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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.

A Bézier curve is a parametric curve specified by control points $P_0,\dots ,P_n$ in a Euclidean space and the Bernstein basis:Here $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$ is a binomial coefficient. Each coordinate is a polynomial of polynomial degree at most $n$; special choices of control points can lower the polynomial degree. The ordered control points form the control polygon. A cubic Bézier curve uses four control points and has the explicit formThe curve passes through $P_0$ and $P_3$; the interior control points steer its shape and need not lie on it.^[ref]

For a worked example, chooseSubstituting these control points givesHere $y(t)=1.5-6(t-1/2{)}^2\le 1.5$, so neither interior control point, both at height $2$, lies on the curve. De Casteljau's algorithm computes the same point using only successive linear interpolations.

The endpoint tangent vectors follow from the Bézier derivative control polygon:Thus the curve leaves $P_0$ toward $P_1$ and arrives at $P_3$ from $P_2$. These nonzero derivatives determine the endpoint tangent lines; if an endpoint derivative vanishes, that formula alone does not determine its tangent direction. Endpoint handles are useful when arranging a parametric first-derivative join of Bézier curves.

The parameter $t$ is usually not arc length. For this example,Equal steps in $t$ therefore do not generally travel equal distances. Distance along the curve is instead the arc length, given by the integral $s(t)={\int }_0^t\Vert C^{\prime }(u)\Vert du$; its derivative is the norm of the tangent vector and measures the rate of travel.

Two geometric facts make Bézier curves especially convenient for design. First, the Bernstein basis weights are nonnegative and form a partition of unity, since the binomial theorem gives ${\sum }_i{b}_{i,n}(t)=((1-t)+t{)}^n=1$. Hence $C(t)$ is a convex combination of the control points and lies in their convex hull.^[ref] The convex hull is the smallest convex set containing the control points, not their axis-aligned bounding box. Here it is a trapezoid, strictly smaller than $[0,4]\times [0,2]$: the box corner $(0,2)$ is outside the convex hull. The curve need not fill or touch every edge of either enclosure. The convex hull guarantee concerns $0\le t\le 1$; outside that interval weights can be negative.

Second, applying an affine map to every control point gives exactly the same result as applying it to the entire curve. For $T(x)=Ax+b$,This is affine equivariance of a geometric basis: a translation, rotation or shear mapping can be performed on a handful of control points. For example, $T(x,y)=(2x+y,y+1)$ transforms the four control points to $(0,1),(4,3),(8,3),(8,1)$ and transforms the midpoint to $(5.5,2.5)$, also obtained from the same weights $(1,3,3,1)/8$.

These editable handles explain the usefulness of Bézier curves in computer aided geometric design, vector graphics paths and outline fonts: a small collection of points controls an entire smooth stroke, and resizing the design preserves its mathematical shape. Cambridge's computer-graphics notes describe their use in graphic design, typography and PostScript outlines. The industrial context includes Pierre Bézier's work at Renault and Paul de Casteljau's work at Citroën.^[ref]

// Target: analysis.bigb

Chaikin subdivision generates a rounded subdivision curve by repeatedly cutting the corners off a control polygon. Each edge contributes two new control points, one a quarter of the way along the edge and the other three quarters of the way along it. Connecting these new control points in edge order replaces the old corners by shorter line segments. This makes Chaikin subdivision a useful way to draw a smooth outline from a small, editable set of control points, without evaluating B-spline basis functions at each drawing step. Joy presents both the corner-cutting rule and its closed version.^[ref]

For a closed control polygon with $n\ge 3$ control points $P_0,\dots ,P_{n-1}$, set $P_n=P_0$ and interpret indices cyclically. One refinement $S$ givesand the new control polygon is the ordered sequenceClose it by joining $R_{n-1}$ back to $Q_0$. Use the old control points for every average in the same refinement; updating them in place would define a different algorithm. One refinement produces $2n$ stored control points, and $k$ refinements produce $2^{k}n$, counting entries even if some coordinates coincide. Open control polygons need an explicit endpoint convention: applying only the interior edge rule discards the original endpoints, whereas retaining endpoints changes the boundary rule. The closed analysis below uses cyclic indices throughout.

