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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 bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)=0$ gives $B(v,w)+B(w,v)=0$, hence $B(v,w)=-B(w,v)$. If the characteristic is not two, the converse follows from $B(v,v)=-B(v,v)$ by dividing by two. In characteristic two, the relation $B(v,w)=-B(w,v)$ is just symmetry and does not force zero self-pairings: $B(x,y)=xy$ on ${\mathbb{F}}_2$ has $B(1,1)=1$.^[ref]

The matrixgives an alternating bilinear form over every field. Over the real numbers, its value is the signed area of the parallelogram with sides $v,w$: $B((2,1),(1,3))=6-1=5$, and swapping the sides gives $-5$. Every self-pairing is zero, yet $\det J=1$, so this bilinear form is nondegenerate. Directly, pairing $(x,y)$ with $(1,0)$ gives $-y$, and pairing it with $(0,1)$ gives $x$; both vanish only when $(x,y)=0$.

In characteristic different from two, a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating bilinear form on a finite odd-dimensional vector space is a degenerate bilinear form. The stronger statement that the rank of an alternating bilinear form is even holds in all characteristics: splitting off a two-dimensional block with matrix $J$ repeatedly proves it. A nondegenerate alternating bilinear form defines a symplectic vector space, the linear setting for signed area rather than squared length.

A bivector is an element of the second exterior power ${\Lambda }^{2}V$. A simple bivector is one exterior product $u\wedge v$; in real Euclidean space, it records the oriented area generated by those two vectors. A general bivector is a sum of such elements and need not describe just one plane. Covariant counterparts are differential two-forms, which take two vectors as inputs rather than belonging to ${\Lambda }^{2}V$.

In ${\mathbb{R}}^3$, chooseExpanding by bilinearity givesThe three coefficients are signed areas of the projections onto the $xy$, $xz$, and $yz$ coordinate planes. They are three components of one bivector, not the three coordinates of either input vector.

With the usual dot product and a fixed orientation of a vector space, the cross product represents this area element by the normal vector $(6,-3,1)$. The order of the components matters: $e_1\wedge e_2$ corresponds to $e_3$, while $e_1\wedge e_3$ corresponds to $-e_2$. The Gram determinant checks the full area:Here the norm on bivectors is the one induced by the Euclidean inner product. The exterior product exists without that metric and in arbitrary dimension; the familiar cross product description relies on the additional three-dimensional geometry.

Keep the same origin and the same vector $p=(2,1{)}^T$, but use the nonperpendicular basis vectors $b_1=(1,0{)}^T$ and $b_2=(1,1{)}^T$. They form a basis because the column matrix $B$ has nonzero determinant:The reconstruction is $p=b_1+b_2$. Its coordinates are read along lines parallel to the other axis, producing a parallelogram. These lines need not be perpendicular to the axis being measured. In contrast, the dot products are $p\cdot b_1=2$ and $p\cdot b_2=3$; treating those as coordinates would reconstruct $2b_1+3b_2=(5,3{)}^T$, a different vector.

The dual basis supplies the right measuring rules for any basis.^[ref] Here these coordinate functionals are $b_1^{\ast }(x,y)=x-y$ and $b_2^{\ast }(x,y)=y$, the rows of $B^{-1}$. They give $b_1^{\ast }(p)=b_2^{\ast }(p)=1$. Equivalently, their representing vectors $d_1=(1,-1{)}^T$, $d_2=(0,1{)}^T$ satisfy $d_i\cdot b_j={\delta }_{ij}$ and extract the coordinates as $d_i\cdot p$. These are Riesz dual basis in an inner product space representatives, rather than the original basis vectors.

The Gram matrix records how the nonperpendicular directions interact:For any real basis represented by a square invertible matrix $B$, writing $p=Bc$ gives these identities immediately. One can recover the coordinates either as $c=B^{-1}p$ or by solving $Gc=B^{T}p$. In this example, if $c=(a,b{)}^T$, the squared norm is $a^2+2ab+2b^2$, not $a^2+b^2$. Oblique coordinates remain useful when natural directions are not perpendicular, but the Gram matrix must accompany them when computing lengths or dot products.

The cat map is the area-preserving toral automorphism induced by the integer matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$. Its eigenvalues are $(3\pm \sqrt{5})/2$, giving one expanding and one contracting eigendirection. Cutting its image parallelogram by the integer grid and translating the pieces modulo one gives a concrete geometric construction on the unit square.

The change of variables formula expresses an integral in coordinates that make its integrand or its region simpler. Three ingredients change together: the region, the function being integrated, and the element of area or volume. The last change is measured by the absolute Jacobian determinant; forgetting it amounts to counting transformed cells as though their sizes had stayed the same.

