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For $\dot{e}=Ae$, differentiation gives $d\Vert e{\Vert }^2/dt=e^T(A+A^T)e$. A uniform negative-definiteness bound with $\kappa >0$ yields exponential contraction by direct integration of this differential inequality. This is a sufficient criterion for asymptotic stability, stronger than requiring all eigenvalues of $A$ to have negative real parts. A merely semidefinite symmetric part needs additional analysis.

A real symmetric matrix is positive definite when its quadratic form is positive on every nonzero vector. By the spectral theorem, this is equivalent to all its eigenvalues being positive. It has a unique symmetric positive-definite square root.

With equally many initially pure habitats of each allele, symmetry preserves $y=1-x$. The heterogeneous equilibrium exists for $\sigma <s/2$. Its full-system eigenvalues are $-s+3\sigma$ in the mean-frequency direction and $-s+2\sigma$ in the antisymmetric direction, so it attracts general nearby perturbations only for $\sigma <s/3$. Exact symmetry can preserve both alleles beyond this robust range; at stronger migration it approaches the unstable uniform midpoint.

For circular reduced deflection $\alpha (r)=r\bar{\kappa }(r)$, a displacement perpendicular to the radial direction is multiplied by $1-\bar{\kappa }$. Vanishing of this eigenvalue gives a tangential gravitational-lensing critical curve. Its source position is $\beta =r-\alpha (r)=0$, so its caustic is the origin in the circular model. Near it, images are stretched tangentially into arcs or an Einstein ring.

With $A_D=S_D{\Lambda }^{ϵ}/(2\pi {)}^D$, the one-loop renormalization-group beta function is ${\beta }_g=-ϵg+(n-2)A_D{g}^2$. For $n>2$ and $ϵ>0$, its two nonnegative renormalization-group fixed points are zero and $g_{\ast }=ϵ/[(n-2)A_D]$. Differentiating the beta function gives the displayed thermal eigenvalues, so the nonzero point has correlation-length critical exponent $\nu =1/ϵ$ to leading order. At $ϵ=0$ the zero eigenvalue requires nonlinear analysis: $g$ is marginally relevant for $n>2$, but the $n=2$ perturbative spin-wave theory has a line of fixed points and omits vortices.

A discrete renormalization group with length factor $b>1$ and positive relevant thermal multiplier ${\lambda }_t>1$ sends $t$ to ${\lambda }_{t}t$ and the correlation length to $\xi /b$. Combining this with $\xi \sim |t{|}^{-\nu }$ gives the displayed correlation-length critical exponent. The corresponding scaling dimension is $y_t=\log {\lambda }_t/\log b=1/\nu$. Relevance of discrete multipliers is determined by magnitude relative to one, unlike the sign criterion for continuous-flow eigenvalues.

The scalar Wilson-Fisher fixed point is attractive within its critical surface, but has a relevant thermal direction. With $r=m^2{\Lambda }^{-2}$, $\lambda =g{\Lambda }^{-ϵ}$, and $K_D={\Omega }_D/(2\pi {)}^D$, its one-loop equations are $\dot{r}=2r+K_D\lambda /[2(1+r)]$ and $\dot{\lambda }=ϵ\lambda -3K_D{\lambda }^2/[2(1+r{)}^2]$. Their stability matrix at $r_{\ast }=-ϵ/6+O({ϵ}^2)$ and ${\lambda }_{\ast }=16{\pi }^2ϵ/3+O({ϵ}^2)$ has the displayed eigenvalues. Thus an unrestricted claim of infrared attraction is false: the bare mass must be tuned to the critical surface.

A real orthogonal matrix $T$ of size three with $\det T=-1$ has an eigenvector $n$ of eigenvalue $-1$. Its orthogonal complement is invariant, and the restriction there is an orientation-preserving plane rotation. Thus $T$ is a rotation about the axis $\mathbb{R}n$ composed with a reflection across the plane perpendicular to $n$. Writing the plane angle as $\varphi$, $\mathrm{tr}T=-1+2\cos \varphi$. A pure plane reflection is the case $\varphi =0$; if $\det (T-I)\ne 0$, it cannot be a pure plane reflection.

A Galerkin method using one convective streamfunction mode and two Cartesian magnetic flux function modes yields the displayed system. With $\alpha =\pi /\sqrt{3}$, use $\psi =4\sqrt{2}a\sin (\alpha x)\sin (\pi z)$ and $A=(3\sqrt{2}/\pi )b\sin (\alpha x)\cos (\pi z)+(\sqrt{3}/\pi )c\sin (2\alpha x)$, time $\tau =(4{\pi }^2/3)t$, $q=9Q/(16{\pi }^2)$ and $r=9R/(64{\pi }^4)$. The origin's planar trace is $\sigma (r-1)-\zeta$ and determinant $\sigma \zeta (1+q-r)$; the third eigenvalue is $-\zeta$. The stationary threshold is $r=1+q$, and the oscillatory threshold is $r=1+\zeta /\sigma$ when $q>\zeta /\sigma$. They meet in a symmetric double-zero point at $(q,r)=(\zeta /\sigma ,1+\zeta /\sigma )$. Higher spatial harmonics generated by the quadratic terms are discarded, so this is a projected approximation rather than an exact invariant subspace of the original equations.

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.

The weighted strain density in Ornstein-Uhlenbeck multiplicative amplification evolves under ${\mathcal{L}}_{n}p=(\kappa /2)p^{\prime \prime }+{\tau }^{-1}(\sigma p{)}^{\prime }+n\sigma p$. Setting $p=e^{-{\sigma }^2/(2\kappa \tau )}\psi$ eliminates its first derivative. Shifting and scaling to $x=(\sigma -\kappa n{\tau }^2)/\sqrt{\kappa \tau }$ gives ${\psi }_{xx}+[1+\kappa n^2{\tau }^3-2\gamma \tau -x^2]\psi =0$. The quantum harmonic oscillator levels yield the displayed eigenvalues. Its ground state is the dominant positive mode; higher sign-changing modes may occur in a transient expansion.

An integer matrix with determinant $\pm 1$ induces an invertible transformation of the real torus, preserving normalized Haar measure. It is a mixing measure-preserving transformation exactly when none of its eigenvalues is a root of unity. Indeed Fourier frequencies evolve by $A^T$: a nonzero integer frequency cannot revisit a fixed frequency unless some power has eigenvalue one. Conversely a root of unity gives a nonzero integer vector fixed by a power of $A^T$, and its nonconstant character obstructs mixing.

For a real $2\times 2$ Jacobian matrix $J$, both eigenvalue have negative real part exactly whenThe linearization stability theorem then makes a hyperbolic equilibrium locally asymptotically stable.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.

For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.

For the pure qubit $\alpha |0⟩+\beta |1⟩$, subtracting the output of dephasing channel leaves a Hermitian matrix with zero diagonal and off-diagonal modulus $|\alpha \beta ||1-c|$. Its two eigenvalues are that number and its negative, giving the displayed trace distance. For real $c=1-2p$, this is $2p|\alpha \beta |$.

A second-rank Cartesian tensor is traceless when the contraction $T_{ii}$ vanishes. For a symmetric second-rank tensor, this is equivalent to its eigenvalues summing to zero. The orientational nematic order parameter is both symmetric and traceless, excluding an isotropic scalar contribution.