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The set of real positive semidefinite matrices with unit diagonal. Equivalently, its members are Gram matrices of unit vectors, by the real spectral theorem. Every two-by-two principal minor gives $|X_{ij}|\le 1$, so the set is bounded; it is also closed, hence compact in finite dimension. This gives attainment of linear objectives used in semidefinite programming.

For three non-collinear points with position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, a nonzero normal vector isThe plane is the set of $\mathbf{r}$ satisfying $(\mathbf{r}-\mathbf{a})\cdot \mathbf{N}=0$. Expanding the cross product and using the scalar triple product givesNon-collinearity is essential: otherwise $\mathbf{N}=0$ and the equation fails to select a unique plane.

The general linear group over a finite field $G{L}_2({\mathbb{F}}_2)$ acts faithfully on the four vectors of ${\mathbb{F}}_2^2$, since a matrix fixing all vectors fixes the standard basis. The zero vector is fixed by every element; the induced action on the three nonzero vectors identifies this six-element group with $S_3$. Swapping the two coordinates fixes $(0,0)$ and $(1,1)$ and exchanges the other two vectors, so its four-point permutation is odd. This action therefore embeds the group into $S_4$ with nontrivial composite sign homomorphism.

For centered, unit-variance jointly Gaussian random variables with correlation coefficient $\rho$, the displayed identity holds. Realize the pair as projections onto unit vectors of angle $\theta =\arccos \rho$. Rotational symmetry of the two-dimensional Gaussian random vector direction makes sign disagreement have probability $\theta /\pi$. The expectation is therefore $1-2\theta /\pi =(2/\pi )\arcsin \rho$. Perfectly correlated and anticorrelated endpoints follow directly; the sign at zero does not affect the result.

The gradient of a scalar-valued differentiable function is the vector of its first partial derivatives.

Under the displayed diagonal-sum convention, set $w(z)=(1,z^{-1},\dots ,z^{-d}{)}^T$ on the unit circle. A Hermitian matrix $M$ then givesConsequently $M⪰0$ proves nonnegativity. Conversely the Fejér–Riesz theorem produces the rank-one spectral-factor Gram matrix. This gives a semidefinite programming representation of nonnegative trigonometric polynomials. The reversed convention $j-i=k$ instead uses the unconjugated monomial vector.

An indefinite Hermitian form takes both positive and negative values on $⟨v,v⟩$ for nonzero vectors. For example, $-|v_0{|}^2+|v_1{|}^2+|v_2{|}^2+|v_3{|}^2$ is indefinite. It is not a positive inner product defining a Hilbert space, and a nonzero vector can have zero norm without being orthogonal to all vectors. A positive semidefinite restricted form becomes a positive inner product only after quotienting its radical of a Hermitian form.

The interior consists exactly of strictly copositive matrices. If $x^{T}Ax$ has positive minimum $\delta$ on the compact nonnegative unit sphere, perturbations of operator norm less than $\delta$ preserve this positive lower bound there. Conversely, if a unit nonnegative vector has zero quadratic form, $A-\epsilon I$ leaves the copositive cone for every $\epsilon >0$. Thus such a matrix is not interior. The identity matrix is a convenient interior point.

The number of nonzero coordinates of a finite vector is its L0 sparsity count. It vanishes only at zero, but is unchanged under multiplication by a nonzero scalar, so it is not a norm. Minimizing it under linear measurement constraints finds a sparse vector with the smallest possible support of a vector. Sparse injectivity of order $s$ guarantees unique recovery of every sparse vector with at most $s$ nonzero coordinates by this objective.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

When $A_S$ is injective, this vector solves $A_S^{\ast }{h}_0=\mathrm{sgn}(x_S)$. Any other solution differs by a vector in $\ker A_S^{\ast }$, orthogonal to the range of $A_S$ containing $h_0$. The Pythagorean identity therefore proves that $h_0$ has the smallest Euclidean norm. It is a strict dual certificate for basis pursuit exactly when each inactive coordinate of $A^{\ast }{h}_0$ has magnitude less than one. Failure of that test does not rule out another certificate: a null space component of $A_S^{\ast }$ can alter those inactive coordinates.

For $0<q\le 1$, this strict inequality for every nonzero $v\in \ker A$ and every $|S|\le s$ is equivalent to unique constrained ${\ell }^q$ recovery of every sparse vector of order $s$. For sufficiency, use $|x_i+v_i{|}^q\ge |x_i{|}^q-|v_i{|}^q$ on the support of a vector. For necessity compare $x=-v_S$ with $v_{S^c}$, which have the same measurements. The statement concerns uniform recovery over the whole sparse class, rather than a single signed vector.

The displayed quantity is homogeneous and separates zero, but generally violates the triangle inequality. Its $q$th power is subadditive because $(a+b{)}^q\le a^q+b^q$ for nonnegative $a,b$. Minimizing it is equivalent to minimizing its $q$th power. On finite vectors it is a standard nonconvex sparsity penalty.

If the Lq null space property holds at $q\le 1$, it holds at every $0<p<q$. Order a nonzero null vector's magnitudes $a_1\ge \cdots \ge a_N$ and put $t=a_s>0$. Since $p-q<0$, the top-$s$ $p$-power sum is at most $t^{p-q}$ times its $q$-power sum, while the tail $p$-power sum is at least that multiple of the tail $q$-power sum. The strict $q$ inequality therefore gives the strict $p$ inequality. The largest $s$ magnitudes are the worst support of a vector, so all other supports satisfy it too.

A real symmetric matrix $A$ is negative definite when $x^{T}Ax<0$ for every nonzero real vector $x$. Equivalently, $-A$ is a positive-definite matrix and every eigenvalue of $A$ is negative.

Two vectors are nonorthogonal when their inner product is nonzero. In the power method, the starting vector must be nonorthogonal to a dominant eigenvector.

For prime $p$, every order-$p$ element of $G{L}_2({\mathbb{F}}_p)$ is conjugate to the displayed nonidentity unipotent matrix. Its cyclic subgroup has vector group orbits of sizes one or $p$; counting all $p^2$ vectors implies at least $p$ fixed vectors. Choose a nonzero fixed vector as first basis vector. The matrix becomes triangular with first diagonal entry one; its other diagonal entry satisfies $d^p=1$ and hence $d=1$ in the prime finite field. The off-diagonal entry is nonzero and can be rescaled to one. All matrices involved have invertible changes of basis.