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For a numerical evolution, numerical stability means that perturbations remain controlled over a fixed finite interval by a bound independent of sufficiently small discretization parameters. With perturbations ${\delta }_n$ to starting data and added step defects $d_n$, a typical estimate iswhere $C_T$ is independent of $h$. For a linear homogeneous recurrence, a uniform bound on powers of its update matrix supplies this control. von Neumann stability analysis and the energy method provide complementary tools.

Finite-interval numerical stability differs from absolute stability at a fixed step size. Applying the Euler method to $y^{\prime }=-2y$ multiplies a perturbation by $1-2h$ each step. At $h=0.2$ this is $0.6$, so perturbations shrink; at $h=1.2$ it is $-1.4$, so they alternate and grow. Nevertheless the Euler method converges as $h\to 0$ for this problem: failure at a coarse step does not contradict the convergence theorem. Absolute stability asks whether a selected step preserves boundedness of the Dahlquist test equation over arbitrarily many steps. MIT's analysis uses strict decay, $|1+h\lambda |<1$, rather than the boundary-inclusive boundedness convention.^[ref]

The subgroup generated by elementary transvections in finite matrix sizes, with identity stabilization. A third distinct index expresses each transvection as a commutator; stable block identities show that this subgroup is normal and that its quotient in the stable general linear group is abelian. Thus it is precisely the commutator subgroup.

The union of finite invertible matrices under $g\mapsto \mathrm{diag}(g,1)$. Every element and relation occurs in a finite matrix size; adding identity blocks permits comparisons between different sizes without changing the stable class.

A stage-structured population model tracks distinct life stages, such as juveniles and adults, with separate reproduction, survival and transition terms. A matrix update describes the linear case; density-dependent reproduction gives a nonlinear discrete dynamical system.

Choose $r$ of the $n$ basis vectors and require their images in the quotient to be a basis. Normalize those columns to the identity; the remaining $r(n-r)$ entries are arbitrary affine coordinates. Between column sets $I,J$, the overlap is the principal open subset where $\det A_J$ is a unit, and the new normalized matrix is $A_J^{-1}A$. These matrix transformations satisfy the cocycle identity. Locally every surjective matrix has an invertible maximal minor, so the charts cover all rank-$r$ locally free quotients. The normalized matrices glue the universal quotient and give the natural identification with its functor of points.

// Target: ringed-space.bigb

The stiffness matrix represents a variational bilinear form $a$ in a finite element basis. A symmetric coercive bilinear form gives a positive-definite matrix. Local support of the piecewise-linear hat functions produces a sparse matrix, allowing efficient numerical solution.

A row-stochastic matrix has nonnegative entries and each row sums to one. For a time-homogeneous Discrete-time Markov chain, its transition matrix has entriesIf a state has zero probability at time $n$, this row is specified by the model's Markov kernel, rather than recovered by conditioning on that zero-probability event. Rows index departure states and columns arrival states. Writing the current probability distribution as a row vector ${\mu }_n$ givesThe row-sum condition ensures that matrix multiplication preserves total probability. Under a column convention the evolution uses the transpose; the convention must accompany the formula.

For the sunny/rainy transition matrix, direct matrix multiplication givesIn particular, $(P^2{)}_{SS}=0.8\times 0.8+0.2\times 0.4=0.72$. The two terms sum the disjoint paths $S\to S\to S$ and $S\to R\to S$. Requiring both subsequent days to be sunny has probability $0.64$; requiring only the second day to be sunny permits the extra path through rain.

Generally $(P^n{)}_{ij}$ is the $n$-step transition probability. The identity $P^{m+n}=P^m{P}^n$ is the discrete-time Chapman-Kolmogorov equation: sum over the intermediate state. This makes matrix powers useful for forecasting many steps without enumerating every path individually.

For a proper cone $K$, suppose a point satisfies $Ax-b\in \mathrm{int}K$ and $p_{\ast }$ is finite. This generalized Slater condition gives strong duality and attainment of the conic dual problem's optimum. It does not by itself assert primal attainment.^[ref]

The two worked examples have strict primal points: $(u,v)=(2,2)$ has slack $(2,2,1)$ in the interior of the nonnegative orthant, and $(t,z)=(2,0)$ has slack $(2,1,0)$ in the interior of the second-order cone. Both optima were already certified by explicit feasible pairs. Slater's condition explains why a matching optimal dual variable can be expected under these hypotheses; checking a supplied zero-gap pair needs only weak duality.

Changing cone geometry preserves the bound mechanism. For a positive semidefinite cone, the dual cone pairing is the Frobenius inner product of two matrices. This connects the same certificate idea to semidefinite programming, while the central path uses strictly interior primal and dual points to approach vanishing gap.