In one dimension, substitution is the integrated chain rule. If $g$ is continuously differentiable on $[a,b]$ and $f$ is continuous on its image, thenThese are directed definite integrals: if $g$ decreases, the transformed endpoints reverse. This identity does not require $g$ to be injective, because an antiderivative composed with $g$ already accounts for any backtracking.^[ref]

For a concrete computation, set $u=1+x^2$, so $du=2xdx$. The endpoints $x=0,1$ become $u=1,2$:The numerator is precisely the derivative of the expression in the denominator. Substitution reveals the familiar natural logarithm hidden inside this rational function. Keeping the original endpoints after changing the integrand would describe a different integral.

For the elementary two-dimensional theorem, let $T:U\to V$ be a $C^1$ diffeomorphism between open sets in the plane, let $S\subset U$ be a bounded region with piecewise smooth boundary, and let $f$ be continuous on $T(S)$, with the displayed integrals finite. ThenHere $DT$ is the Jacobian matrix. The diffeomorphism assumption means a bijection with a continuously differentiable inverse, and ensures a nonzero Jacobian determinant. This excludes overlapping sheets; the more general area formula handles multiplicity.^[ref]

Take $S=[0,1{]}^2$ and the linear mapIts inverse is $(u,v)=((x-y)/2,y)$. The square becomes a parallelogram with vertices $(0,0),(2,0),(3,1),(1,1)$: base two, height one, area two. Consequently ${\iint }_{T(S)}1dxdy={\iint }_{S}2dudv=2$. A nonconstant integrand changes too: ${\iint }_{T(S)}xdxdy={\int }_0^1{\int }_0^1(2u+v)2dudv=3$.

The direction of the map matters when selecting the factor. Here $T$ goes from the new coordinates to the original coordinates, so the factor is two. Its inverse has Jacobian determinant $1/2$, appropriate for converting area in the opposite direction. Writing down the map before computing its Jacobian matrix prevents a reciprocal-factor error.

Now reflect the image vertically: $\tilde{T}(u,v)=(2u+v,-v)$ has Jacobian determinant $-2$, but still produces area two. The sign records reversal of the orientation of a vector space; ordinary area uses its absolute value. For oriented differential forms, the pullback of a differential form instead retains the signed determinant.^[ref]

The geometric beauty is that a nonlinear differentiable map has a local linear approximation: tiny rectangles behave approximately like these parallelograms. Summing their scaled contributions gives the change of variables formula. In polar coordinates this factor is $r$, explaining $dxdy=rdrd\theta$; equal angular steps cover more area farther from the origin. Choosing suitable coordinates can simplify both the boundary and the integrand.^[ref]

For a real three-dimensional linear map $A$, preservation of every surface area is equivalent to being an orthogonal matrix. For necessity, diagonalize the symmetric matrix $A^{T}A$ in an orthonormal basis, with nonnegative eigenvalues ${\lambda }_i$. The three unit coordinate parallelograms have image areas $\sqrt{{\lambda }_1{\lambda }_2}$, $\sqrt{{\lambda }_1{\lambda }_3}$, and $\sqrt{{\lambda }_2{\lambda }_3}$. Setting all three equal to $1$ forces every ${\lambda }_i=1$. Sufficiency follows from preservation of dot products and $\Vert \mathbf{u}\times \mathbf{v}{\Vert }^2=\Vert \mathbf{u}{\Vert }^2\Vert \mathbf{v}{\Vert }^2-(\mathbf{u}\cdot \mathbf{v}{)}^2$. This specifically uses dimension three; in dimension two, preservation of planar area alone only requires $|\det A|=1$.

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]

Two side vectors span a parallelogram whose signed area is their determinant. Its two congruent triangles give the factor one-half, and the absolute value removes orientation. Reversing the vertex order changes the sign but not the area; collinear vertices give zero.

A Euclidean vector is a vector in a finite-dimensional real inner product space. Choosing an orthonormal basis identifies it with its real vector components in ${\mathbb{R}}^n$. A geometric arrow represents a Euclidean vector by a length and a direction; translating the arrow without changing either represents the same Euclidean vector. The zero vector has length zero and no distinguished direction.^[ref]

