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A finite-dimensional quantum channel is primitive when some power maps every nonzero positive semidefinite operator to a positive-definite operator. Repeated application then forgets the initial state and approaches a unique stationary density operator with full rank.

An equivalent spectral criterion requires a full-rank stationary state, a simple eigenvalue one and no other eigenvalue on the unit circle. The full-rank condition is essential: an amplitude damping channel can have a unique stationary pure ground state and all other eigenvalues inside the unit circle without being primitive. A state starting in that ground state never becomes positive definite. Primitivity is therefore stronger than merely having a unique attractive stationary state, and its use in relaxation arguments must specify the finite-dimensional setting.

A quantum channel is strictly contractive in trace distance when it reduces the distance between every pair of distinct density operators by a uniform factor smaller than one.

A quantum channel is unital when $T(I)=I$. Every random unitary channel is unital, while the converse fails in dimension at least three.

A random unitary channel has $T(\rho )={\sum }_j{p}_j{U}_j\rho U_j^{†}$. Its Choi matrix is a convex combination of maximally entangled pure states.

For normalized maximally entangled $|\varphi ⟩=d^{-1/2}{\sum }_j|jj⟩$, the Choi matrix of $T$ is $C_T=(T\otimes \mathrm{id})(|\varphi ⟩⟨\varphi |)$. Choi's theorem says $T$ is completely positive exactly when $C_T\ge 0$.

Every finite-dimensional completely positive map has $T(\rho )={\sum }_j{K}_j\rho K_j^{†}$. It is trace preserving exactly when ${\sum }_j{K}_j^{†}{K}_j=I$.

For hypotheses ${\rho }_0,{\rho }_1$ with priors $p,1-p$, the optimal success probability isAn optimal measurement projects onto the positive and negative spectral subspaces of the weighted difference.

This is the minimum-error optimum in binary quantum state discrimination.

A dense linear order without endpoints is a linear order in which a third point lies strictly between every two distinct points and every point has points both below and above it.

The back-and-forth method constructs an isomorphism by alternately extending finite partial isomorphisms to include elements from each structure.

For a measure $\mu$ with moments $m_j=\int x^{j}d\mu (x)$, a quadrature rule $Q(f)={\sum }_i{w}_{i}f(x_i)$ is exact through polynomial degree $d$ precisely whenNecessity follows by testing each monomial; sufficiency follows by linearity for any polynomial. These equations turn an integration requirement into algebraic constraints on the quadrature nodes and quadrature weights.

For ordinary integration on $[-1,1]$, the first moments are $m_0=2$, $m_1=0$, $m_2=2/3$ and $m_3=0$. Symmetric nodes $\pm a$ with equal quadrature weights $w$ therefore require $2w=2$ and $2w{a}^2=2/3$, giving $w=1$ and $a=1/\sqrt{3}$. Symmetry supplies the odd moments, producing the two-node Gaussian quadrature rule exact through degree three.

If each $(n+1)$-node quadrature rule is exact through degree $n$ and has positive weights, then it converges on every continuous function. The Weierstrass approximation theorem reduces the error to the uniform polynomial-approximation error, while exactness on constants controls the sum of the weights.

If an $(n+1)$-node quadrature rule for a positive interval weight is exact through degree $2n$, every weight is positive. Indeed, applying the rule to the square of the corresponding degree-$n$ Lagrange cardinal polynomial isolates that weight.

No $(n+1)$-node quadrature rule for a positive interval weight can be exact through degree $2n+2$, because it evaluates the nonnegative nodal square $Q_{n+1}^2$ as zero although its integral is positive.

Let an $(n+1)$-node quadrature rule already be exact on all polynomials of polynomial degree at most $n$, for integration against a specified measure $\mu$. Its nodal polynomial isFor an integer $k\ge 0$, the rule is exact through polynomial degree $n+1+k$ if and only iffor every polynomial $r$ of polynomial degree at most $k$, assuming the needed moments exist. Indeed, polynomial division expresses every candidate $p$ as $p=Q_{n+1}r+s$, with $\mathrm{deg}s\le n$. The quadrature rule sees only $s$, since $Q_{n+1}$ vanishes at every quadrature node; exactness on $s$ leaves precisely the displayed orthogonality condition. Conversely, exactness applied to each product $Q_{n+1}r$ forces that condition. This is the mechanism behind the higher precision of Gaussian quadrature.

If $Q$ and $N$ are the nonzero square and nonsquare classes modulo eleven and $\omega =e^{2\pi i/11}$, thenTheir sum is $-1$, and their difference is the quadratic Gauss sum $i\sqrt{11}$.

Over a field of characteristic different from two, the roots of $a{x}^2+bx+c=0$, where $a\ne 0$, are

For symmetric $A$, the quadratic form $x^{T}Ax$ has gradient $2Ax$.

A vector $v$ is isotropic for a quadratic form $q$ when $q(v)=0$.

A subspace $E$ is totally isotropic for $B$ when $B(u,v)=0$ for all $u,v\in E$. For a real nondegenerate bilinear form of inertia $(p,q)$, its dimension is at most $\min (p,q)$.

Every real quadratic form is congruent to a diagonal form with entries $+1,-1,0$, and the numbers of entries of each kind are invariant.