In 1995, Lieven Vandenberghe and Stephen Boyd presented a primal-dual interior-point method for convex optimization with matrix inequalities. Their algorithmic study included exploiting problem structure and allowing incomplete conjugate gradient method solves within an iteration. This illustrates how the primal-dual viewpoint guides computation as well as checking a final answer.^[ref]

A first-order system $U_t+A(U)U_x=C(U)$ is strictly hyperbolic at a state when $A$ has distinct real eigenvalues. These separate its characteristic families and permit independent compatibility equations from left eigenvectors. At a repeated eigenvalue strict hyperbolicity fails; if the matrix is also defective, even hyperbolicity in the complete-eigenvector sense fails. For $A=\left(\begin{array}{cc}u& 2\\ 2v& u\end{array}\right)$ the eigenvalues $u\pm 2\sqrt{v}$ are distinct precisely when $v>0$.

The four-by-four unitary matrices of determinant one form this special unitary group. Its two-index antisymmetric representation has complex dimension six and the real structure of the SU4 exterior square realizes it on a six-dimensional real space.

A submatrix is obtained from a matrix by retaining selected rows and selected columns. A principal submatrix of a square matrix retains the same index set for its rows and columns.

If idempotent matrices $P_1,\dots ,P_k$ satisfy $P_i{P}_j=0$ whenever $i\ne j$, then $({\sum }_i{P}_i{)}^2={\sum }_{i,j}{P}_i{P}_j={\sum }_i{P}_i$. Thus the sum is idempotent. For a real orthonormal set $q_i$, the matrices $P_i=q_i{q}_i^T$ meet this condition. Their sum is the orthogonal projector onto their span; the symmetric matrix property follows by transposing the sum. Arbitrary sums of idempotent matrices need not be idempotent.

A $(j,k)$ tensor has one fundamental representation factor for every upper index and a dual factor for every lower index. Under $A\in SU(n)$, upper factors transform by $A$ and lower factors by $A^{-1}$; unitarity identifies the latter with conjugate fundamental matrices. Thus complex conjugation exchanges $(j,k)$ and $(k,j)$. Index permutations and contraction with an invariant tensor commute with the action, producing invariant subspaces and decomposition maps.

Let $S$ be a real symmetric positive definite matrix, $C>0$ and $q\ne 0$. The Cauchy-Schwarz inequality in the $S$-weighted inner product gives $q^{T}x=(S^{-1}q{)}^{T}Sx\le \sqrt{q^T{S}^{-1}q}\sqrt{x^{T}Sx}$. Equality is attained by a positive multiple of $S^{-1}q$. Thus the displayed maximum holds. For a translated ellipsoid, add $q^{T}a$ for its centre $a$. This computes a coordinate extremum without constructing a rotation matrix.

For a heterogeneous SIS model with common recovery rate and a positive endemic equilibrium $I_{\ast }$, force balance gives $\mathrm{diag}(s_{\ast })K{I}_{\ast }=I_{\ast }$. When the transmission matrix is irreducible and nonnegative, the positive equilibrium vector and the Perron–Frobenius theorem imply the displayed spectral radius identity. It replaces the homogeneous scalar relation $s_{\ast }=1/R_0$: different recipient types generally have different susceptible fractions.

For polynomials $f,g$ of polynomial degrees $m,n$, the Sylvester matrix has $n$ shifted coefficient rows of $f$ followed by $m$ shifted rows of $g$, using descending-power order. It is a square matrix of size $m+n$. Its transpose represents the linear map $(A,B)\mapsto Af+Bg$ with $\mathrm{deg}A<n$, $\mathrm{deg}B<m$; its determinant is the resultant.

For an original parameter example, choose $f(x)=x^2-1$ and $g_t(x)=x-t$. ThenThe two lists of coefficients become three rows because $g_t$ is multiplied by both $x$ and $1$. Thus the common-root parameter values are $t=-1$ and $t=1$. At $t=1/2$, the resultant is $-3/4$, so the two polynomials have no common root.

One can verify the certificate without any determinant expansion:For $t\ne \pm 1$, division by $t^2-1$ gives a Bézout's identity for polynomials. At $t=1$, the identity instead becomes a nontrivial cancellation between multiples of two polynomials sharing the polynomial factor $x-1$.

The general proof needs only a short linear algebra outline. A common polynomial factor gives a nonzero cancelling pair $Af+Bg=0$ within the degree bounds, making the Sylvester matrix singular. Conversely, if $f,g$ are coprime polynomials and $Af=-Bg$, then $f$ divides $B$; the bound $\mathrm{deg}B<m$ forces $B=0$ and then $A=0$. Thus the linear map has zero kernel of a linear map exactly in the coprime case. Over an algebraic closure, a nonconstant common polynomial factor is equivalent to a common root.