For points $P=(1,1)$ and $Q=(4,5)$ in a Cartesian coordinate system, the displacement is $Q-P=(3,4)$. The arrow from $(-2,0)$ to $(1,4)$ represents the same Euclidean vector, even though its endpoints differ. An endpoint is a position; their difference is a vector. In an orthonormal basis $({\mathbf{e}}_1,{\mathbf{e}}_2)$,The Euclidean norm comes from the inner product: $\Vert \mathbf{v}{\Vert }_2^2=⟨\mathbf{v},\mathbf{v}⟩$. The sum of squared vector components is valid in an orthonormal basis; an arbitrary basis requires the appropriate Gram matrix. This distinction lets the geometric vector stay the same while its coordinate vector changes.^[ref]

Vector addition is componentwise. With $\mathbf{u}=(2,1)$ and $\mathbf{w}=(1,3)$,Geometrically, translate the second arrow so its tail starts at the first arrow's head. The vector sum joins the starting point to the final endpoint. Reversing the order gives the other route around the same parallelogram, so vector addition is commutative. This is the useful bookkeeping behind adding successive displacements, velocities, or forces: direction information survives the calculation. For a force, the Euclidean vector alone does not specify its point of application, which also affects the torque.^[ref]

The magnitude of a vector sum is generally different from the sum of the magnitudes. Here $\Vert \mathbf{u}+\mathbf{w}{\Vert }_2=5$, whereas $\Vert \mathbf{u}{\Vert }_2+\Vert \mathbf{w}{\Vert }_2=\sqrt{5}+\sqrt{10}>5$. The triangle inequality explains why the direct displacement cannot exceed the length of this two-segment route. A return displacement $(-3,-4)$ makes the net vector sum zero, although the lengths of the three travelled segments sum to a positive number.

The dot product exposes direction quantitatively: for nonzero Euclidean vectors, $⟨\mathbf{u},\mathbf{w}⟩=\Vert \mathbf{u}{\Vert }_2\Vert \mathbf{w}{\Vert }_2\cos \theta$. For the displayed pair, the dot product is $2\cdot 1+1\cdot 3=5$, so $\cos \theta =1/\sqrt{2}$ and the angle is $45^{\circ }$. For orthogonal vectors, that dot product is zero. Dividing by magnitudes separates direction from size, which is the purpose of vector normalization.

The exterior product, also called the wedge product, combines oriented elements through the multiplicationFor two vectors, it is bilinear and alternating:These rules imply $v\wedge u=-u\wedge v$. They distinguish the exterior product from both the symmetric dot product and the unconstrained tensor product.^[ref]

In an oriented orthonormal plane, write $u=(a,b)$ and $v=(c,d)$. Expanding in its standard basis givesThe determinant $ad-bc$ is the signed area of the parallelogram formed by the two inputs. Its absolute value is the ordinary area. Reversing the input order reverses the sign; parallel inputs produce zero.

For the original example $u=(3,1)$ and $v=(1,2)$,Replacing $v$ by $v+u=(4,3)$ changes the shape but preserves the product: $3\cdot 3-1\cdot 4=5$. Both parallelograms have the same oriented area element. If $v=2u$, the exterior product is zero, since $u\wedge (2u)=2(u\wedge u)=0$.

The same rule adds directions in higher dimensions. With $w=(2,-1,1)$ and the three-dimensional inputs under bivector,Its coefficient is the oriented volume of the three-dimensional parallelepiped. Repeating or linearly depending on earlier inputs collapses the volume, giving zero.

For homogeneous elements of a graded algebra of degrees $p,q$, the general commutation rule isSwapping a bivector with another bivector therefore produces a positive sign, since $pq=4$. It is the number of individual vector swaps that controls the sign, not a rule that every pair of objects must anticommute.

Let $V$ be a vector space over a specified field $F$, and let $v_1,\dots ,v_n\in V$. A linear combination is a finite sumThe coefficients $a_i$ are scalars from that same field; each term uses scalar multiplication, and the terms are joined by vector addition. For coefficients in the real numbers, negative values reverse arrows and fractional values scale them by fractional amounts. The collection of attainable vectors is the linear span. The column picture in Strang's MIT notes interprets solving a linear system as choosing coefficients to reach a target vector.^[ref]

For a concrete calculation over the real numbers, takeThen $2v_1+v_2=(2,2)+(1,-1)=(3,1)=w$. More generally, solving $a+b=x$ and $a-b=y$ givesEvery vector in the Euclidean plane is therefore a linear combination of $v_1,v_2$, so their linear span is all of ${\mathbb{R}}^2$. The target $(3,1)$ has coefficients $(2,1)$; the target $(0,2)$ has coefficients $(1,-1)$, since $(1,1)-(1,-1)=(0,2)$. The displayed formula also makes the coefficients unique. Consequently $(v_1,v_2)$ is an ordered basis, with coordinate vector $[(x,y){]}_{(v_1,v_2)}=((x+y)/2,(x-y)/2{)}^T$.

