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An affine combination is a weighted point $P={\sum }_{i=0}^k{\lambda }_i{P}_i$ with ${\sum }_{i=0}^k{\lambda }_i=1$. The coefficients can be negative. The sum is meaningful in an affine space without selecting a preferred origin: relative to any point $O$, it means $P=O+{\sum }_i{\lambda }_i(P_i-O)$. Affine maps preserve these weighted points. ^[ref]

For an original example in the Euclidean plane, take $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$. Then $P=\frac{1}{4}A+\frac{1}{2}B+\frac{1}{4}C=(1,\frac{1}{2})$ lies inside their triangle. Its coefficients are nonnegative, making it a convex combination. In contrast, $R=-\frac{1}{2}A+\frac{3}{2}B=(3,0)$ is beyond $B$: it is an affine combination but not a convex combination of these three points. Negative weights allow extrapolation, rather than only interpolation inside the convex hull.

For the affine map $T(x,y)=(2x+y+1,y-2)$, the same weights produce $T(P)=(\frac{7}{2},-\frac{3}{2})$ from $T(A),T(B),T(C)$, and produce $T(R)=(7,-2)$ from $T(A),T(B)$. Thus one can compute a weighted point before transforming the vertices or afterwards, obtaining exactly the same point.

This is useful whenever an object is built from weighted control points: its construction commutes with an affine map, so transforming the control points transforms the constructed object as well. The sum of the weights is the essential condition; no assumption about lengths or angles is needed.

An affine map combines a linear map with a translation. After choosing coordinate origins, it has the form $T:{\mathbb{R}}^p\to {\mathbb{R}}^q$, $T(x)=Mx+b$, where the $q\times p$ matrix $M$ is its linear part of an affine map and $b$ is a fixed vector. Equivalently, it preserves every affine combination. ^[ref]

An original planar example isIt stretches horizontally, applies a shear mapping, and then translates. The vertices $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$ become $A^{\prime }=(1,-2)$, $B^{\prime }=(5,-2)$ and $C^{\prime }=(3,0)$. A square grid becomes a grid of parallelograms; straight edges and their weighted points still fit together.

This map is not a linear map with respect to the displayed origins, because $T(0,0)=(1,-2)$ rather than zero. It also is not a Euclidean similarity: $AB$ has length $2$ and $A^{\prime }{B}^{\prime }$ has length $4$, while the right angle at $A$ becomes $45^{\circ }$. Its determinant is $2$, so it doubles planar area. These metric changes explain why affine geometry studies straightness, incidence and weighted points rather than treating every transformation as an isometry.

The linear part is an invertible matrix, giving the concise inverse $T^{-1}(u,v)=((u-v-3)/2,v+2)$. More generally an affine map is bijective exactly when its linear part of an affine map is bijective. Between equal finite-dimensional spaces, this means an invertible matrix; the translation cannot repair a loss of rank of a linear map.

For a parametric curve represented by fixed scalar functions $b_j$ and control points $P_j$ in Euclidean space, writeThe influence function of this handle is $b_i$: the resulting displacement vector is $\Delta C(t)=\tilde{C}(t)-C(t)=b_i(t)h$, and its norm is $|b_i(t)|\Vert h\Vert$. These are exact finite changes, not only a first-order estimate. Where $b_i$ is negative the displacement points opposite to $h$. Where $b_i(t)=0$, the curve is unchanged at that parameter.

For the cubic Bézier curve above, the Bernstein basis influence function is $b_{1,3}(t)=3t(1-t{)}^2$. Its derivative isIt increases up to $t=1/3$ and decreases thereafter on $[0,1]$. Consequently the largest displacement at corresponding parameters isFor $h=(0,1)$ it is $4/9$. This is a comparison at the same parameter. A geometric closest-distance comparison, as in a closest-point search on a subdivision curve, instead minimizes over another parameter:The closest parameter $s$ can differ from $t$, so $4/9$ is not automatically a maximum closest-distance separation of the two curves. Neither parameter is assumed to be arc length.

The support of a function $b_i$ is the closure of the parameter values where $b_i\ne 0$. It bounds where a nonzero handle move can act, but may include boundary points where the actual displacement is zero. Here the support of a function $b_{1,3}$ is all of $[0,1]$ relative to the parameter domain, while the endpoints are unaffected. In contrast, a B-spline of order $k$ has support within $[t_i,t_{i+k}]$ in its spline knot sequence, so its handle affects only that parameter range; the support and Gram bandwidth of B-splines describes the same locality in another application. Changing the spline knots or the basis along with the handle requires a different comparison.