In 1840, James Joseph Sylvester described forming the shifted coefficient rows and taking a signed sum of products selected by permutations. In modern notation, this is the Sylvester matrix and its determinant. The construction turns the nonlinear common-root of a polynomial question into linear algebra. His paper's word derivatives concerns elimination, rather than the formal derivative used below.^[ref]

A Hermitian matrix $A$ is a positive-definite matrix exactly when its leading principal minorsare all positive. The leading principal minor uses the first $k$ rows and the first $k$ columns in the same ordering. For the two-by-two Cholesky decomposition example, ${\Delta }_1=4$ and ${\Delta }_2=4\cdot 5-2^2=16$, so the criterion proves positive definiteness without calculating eigenvalues.

The successive pivot elements before taking square roots are ${\Delta }_k/{\Delta }_{k-1}$, with ${\Delta }_0=1$. Here they are $4$ and $16/4=4$. In general, a leading triangular matrix factor has determinant equal to the product of its diagonal entries, soConversely, positive leading principal minors make each successive Schur complement pivot element positive, allowing the Cholesky decomposition to be constructed. The identity $u^{T}Au=\Vert L^{T}u{\Vert }_2^2$ then proves positive definiteness; for complex matrices, replace the transposes by conjugate transposes. This gives a short inductive proof of the criterion.

For $B=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$, the first minor is positive but $\det B=-3$. The second Cholesky radicand would be $1-2^2=-3$, and $v=(1,-1{)}^T$ directly gives $v^{T}Bv=-2<0$. Thus positive diagonal entries are insufficient. Replacing “positive” by “nonnegative” in just the leading-minor test is also insufficient for positive semidefiniteness: $\mathrm{diag}(0,-1)$ has leading minors $0,0$, but its quadratic form is negative on $(0,1{)}^T$.

Over a field whose characteristic is not two, a bilinear form has a unique decomposition into a symmetric bilinear form $S$ and an alternating bilinear form $K$:Adding and subtracting the equations $B(v,w)=S(v,w)+K(v,w)$ and $B(w,v)=S(v,w)-K(v,w)$ both proves uniqueness and gives these formulas. In matrix notation, $S$ has matrix $(A+A^T)/2$ and $K$ has skew-symmetric matrix $(A-A^T)/2$.^[ref]

The diagonal values $q(v)=B(v,v)$ alone do not determine a general bilinear form, because $K(v,v)=0$ for every vector. For the matrix of a bilinear form in the example above,Thus two different bilinear forms, represented by $A$ and its symmetric matrix part, have exactly the same quadratic form. To recover a bilinear form from its diagonal, one needs extra information such as the hypothesis that it is a symmetric bilinear form.

A bilinear form $B$ is symmetric when $B(v,w)=B(w,v)$ for every pair of vectors $v,w$. Equivalently, its matrix of a bilinear form in any basis is a symmetric matrix, since $A_{ij}=B(e_i,e_j)=B(e_j,e_i)=A_{ji}$. Symmetry makes the relation $B(v,w)=0$ reciprocal, giving the bilinear orthogonal complement a geometric interpretation.

Over the real numbers, set $q(v)=B(v,v)$. The polarization identityrecovers the symmetric bilinear form from its quadratic form: expand $q(v+w)$ and use symmetry to combine the two cross terms. For $q(x,y)=x^2-y^2$, take $v=(2,1)$ and $w=(1,3)$. Then $q(v+w)=-7$, $q(v)=3$, and $q(w)=-8$, so $B(v,w)=(-7-3+8)/2=-1=2\cdot 1-1\cdot 3$. The same formula works whenever the characteristic is not two.^[ref]

Three real diagonal matrices show how much variety symmetry allows. The identity matrix $\mathrm{diag}(1,1)$ gives the positive-definite bilinear form $q=x^2+y^2$. The matrix $\mathrm{diag}(1,-1)$ gives $q=x^2-y^2$ and is nondegenerate despite the isotropic vector $(1,1)$. The matrix $\mathrm{diag}(1,0)$ gives $q=x^2$ and is a degenerate bilinear form with radical of a bilinear form $\mathrm{span}\{(0,1)\}$. Their level sets make the distinction visible: a circle, two branches of a hyperbola, and two parallel lines for $q=1$.

The inertia of a bilinear form records the numbers of positive, negative, and zero directions after matrix congruence diagonalization. By Sylvester's law of inertia, these counts survive change of basis; the three examples have inertia of a bilinear form $(2,0,0)$, $(1,1,0)$, and $(1,0,1)$. Historically, Sylvester stated this invariance in an 1852 paper on reducing real quadratic forms to sums of positive and negative squares. Ghys and Ranicki reproduce his statement and distinguish it from his subsequent 1853 proof.^[ref]

Congruence $X\mapsto AX{A}^T$ on real symmetric two-by-two matrices preserves the quadratic form $-\det X$ of signature $(2,1)$ when $\det A=1$. Its kernel is $\{\pm I\}$ and its derivative is injective: $aX+X{a}^T=0$ for every symmetric $X$ implies $a=0$. Both Lie groups have dimension three, so the connected image is the full identity component. Also $A^{T}JA=(\det A)J$ identifies $Sp(2,\mathbb{R})$ with $SL(2,\mathbb{R})$.