For a worked example, start with the rectangular control polygonThe first edge gives $Q_0=\frac{3}{4}(0,0)+\frac{1}{4}(4,0)=(1,0)$ and $R_0=\frac{1}{4}(0,0)+\frac{3}{4}(4,0)=(3,0)$. Applying Chaikin subdivision to all four edges, including the closing edge, gives exactly eight control points in this order:Thus the rectangle becomes an octagonal control polygon. Two refinements have $16$ control points, and five have $128$. The illustration shows finite control polygons at stages $0,1,2,5$; the last is already visually rounded but still consists of line segments.

The convex hull property follows directly from the weights. Each new control point is a convex combination of the endpoints of one old edge, because the weights are nonnegative and sum to one. Consequently, if $H_k$ is the convex hull of the stage-$k$ control points, then $H_{k+1}\subseteq H_k$. Every later line segment, and every limit point, stays in the original convex hull. This gives a useful geometric enclosure. It also explains why Chaikin subdivision is approximating rather than an interpolation rule: the rectangle's four original corners are outside the first octagon's convex hull, so the limiting curve cannot pass through them. A refinement control polygon generally lies off the limiting curve as well.

For the cyclic uniform rule, the closed limit is the periodic quadratic uniform B-spline curve determined by the original control points. Uniform refers to equal spacing in the spline knot sequence, not equal physical lengths of edges of the control polygon. An infinite uniform control sequence has the corresponding local description; arbitrary open endpoint conventions need separate treatment. Joy derives the quarter-point refinement by splitting quadratic uniform B-spline spans.^[ref]

Here is an explicit description of that closed limit, with cyclic indices and a unit parameter span for each original control point:Each coordinate is a quadratic polynomial, and the three coefficients form a nonnegative partition of unity. The span runs from $(P_{i-1}+P_i)/2$ to $(P_i+P_{i+1})/2$. Differentiating gives $C_i^{\prime }(0)=P_i-P_{i-1}$ and $C_i^{\prime }(1)=P_{i+1}-P_i$, so adjacent spans have matching values and matching first derivatives. Their second derivatives generally differ. Thus the parameterized curve is $C^1$, generally not $C^2$; degenerate control points can still make its tangent vector vanish. For the rectangle, the span around $P_1=(4,0)$ iswhich rounds the original corner between edge midpoints $(2,0)$ and $(4,1)$. For equally spaced abscissae, the centered Chaikin basis function gives the scalar graph version of this quadratic B-spline representation and has support of width three original grid intervals.

Another useful property is affine equivariance of a geometric basis: refining and applying an affine map commute. For $T(x)=Ax+b$,and the same calculation holds for $R_i$. The sum-to-one weights account for the translation. Therefore $T(S(P))=S(T(P))$ at every refinement, and also for the limit. A designer can apply a shear mapping, translation, or other affine map to the original control polygon and obtain exactly the transformed outline.

For example, compose the shear mapping $(x,y)\mapsto (x+y/2,y)$ with the translation $(1,-1)$ to obtain $T(x,y)=(x+y/2+1,y-1)$. The rectangle's control points become $(1,-1),(5,-1),(6,1),(2,1)$. On its right edge, cutting first gives $Q_1=(4,1/2)$ and then $T(Q_1)=(21/4,-1/2)$. Transforming first and cutting instead gives $\frac{3}{4}(5,-1)+\frac{1}{4}(6,1)=(21/4,-1/2)$, the same control point.