When the basis functions form a partition of unity, ${\sum }_j{b}_j(t)=1$, each evaluation is an affine combination of the control points. Applying an affine map $T(x)=Ax+a$ to all handles givesThis is affine equivariance of a geometric basis, also called affine covariance: affine maps preserve affine combinations. In particular, translating every handle by $h$ translates the whole curve by $h$, whereas moving just one handle produces $b_i(t)h$. The Bernstein basis has this partition of unity property.^[ref] Nonnegative weights further make evaluations convex combinations, giving the convex hull bound illustrated by the control polygon; nonnegativity is not needed for affine covariance.

For a proof outline of the influence formula, subtract the two basis representations: every unchanged coefficient cancels, leaving $b_i(t)(P_i+h)-b_i(t)P_i=b_i(t)h$. The same subtraction with several moved handles gives $\Delta C(t)={\sum }_i{b}_i(t)h_i$. No claim about interpolation, closest-point search on a subdivision curve or endpoint motion follows until the actual basis values are examined.

Estimating values beyond the range of the supplied sample points. For two distinct points $P,Q$, the affine combination $(1-t)P+tQ$ lies beyond an endpoint of their line segment when $t<0$ or $t>1$. The construction specifies an extension of the chosen model; it does not show that an unknown underlying function follows that extension.

Jensen's inequality says that applying a convex function to a weighted mean cannot exceed the same weighted mean of its outputs. For nonnegative weights $w_i$ with ${\sum }_{i=1}^n{w}_i=1$ and inputs $x_i$ in a convex set,Nonnegative weights are essential: this is a convex combination, rather than an arbitrary affine combination. For two inputs it is precisely the defining inequality for a convex function; repeatedly grouping inputs gives the finite weighted version.

Take $f(x)=x^2$, weights $1/4$ and $3/4$, and inputs $1$ and $3$. Their weighted mean is $2.5$, so the two sides areThe Jensen gap is $0.75$. The inequality measures the effect of replacing distinct inputs by one representative input before applying a nonlinear function.

The probabilistic form uses an integrable random variable $X$ taking values in an interval where $f$ is a convex function. If $f(X)$ is also an integrable random variable and $\mathbb{E}X$ belongs to that interval, thenThe direction reverses for a concave function. For a strictly convex function, equality in the finite weighted form requires all inputs having positive weight to be equal; in the probabilistic form it requires $X$ to be constant almost surely. An affine function gives equality for any admissible distribution. ^[ref].

For the quadratic function $f(x)=x^2$ and a random variable with a finite second moment, the difference between the two sides is exactly its variance:Consider two equally likely values $2-d$ and $2+d$. Their expected value stays at $2$ as $d$ increases, but their variance is $d^2$. Thus the expected value of the square grows from $4$ to $4+d^2$. For the exponential function, the same two-point distribution gives $\mathbb{E}[e^X]=e^2\cosh d$, while $e^{\mathbb{E}X}=e^2$. Both are examples of Jensen's inequality, but their Jensen gaps depend differently on the spread of the inputs.

Another useful choice is the concave function $\log x$ on positive inputs. Its version of Jensen's inequality gives ${\sum }_i{w}_i\log x_i\le \log ({\sum }_i{w}_i{x}_i)$. Applying the increasing exponential function shows that the weighted geometric mean cannot exceed the weighted arithmetic mean. For the equally weighted inputs $1$ and $9$, these are $3$ and $5$. This recovers the arithmetic-geometric mean inequality and illustrates how choosing one function can generate an entire family of bounds.

A short proof idea explains the probabilistic form. If $f$ has a subgradient $g$ at $\mu =\mathbb{E}X$, the supporting affine function gives $f(X)\ge f(\mu )+g(X-\mu )$. Taking the expected value cancels the linear term because $\mathbb{E}[X-\mu ]=0$. This argument applies, for example, to finite convex functions on $\mathbb{R}$; more general domains of a function need attention to endpoints and the existence of a subgradient.

Johan Jensen's 1906 paper “Sur les fonctions convexes et les inégalités entre les valeurs moyennes” is an early source for the connection between convex functions and inequalities for means. ^[ref].

An invertible $4\times 4$ matrix $M$, modulo scalar multiples, acts linearly on homogeneous coordinates. In an affine chart its projective transformation has the formwhere the denominator must be nonzero for a finite image. The important nested groups are Euclidean motions, Euclidean similarities, invertible affine maps, and general projective transformations. Their defining preserved structures become progressively weaker.

A Euclidean motion has $x^{\prime }=Rx+b$ with $R^{T}R=I$. It preserves all distances and angles, hence lengths, areas and volumes. Proper motions have $\det R=1$ and preserve orientation as well; allowing reflections preserves unsigned measurements but can reverse signed ones. A Euclidean similarity has $x^{\prime }=sRx+b$ with $s>0$. It preserves angles and all ratios of lengths; lengths, areas and volumes scale by $s,s^2,s^3$. Thus rigid shape is preserved up to one overall scale.

