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For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.

Two separated short-range scatterers are coupled because the outgoing wave from either centre supplies part of the regular incoming field at the other. Applying the Bethe-Peierls boundary condition at both centres gives the displayed two-by-two linear system, where $s_j$ is the incident plane-wave phase. matrix inversion sums repeated scattering between the centres. Near a small determinant, effective-range corrections may matter even if the nominal microscopic range is short.

For planar vectors $u=(u_x,u_y)$ and $v=(v_x,v_y)$, the scalar $u\times v=u_x{v}_y-u_y{v}_x$ is the determinant of their coordinate columns. It equals the perpendicular component of the three-dimensional cross product after embedding the plane in three dimensions. Its sign records orientation, and its absolute value is twice the area of the triangle spanned by $u,v$.

A real $2\times 2$ Jacobian matrix has both eigenvalues in the open left half-plane exactly when its trace is negative and its determinant is positive. At zero trace the test no longer determines nonlinear stability.

For a unit-trace two-dimensional reduced state of a bipartite pure state, determinant $D$ gives eigenvalues $(1\pm \sqrt{1-4D})/2$. Their binary entropy quantifies entanglement. Natural logarithms give a maximum $\ln 2$; base-two logarithms give one bit.

A unimodular matrix is a square integer matrix whose determinant is $1$ or $-1$. The decisive property is that its inverse matrix also has integer entries: multiplication by $U$ gives a bijection of the integer lattice ${\mathbb{Z}}^n$ onto itself. Thus an invertible linear map can preserve not only real coordinates but the requirement that every coordinate is an integer. These matrices form the integral general linear group.^[ref]

For an original two-dimensional example, takeIts determinant is one, and it maps $(a,b)$ to $(a+b,b)$. Conversely, the integer point $(x,y)$ comes from $(x-y,y)$. For instance, $(3,2)$ has new coordinates $(1,2)$ relative to the columns $(1,0)$ and $(1,1)$. No integer point disappears, although the basis vectors and their fundamental cell change shape.

This is useful whenever coordinates must remain integral. A unimodular change of lattice basis can give a shorter or more convenient description without changing the objects represented. Smith normal form and Hermite normal form use controlled integer operations to simplify integer matrices; ordinary Gaussian elimination may divide by numbers that destroy integrality. The geometric freedom is substantial: a sheared basis may look completely different while describing exactly the same infinite set of integer points.

A determinant of $-1$ is allowed too. Swapping two basis columns reverses orientation while preserving the integer lattice. The condition is therefore preservation of the complete integer structure, rather than a requirement to preserve orientation.

The group of invertible linear maps preserving a nondegenerate Hermitian form over a quadratic finite field is, in an orthonormal basis, the displayed matrix group. Here $q^2$ denotes the size of the field on which the matrices act; some authors denote this same group by $U_n(q)$. Imposing determinant one gives the special unitary subgroup, and subsequently quotienting its scalar center gives the projective special unitary group over a finite field.

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.

A lower triangular matrix with every diagonal entry equal to one. Its determinant is one, and its inverse is again unit lower triangular. Such matrices store the elimination multipliers in an LU decomposition.

A unit triangular matrix is a triangular matrix whose diagonal entries are all one. It is an invertible matrix with determinant one. In the convention used for LU decomposition, $L$ is a unit lower triangular matrix, so forward substitution needs no division by its diagonal entries.

A Hartree-Fock ansatz allowing different spatial orbitals for alpha and beta spin. This can lower the energy of a stretched bond, but the determinant need not be an eigenstate of total spin squared.

The nonzero-sum two-cycles of the cubic swap map exist for $a>3$. Their two-step characteristic polynomial has trace $T$ and determinant $D$ as displayed. Since $1-T+D=-2(a-3)(a+1)<0$, one periodic-orbit multiplier exceeds one, proving instability throughout. The other crosses minus one at $a=(1+\sqrt{29})/2$; the cycles are saddles before that threshold and repelling afterwards. At $a=3$ they meet the fixed branches in an unstable subcritical-flip branch.

An upper triangular matrix has zero entries below its main diagonal. Its determinant is the product of its diagonal entries, and an invertible upper triangular linear system is solved by backward substitution.

Write $S=\frac{1}{2}\varphi A\varphi$ and use the vacuum prescription $A\mapsto A+i0$. Completing the Gaussian path integral gives $Z[J]/Z[0]=\exp [-iJ{A}^{-1}J/2]$, hence $W=-J{A}^{-1}J/2$, $\phi =-A^{-1}J$ and $\Gamma =\phi A\phi /2=S[\phi ]$. The source-independent determinant cancels in vacuum normalization. Without that normalization, the equality holds up to a field-independent vacuum constant.

// Target: perturbative-quantum-field-theory.bigb

Evaluating arbitrary monic polynomials of successive degrees $0,\dots ,m-1$ gives the same determinant as evaluating the monomials of those degrees. The basis-change matrix is triangular with diagonal entries one, so its determinant is one. This permits an orthogonal polynomial basis to replace the monomials in eigenvalue densities.