Historically, George M. Chaikin published “An algorithm for high-speed curve generation” in Computer Graphics and Image Processing 3(4), December 1974, pages 346–349. Its emphasis was practical curve generation through a recursive algorithm using simple arithmetic, with applications including machine control and three-dimensional curves. The lasting appeal of Chaikin subdivision is how an elementary local average creates an entire smooth B-spline curve.^[ref]

A matrix is defective when an eigenvalue has geometric multiplicity smaller than its algebraic multiplicity. Over a field in which the characteristic polynomial splits, this is precisely the obstruction to an eigenbasis: at least one Jordan block has size greater than one, and generalized eigenvectors are needed to complete a basis. The shear mapping in diagonalizable matrix is the simplest example.

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

All the transformations below fix the origin. A planar rotation preserves lengths and angles and turns every vector through one common oriented angle. Its rotation matrix isAn orthogonal reflection in a line fixes that line and reverses its perpendicular direction; reflection in the horizontal axis has reflection matrix $\mathrm{diag}(1,-1)$. A uniform dilation multiplies every vector by the same positive scale $s$, represented by $sI$; for example $2I$ doubles lengths. A shear mapping fixes one line pointwise and shifts vectors parallel to that line by an amount proportional to their transverse coordinate. A horizontal shear mapping has matrixwith $S_1$ a nontrivial example.

The first given matrix is the clockwise planar rotation through $\pi /4$, namely $A=R_{-\pi /4}$. Direct multiplication givesThus $C$ is a horizontal shear mapping of strength $2$. Because $A^{-1}=R_{\pi /4}$,So $B$ is the composition of that shear mapping followed by an anticlockwise planar rotation through $\pi /4$; order matters. It is not individually one of the four elementary types under these definitions. In particular, its determinant is $1$, but it is not an orthogonal matrix, excluding both a planar rotation and an orthogonal reflection. It is not a scalar matrix, excluding a uniform dilation; its eigenvalues $\sqrt{2}+1$ and $\sqrt{2}-1$ exclude a pure shear mapping, whose eigenvalues are both $1$. As a symmetric positive-definite matrix, $B$ can also be described as stretching two perpendicular principal directions by these reciprocal factors, preserving area.

Expand $(I+B{)}^2$ in components:If $B^2=0$, all powers $B^k$ for $k\ge 2$ vanish. The matrix binomial theorem, valid here because $I$ commutes with $B$, therefore yieldsThese powers of the identity plus a square-zero matrix can also be proved by induction: $(I+nB)(I+B)=I+(n+1)B$. For the displayed example, $B^2=0$ directly, andThis is the repeated function composition of a shear mapping. The argument gives every nonnegative integer $n$; since $(I+B{)}^{-1}=I-B$, the same formula also holds for negative integers.

For the specified coordinate vectors,Choose the orthogonal matrix representing a quarter-turn,Then $S^{T}S=I$ and direct matrix multiplication gives $S^{T}LS=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=M$. Thus these two shear mappings are orthogonally similar.

Rotate the specified line onto the horizontal axis, stretch horizontally by two while leaving the orthogonal component unchanged, then rotate back. With $\theta =\pi /3$ and counterclockwise rotation matrix $R(\theta )$, the requested product isThe rightmost rotation acts first. In particular, $R(\theta )=\left(\begin{array}{cc}1/2& -\sqrt{3}/2\\ \sqrt{3}/2& 1/2\end{array}\right)$ and $R(-\theta )=R(\theta {)}^T$. For the unit direction $u=(1/2,\sqrt{3}/2{)}^T$, the same transformation is $T=I+u{u}^T$. It sends $u$ to $2u$ and fixes every orthogonal vector, verifying that the product is a directional stretch rather than a shear mapping or scaling in every direction. The constructed second panel in the geometric figure in Question 5 illustrates this separation.

The binomial theorem for the commuting matrices $I,N$ proves the formula for nonnegative integers: every term containing $N^2$ vanishes. Since $(I+N)(I-N)=I$, the matrix inverse is $I-N$; applying the same argument to its powers gives the negative integers as well. The same identity shows that $I+N$ is invertible even when the nilpotent matrix $N$ is nonzero. In dimension two, the example $N=\left(\begin{array}{cc}0& p\\ 0& 0\end{array}\right)$ describes repeated shear mappings, whose displacement parameters add under composition.

Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.

For any fixed real $n$-by-$n$ matrix $A$, as the real parameter $t$ tends to zero,The big O notation means the remainder is bounded in absolute value by $C{t}^2$ for some constant $C$ near zero. The determinant stays positive for sufficiently small $t$, so its oriented volume factor is also its ordinary volume factor. Thus the trace is the first-order fractional volume change, measured from the identity matrix.

A proof follows directly from multilinearity of the determinant. Each column of $I+tA$ is $e_j+tA{e}_j$. The constant term chooses all the standard basis columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $A{e}_j$; expanding that vector in the standard basis leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\mathrm{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the derivative of the determinant at the identity matrix.

This coefficient has a direct visual meaning. At the point $e_j$, the linear map $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional area or volume change; the off-diagonal entries initially change shape rather than contributing to this first-order total.

For the constant-coefficient linear system of differential equations $x^{\prime }=Ax$, the exact linear flow is $x(t)=e^{tA}x(0)$, using the matrix exponential. Its volume factor isThis is the matrix exponential determinant identity. One can derive it without diagonalization: set $E(t)=e^{tA}$, whose matrix inverse is $E(-t)$, and use $E^{\prime }=AE$ in Jacobi's formula to get $V^{\prime }=\det E\mathrm{tr}(E^{-1}AE)=(\mathrm{tr}A)V$. The scalar differential equation with initial value one gives the displayed exponential function. Thus positive trace causes volume expansion, negative trace causes contraction, and zero trace preserves volume, even when individual directions behave differently.

In two dimensions, compare four generators at $t=0.4$:The first and third diagonal matrices expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged area. For the last nilpotent matrix, $A^2=0$, so its matrix exponential is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a shear mapping. Zero trace allows substantial shape change and does not mean the flow map is stationary.

The first-order statement and the exact flow map should be distinguished. For $A=\mathrm{diag}(1,-1)$, $\det (I+tA)=1-t^2$, whereas $\det (e^{tA})=1$ for all real $t$. Zero trace eliminates the first-order change in the approximate step, and it preserves volume exactly in the matrix exponential flow. The examples use an abstract parameter $t$ and coordinate vectors; no physical units are assumed.

Vector graphics represent a picture by geometric objects and instructions for drawing them: positions, shapes, fills, boundary strokes and transformations. A circle can be stored as a centre and radius; a rectangle as a corner, width and height; a polygon as an ordered list of vertices. These geometric primitives describe whole objects rather than assigning a colour to every pixel. SVG provides elements for such basic shapes and for paths assembled from line segments and curves.^[ref]

The useful separation is between the stored geometry and the finite output. A circle of radius 2 remains the same geometric description whether rendered across twenty or two thousand output pixels. This is the sense in which vector graphics are resolution independent: increasing the output grid can evaluate the original geometry again. The resulting raster image still has finite spatial resolution, and numerical coordinates have finite precision. Enlarging a previously rendered raster image through interpolation instead works with its existing samples. Neither operation guarantees that arbitrarily thin details remain visible at every size.

An SVG path uses a current point and drawing commands. For example, M 1 4 L 5 4 L 3 1 Z moves to $(1,4)$, draws two line segments, then closes back to the starting point, giving a triangle. Uppercase commands use absolute coordinates; the corresponding lowercase commands use offsets from the current point. Thus M 1 4 l 4 0 l -2 -3 z describes the same triangle. The coordinates belong to the current user Cartesian coordinate system, not necessarily to output pixels.^[ref]

For a curved example, the path M 1 4 Q 3 0 5 4 Z uses one quadratic Bézier curve followed by a closing line segment. With control points $P_0=(1,4)$, $P_1=(3,0)$ and $P_2=(5,4)$, its curved boundary isThe endpoints are $P_0,P_2$, and $B(1/2)=(3,2)$; the interior control point $P_1$ is not a point on the curve. A fill paints the enclosed region, while a stroke paints a band around its boundary with a chosen width. One can use either or both: geometry and paint instructions are distinct.^[ref][ref]

