areFor the last equation the left eigenvectors are $(\pm c/h,1,\pm h/(2c))$, and multiplying the quasilinear equations by them gives the displayed source terms. This is the sedimenting rectangular-channel characteristic compatibility relation; it supplies one differential relation on each wave family, rather than three independently closed scalar equations there. The contact eigenvector has $du=0$ and $d(b{h}^2)=0$. For reference, an equivalent form isOnly for constant $b$ without settling do $u\pm 2c$ become genuine Riemann invariants. With settling, the spatial buoyancy gradient cannot be silently omitted.

Expanding the cross product component by component gives its cross-product matrix:It is a real skew-symmetric matrix and annihilates $\mathbf{n}$. By the vector triple product,so $A^2=\mathbf{n}{\mathbf{n}}^T-I$ for a unit vector $\mathbf{n}$. Choose a unit vector $\mathbf{u}\perp \mathbf{n}$ and put $\mathbf{v}=\mathbf{n}\times \mathbf{u}$. Then $(\mathbf{u},\mathbf{v},\mathbf{n})$ is an orthonormal basis and $A\mathbf{u}=\mathbf{v}$, $A\mathbf{v}=-\mathbf{u}$, $A\mathbf{n}=0$. Thus the characteristic polynomial is $\lambda ({\lambda }^2+1)$ and the cross-product matrix spectrum isOver the complex numbers, corresponding eigenvectors are $\mathbf{n}$, $\mathbf{u}-i\mathbf{v}$, and $\mathbf{u}+i\mathbf{v}$. Only the zero eigenvalue has real eigenvectors.

For the specified third coordinate axis, direct multiplication givesIts zero eigenspace is $\mathrm{span}\{(0,0,1)\}$, and its $-1$ eigenspace is the whole plane $\mathrm{span}\{(1,0,0),(0,1,0)\}$. Hence the eigenvalues of $A^2$ are $-1,-1,0$, with every nonzero vector in the appropriate eigenspace an eigenvector.

At a critical point $p$ of a twice continuously differentiable real function, $\nabla f(p)=0$ and the Taylor expansion isThe real symmetric Hessian matrix has an orthonormal eigenbasis, so writing $h={\sum }_i{a}_i{u}_i$ gives $h^{T}Hh={\sum }_i{\lambda }_i{a}_i^2$. If every eigenvalue is positive, this quadratic form is at least ${\lambda }_{\min }|h{|}^2$, and the remainder cannot change its sign for sufficiently small nonzero $h$. Thus $p$ is a strict local minimum. If every eigenvalue is negative, the analogous argument gives a strict local maximum. If positive and negative eigenvalues both occur, displacement along the corresponding eigenvectors gives values above and below $f(p)$, so $p$ is a saddle point of a scalar function.

The second-derivative test can be inconclusive when the Hessian matrix is singular and semidefinite: higher-order terms along directions in the kernel of a linear map matter. For example $x^2+y^4$ and $x^2-y^4$ have the same Hessian $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle point of a scalar function. A zero eigenvalue does not prevent a saddle conclusion if other eigenvalues already have both signs.

For the given function write $r^2=x^2+y^2$. Its first derivatives areThe origin is a critical point for every real $\alpha$, withIts eigenvalues are $1,-1$, so it is always a saddle point of a scalar function. If one coordinate at a critical point is zero, the other must also be zero. If both are nonzero, the derivative equations force $x^2=y^2=1/(2\alpha )$, possible only for $\alpha >0$. Hence for $\alpha \le 0$ the origin is the only critical point.

For $\alpha >0$ there are four additional critical points,At such a point,Thus the two equal-sign points are strict maxima, and the two opposite-sign points are strict minima. They are global: $2|xy|\le r^2$ implieswhere the last maximum occurs at $r^2=1/\alpha$ and equality also requires $x^2=y^2$.

As $\alpha \downarrow 0$, the four nonzero critical points have distance ${\alpha }^{-1/2}$ from the origin and escape to infinity, rather than merging with the origin. Their values diverge as $\pm 1/(2e\alpha )$. As $\alpha \uparrow 0$ from below there are no extra critical points, while the origin stays a saddle throughout. This is critical points escaping to infinity in a Gaussian-weighted bilinear function.