An invertible affine map has $c=0$ and preserves the plane at infinity. It preserves incidence, parallelism, collinear ratios, affine combinations, and ratios of volumes. Absolute volume scales by $|\det A|$; the volume-preserving affine subgroup has $|\det A|=1$. Lengths and angles are not affine invariants. A general projective transformation preserves incidence, collinearity, concurrence, tangency, and the cross-ratio of four collinear points, but can move the plane at infinity. The cross-ratio, rather than a three-point affine ratio, survives arbitrary projective changes. For a homogeneous quadric hypersurface $X^{T}QX=0$, its transformed matrix is $M^{-T}Q{M}^{-1}$, so its rank and projective degeneracy are preserved.

For surface classes, rather than a fixed numerical radius or curvature, the consequences are:

For example, the affine stretch $(x,y,z)\mapsto (2x,y,z)$ changes a unit sphere to $x^{\prime 2}/4+y^{\prime 2}+z^{\prime 2}=1$ and a unit circular cylinder to $x^{\prime 2}/4+y^{\prime 2}=1$. A general conical surface written $P+tD(s)$ transforms affinely into $T(P)+tAD(s)$, so every generator still passes through a finite apex. A nonsingular elliptic paraboloid or hyperbolic paraboloid has quadratic part of rank two and a nonzero linear component in its null direction. An invertible affine map preserves that rank, inertia and nonzero null-direction coupling, so translation and linear coordinate changes again put it in elliptic or hyperbolic paraboloid form.

Under a genuinely projective change, these Euclidean classifications depend on which plane is designated as infinity. For the map $(x,y,z)\mapsto (x,y,z)/(1+\alpha z)$, a circular cylinder $x^2+y^2=1$ becomesa cone whose apex is finite. The inverse map sends that apex to infinity and produces the cylinder. Thus a cone is projectively invariant as a family of concurrent generators if an ideal apex is allowed, but finite-apex cones and cylinders are not separately invariant. Likewise a projective image of a sphere or either type of paraboloid remains a nonsingular quadric hypersurface, but need not retain its affine or metric type. The distinction between general cones and circular cones matters: circularity is a metric property and is not preserved by arbitrary affine changes.

A credibility estimate combines a risk's observed experience with collective or prior information, typically as $Z\overline{X}+(1-Z)m_0$. The credibility factor $Z\in [0,1]$ is the weight assigned to the individual experience: more reliable or more extensive data generally increases it. Bayesian credibility treats the risk parameter as random with a prior distribution and predicts future claims by averaging the conditional claim mean under its posterior distribution. Under squared-error loss this posterior predictive mean is optimal. It need not be an affine combination in every Bayesian model; the Poisson-gamma conjugacy calculation below yields that credibility form exactly, rather than as a best-linear approximation.

Use degree elevation of Bernstein coefficients. If $b_j$ are degree-$p$ Bézier curve controls, the identical curve has degree-$(p+1)$ controlsThis follows by multiplying each degree-$p$ Bernstein basis polynomial by $(1-t)+t$ and collecting the degree-$(p+1)$ terms, so the operation changes the representation, not the curve.

For the quadratic controls in part (a), the cubic controls areRaise these cubic Bézier curve controls once more. ThenThus the requested quartic Bézier curve controls, expressed in the original polygon, areEach expression is an affine combination whose coefficients sum to one. Endpoint positions and derivatives are preserved at each degree-raising step, as is every point of the span.

Take the four-point setthe vertices of a polygon of a rectangle whose side lengths are four and two. Three choices of structure give three different symmetry groups on these same points.

As an unstructured set, $R$ has all $4!=24$ permutations as symmetries, forming the symmetric group $S_4$. If Euclidean distance is part of the data, the allowed maps are only the identity, a half-turn rotation, and the reflections across the two coordinate axes. They form a Klein four-group of group order four. To see that nothing is missing, fix the image of one vertex of a polygon: its three other distances, $2$, $4$, and $\sqrt{20}$, are distinct, so they uniquely determine the images of the remaining vertices.

Preserving affine structure permits more. The affine mapcycles all four points of $R$. It sends a horizontal side of length four to a vertical side of length two, so it is not an isometry. Yet it preserves the rectangle's vertex set and all affine combinations among those points.

Let $A=\mathrm{diag}(2,1)$ and let $r(x,y)=(-y,x)$ be the square's quarter-turn rotation. ThenConjugating every square symmetry by $A$ gives eight affine maps preserving $R$. This group isomorphism makes the affine symmetry group a dihedral group, although some of its maps do not preserve the original Euclidean distances. There are no further affine symmetries: an affine map preserves the common midpoint of the diagonals, so it must preserve opposite pairs. The image of one vertex has four choices and the image of an adjacent vertex has two; those choices determine the map through three noncollinear points.

ThusAn arbitrary permutation need not be an affine map. For example, exchanging the two top vertices while fixing the two bottom vertices sends the common diagonal midpoint to two different points, which no single affine map can do. The phrase "the symmetry of these points" becomes unambiguous only after stating which structure matters.