The first illustration uses increasing $y$ downward, following the initial SVG coordinate convention. Its dashed control polygon guides the eye; it is not part of the filled path. Apply the affine mapThis combines horizontal scaling, a shear mapping and a translation. The transformed control points are $(4.5,5)$, $(5.5,1)$ and $(10.5,5)$, giving the path M 4.5 5 Q 5.5 1 10.5 5 Z. Its curved midpoint is $T(3,2)=(6.5,3)$. The affine equivariance of a geometric basis explains why transforming only the control points transforms the entire Bézier curve: its weights sum to one, so the translation is added exactly once. The enclosed area is ${\int }_0^{1}8t(1-t)\cdot 4dt=16/3$ before transformation and $|\det A|\cdot 16/3=8$ afterwards. These checks concern the filled region, excluding the width of any stroke.

An SVG viewBox specifies a rectangle in user coordinates to map into a display region. For example, viewBox="0 0 6 6" in a square viewport of 120 by 120 display units gives a scale of 20 display units per user unit; in a square viewport of 600 by 600 it gives 100. The path's control points need not change. For unequal proportions, preserveAspectRatio controls whether the mapping preserves proportions or stretches the axes separately. These display units need not correspond one for one to physical display pixels.^[ref][ref]

Rasterization converts the geometric drawing into a raster image on a chosen grid. Deciding whether each pixel centre lies inside a filled region is a simple rule, but a moving diagonal boundary can make entire pixels switch abruptly. Antialiasing reduces these sampling artifacts by accounting for a footprint around each sample, rather than treating every pixel as a single inside-or-outside decision. This connects vector graphics to raster sampling and reconstruction and aliasing; a finer grid and a better sampling rule solve different parts of the problem.

For an exact analytical model, fill the region $R=\{(x,y)\in [0,4{]}^2:y\le x+1/4\}$. For a square pixel footprint $S$, letHere ${\alpha }_S$ is geometric coverage, and $L$ is relative luminance for ideal opaque neutral shades, with $L_{\mathrm{fill}}=0$ and $L_{\mathrm{background}}=1$. The intersection and area determine how much of the footprint is covered. This is a specified area-average model, not a claim that every SVG renderer uses exact coverage. The SVG shape-rendering property supplies rendering hints and permits implementation tradeoffs, including disabling antialiasing for some choices.^[ref]

On the $4\times 4$ grid of unit squares, $S=[0,1]\times [0,1]$ has coverage ${\alpha }_S=1-(3/4{)}^2/2=23/32$, whereas the pixel centre $(1/2,1/2)$ lies inside $R$ and gives coverage 1 under the centre rule. The adjacent square $[0,1]\times [1,2]$ has coverage $1/32$ but its centre lies outside. The entire filled area is $16-(15/4{)}^2/2=287/32$. Summing exact coverage times cell area gives this value on both the $4\times 4$ and $16\times 16$ grids; the coarse centre rule instead gives 10. Coverage values are displayed after sRGB encoding of the modeled relative luminance. The thin coloured line is a geometry overlay, not an extra sampled stroke. Enlarging the middle panel by nearest-neighbor interpolation would repeat its sixteen values; re-rendering the geometry produces the 256 independently evaluated footprints in the right panel. Antialiasing reduces abrupt edges but cannot make a finite grid preserve every geometric detail.

The attractive feature of vector graphics is editability at the level of intent: move one control point to bend a boundary, change a circle's radius, or transform a whole diagram while keeping its parts aligned. Logos, technical diagrams and outline fonts benefit because their structure can be expressed with relatively few shapes and curves. A complicated photograph may require enormous geometric detail, so vector graphics do not imply a smaller file than raster graphics. Nor does an SVG container guarantee that every object is vector geometry: its image element can embed or reference a PNG raster image. Transforming that embedded raster image changes its placement and reconstruction, but does not recover missing original detail.^[ref]