The payoff is a reusable language for synthesis: linear combinations build vectors from chosen ingredients, linear span decides which targets are reachable, and linear independence decides whether the coefficients are unique. The ingredients can also be polynomials or functions, as long as they belong to a vector space over the specified field.

For an invertible matrix $A\in {\mathbb{R}}^{n\times n}$ and real vectors $u,v\in {\mathbb{R}}^n$, the matrix determinant lemma statesThe expression $v^T{A}^{-1}u$ is a scalar, so a rank-one matrix update changes the determinant by a single multiplicative factor. One need not expand the updated determinant entry by entry.

For the preceding example with $A=I$, the dot product is $v^{T}u=3-2=1$. Hence the updated determinant is $1+1=2$. Direct evaluation confirmsMore generally, let $B(t)=I+tu{v}^T$ for a real scalar $t$. The same calculation gives $\det B(t)=1+t$. The determinant is one at $t=0$, two at $t=1$, and zero at $t=-1$.

A compact proof explains the scalar reduction. Put $w=A^{-1}u$ and factor $A+u{v}^T=A(I+w{v}^T)$. If $w\ne 0$, choose a basis starting with $w$. In that basis, $(I+w{v}^T)w=(1+v^{T}w)w$, while every other basis vector changes only by a multiple of $w$. Its matrix is triangular with diagonal $1+v^{T}w,1,\dots ,1$, giving the required determinant. If $w=0$, the correction vanishes. This uses multiplicativity of the determinant to restore the factor $\det A$.

The denominator in the Sherman–Morrison formula is exactly this factor. The rank-one update singularity criterion therefore identifies when the updated matrix inverse ceases to exist. Batson, Spielman and Srivastava use both identities for the symmetric update $A+v{v}^T$ in Twice-Ramanujan Sparsifiers, section 2.3, Lemmas 2.1–2.2.

For an $n$-by-$n$ square matrix $A$ over a field, its matrix inverse is an $n$-by-$n$ matrix $A^{-1}$ satisfying both identitieswhere $I_n$ is the identity matrix. It exists exactly when $A$ is an invertible matrix, equivalently when its matrix rank is $n$ or its determinant is nonzero. It reverses the linear map: applying $A$ and then $A^{-1}$ recovers the input, while applying $A^{-1}$ and then $A$ recovers the output.^[ref]

The matrix inverse is unique. If $B$ and $C$ both satisfy the identities, then $B=B(AC)=(BA)C=C$. Although the definition is two-sided, for finite square matrices either identity alone suffices: $BA=I$ forces a trivial kernel, whereas $AB=I$ forces surjectivity; either makes $A$ an invertible matrix by the rank-nullity theorem. Rectangular matrices may have only one of these properties.

For a concrete example, takeDirect matrix multiplication givesConsequently the vector $v=(1,2{)}^T$ follows a reversible journey:The columns $(2,1{)}^T$ and $(1,1{)}^T$ are the images of the standard basis vectors. They turn the unit square into a parallelogram; the matrix inverse restores both the square and its surrounding grid. Since $\det A=1$, this example changes shape while preserving area, illustrating that invertibility does not require preserving lengths or angles.

For comparison, the singular matrixcannot have a matrix inverse: the output $(1,0{)}^T$ cannot specify which of the two inputs occurred. Their difference $(1,-1{)}^T$ belongs to its kernel. Indeed $S(x,y{)}^T=(x+y,0{)}^T$ collapses the plane to a line, its matrix rank is one, and its determinant is zero. No subsequent function of the output alone can reconstruct every input.

An important use of the matrix inverse is describing the unique solution of a linear system:For $b=(4,3{)}^T$, the preceding computation solves $2x_1+x_2=4$ and $x_1+x_2=3$, giving $x=(1,2{)}^T$. In computation, usually solve the linear system using Gaussian elimination and a factorization such as an LU decomposition, rather than forming every entry of the matrix inverse. The inverse formula specifies the answer; a factorization provides a reusable way to compute it for each right-hand side.^[ref]

Here the LU decomposition is especially simple:Solve $Ly=b$ by forward substitution in a triangular system: $y_1=4$ and $y_2=3-\frac{1}{2}\cdot 4=1$. Then solve $Ux=y$ by backward substitution in a triangular system: $x_2=2$ and $x_1=(4-2)/2=1$. The same factors work when $b$ changes. If the whole matrix inverse is actually needed, its $j$th column is the solution of $Ax=e_j$, where $e_j$ is the $j$th standard basis vector.

Reversing several operations means undoing the last one first. For invertible matrices of the same size,because $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=I$, and the product in the other order is also $I$. This is why a change-of-basis matrix uses its matrix inverse to reverse the direction of coordinate conversion.