For a nondegenerate metric tensor whose signature is fixed during the variation, the Jacobi formula gives $\delta \det g=(\det g)g^{ab}\delta g_{ab}$. Differentiating the square root of the absolute determinant proves the displayed identity. Expressed through the variation of inverse metric, the formula is $-\sqrt{|\det g|}{g}_{ab}\delta g^{ab}/2$. Absolute values accommodate either Riemannian or Lorentzian signature.

Variation of parameters replaces the constants in a complementary solution by functions to construct a particular solution.

For the inhomogeneous linear differential equation $y^{\prime \prime }+p{y}^{\prime }+qy=r$, assume $p,q,r$ are continuous functions on an interval $I$, and $y_1,y_2$ are linearly independent homogeneous solutions there. With $W=W(y_1,y_2)$ and $x_0\in I$, variation of parameters constructs the particular solutionThe Wronskian has no zeros on $I$, so these quotients are defined. For $A{y}^{\prime \prime }+B{y}^{\prime }+Cy=F$, normalize first: $r=F/A$, not $F$.^[ref]

To derive the formula, write $y_p=u_1{y}_1+u_2{y}_2$ and impose $u_1^{\prime }{y}_1+u_2^{\prime }{y}_2=0$. This auxiliary choice removes extra terms from the first derivative. Substitution into the inhomogeneous linear differential equation leaves $u_1^{\prime }{y}_1^{\prime }+u_2^{\prime }{y}_2^{\prime }=r$. The two linear equations for $u_1^{\prime },u_2^{\prime }$ have determinant $W$, and Cramer's rule givesIntegrating from $x_0$ makes $y_p(x_0)=y_p^{\prime }(x_0)=0$. Arbitrary integration constants contribute only homogeneous solutions; adding $c_1{y}_1+c_2{y}_2$ supplies the general solution.

For a worked example, use the same homogeneous solutions $e^{-x},e^{2x}$ and the forcing term $r=6e^{3x}$ in $y^{\prime \prime }-y^{\prime }-2y=6e^{3x}$. With $x_0=0$ and $W=3e^x$,Consequently the particular solution selected by zero initial conditions isIts value and first derivative vanish at zero. Direct substitution confirms the forcing term: the differential operator $L[y]=y^{\prime \prime }-y^{\prime }-2y$ annihilates the last two terms and maps $e^{3x}$ to $4e^{3x}$. The two integrals therefore build a response from known homogeneous solutions, even when the forcing term has no convenient elementary guess.

The wedge product combines a degree-$p$ differential form and a degree-$q$ differential form into one of degree $p+q$. It is associative and graded-commutative:In particular, $dx\wedge dy=-dy\wedge dx$ and $dx\wedge dx=0$. For planar vectors $v,w$,This determinant is the signed area of their parallelogram. Exchanging the edges reverses the sign; parallel edges give zero. Thus antisymmetry records the geometry of oriented area rather than adding an arbitrary algebraic convention. The wedge product of differential forms requires no Riemannian metric to define it. It obeys the graded Leibniz rule under the exterior derivative.

Take labels $E=\{a,b,c\}$ and vectors in the real vector space ${\mathbb{R}}^2$:Here is the complete independence check for all eight subsets. The empty set is independent. Each of $\{a\},\{b\},\{c\}$ is independent because its vector is nonzero. Every pair is independent: the determinants of the two-column matrices $[v_a{v}_b]$, $[v_a{v}_c]$, and $[v_b{v}_c]$ are respectively $1$, $1$, and $-1$. The triple is dependent because $v_c-v_a-v_b=0$. ThereforeDeleting elements from any listed subset stays in this list, verifying heredity. Every exchange comparison has sizes $(0,1)$, $(0,2)$, or $(1,2)$. From the empty set, any element of the larger matroid independent set works: this covers three singleton targets and three pair targets. From any singleton to any pair, every element of the pair outside the singleton works, since every resulting pair is independent. This covers all nine singleton-pair comparisons, including disjoint ones. There are no other comparisons: all fifteen ordered size-increasing pairs satisfy exchange.

The three matroid bases are $\{a,b\}$, $\{a,c\}$, and $\{b,c\}$, all of cardinality two, with weights $13$, $14$, and $11$. The matroid greedy algorithm considers $a$, then $c$, then $b$: it admits $a$, admits $c$ because their determinant is nonzero, and rejects $b$ because the triple is dependent. The resulting matroid basis $\{a,c\}$ has weight $8+6=14$, the largest of the three totals. Every singleton can grow, so none is maximal; the three pairs are both maximal and maximum in cardinality.

For a nonnegative weight $w$ and monic polynomials $p_j$ of degree $j$, this determinant is ${\prod }_i\sqrt{w(x_i)}{\prod }_{i<j}(x_j-x_i)$. Its square is the product of the weights times the squared Vandermonde determinant. For normalized weighted polynomials it differs by the product of their norm factors.