For projection of a planar region with unit normal $\mathbf{n}$ onto a plane with unit normal $\mathbf{k}$, the projected area is $|\mathbf{n}\cdot \mathbf{k}|$ times the original area. Indeed, the signed projected area of the parallelogram spanned by tangent vectors $\mathbf{u},\mathbf{v}$ is $(\mathbf{u}\times \mathbf{v})\cdot \mathbf{k}$, while $\mathbf{u}\times \mathbf{v}$ is parallel to $\mathbf{n}$. Taking magnitudes proves the factor. The same pointwise formula applies to a parametrized surface; adding projected patches counts multiplicity unless their images overlap only in sets of zero area.

Write vectors in a symplectic basis as $v={\sum }_i(q_i{e}_i+p_i{f}_i)$ and $w={\sum }_i(Q_i{e}_i+P_i{f}_i)$. Their coordinate vectors and the matrix of a bilinear form areHere $I_n$ is the identity matrix, and $T$ denotes transpose. The ordering matters: interleaving the basis as $(e_1,f_1,\dots ,e_n,f_n)$ instead gives a block diagonal matrix with $2\times 2$ blocks $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Some conventions reverse every $f_i$, giving $-J$ instead. Always check the displayed pairing convention.^[ref]

Over the real numbers, each summand is the signed area of a parallelogram in the $(q_i,p_i)$ coordinate plane. For $a=(2,1,1,1{)}^T$ and $b=(1,1,0,2{)}^T$, the two signed areas are $-1$ and $1$, so $\omega (a,b)=0$ although the vectors are linearly independent. The cancellation concerns the symplectic form; it does not assert that their dot product vanishes. Over a general field, the same determinant formula is algebraic, without a signed geometric interpretation.

A parallelepiped is a polyhedron with six parallelogram faces in three opposite pairs. Three generating vectors determine its edges; the absolute value of their determinant gives its volume in orthonormal Cartesian coordinates.

Use direction $\mathbf{u}=(-3,-2,-1)$ and displacement $\mathbf{w}=(1,2,3)-(6,5,4)=(-5,-3,-1)$. The point-to-line distance is the height of the parallelogram spanned by these vectors:Equivalently the closest line parameter is $t=(\mathbf{w}\cdot \mathbf{u})/|\mathbf{u}{|}^2=11/7$, giving the foot $(9/7,13/7,17/7)$. Its separation from the point is orthogonal to $\mathbf{u}$ and has the same length.

The proposed linear change of variables makes the two pairs of boundary lines constant-coordinate lines: $v=1,4$ and $u=0,7$. Thus the image of the parallelogram is the rectangle $0\le u\le 7$, $1\le v\le 4$.

The forward Jacobian matrix and its determinant areInverting gives $x=(u+3v)/7$, $y=(v-2u)/7$, so the area element is $dxdy=\frac{1}{7}dudv$. The absolute inverse Jacobian determinant, rather than the forward determinant, belongs in the transformed area integral. Consequentlywhich yields $\boxed{\frac{7}{2}\log 4=7\log 2}$. The transformed denominator is bounded below by one throughout the region, so no singular or improper integral is involved.

The Minkowski convex body theorem states that a centrally symmetric convex measurable set in ${\mathbb{R}}^d$ of volume greater than $2^d\mathrm{covol}(\Lambda )$ contains a nonzero point of the full lattice $\Lambda$. For a compact convex body, the weak inequality also suffices: apply the strict version to $(1+\epsilon )C$, let $\epsilon \downarrow 0$, and choose a constant subsequence of the resulting lattice points in the bounded set $2C$. Its nonzero value belongs to $C$ by closedness.

For $Q>1$, apply this borderline version with $\Lambda ={\mathbb{Z}}^2$ toThe linear transformation $(u,v)\mapsto (u,xu-v)$ has determinant of absolute value one, so this parallelogram has area four. A nonzero lattice point $(n,m)$ satisfies $|n|\le Q$ and $|nx-m|\le Q^{-1}$. If $n=0$, then $|m|<1$, giving the forbidden zero point; hence $n\ne 0$. It follows thatIf $x$ is irrational, these pairs cannot remain in a finite set as $Q\to \infty$: the positive errors $|nx-m|$ of any such finite set have a positive minimum, contradicting their bound $Q^{-1}$. If $x=p/q$ is rational in lowest terms with $q>0$, the pairs $(n,m)=(jq,jp)$ for $j\ge 1$ have zero error and are infinitely many distinct pairs. This proves the assertion for every real $x$.