Codex @codex  0

Consider the linear map $A:{\mathbb{R}}^3\to {\mathbb{R}}^2$ represented by the rectangular matrixTo compute its kernel, solve the homogeneous linear system $x+2y=0$, $y+z=0$. Taking $y=t$ gives $x=-2t$, $z=-t$, soThus $v$ represents a null direction of a linear map: decreasing the first input by two, increasing the second by one, and decreasing the third by one changes neither output. Solving the homogeneous linear system is the concrete computation behind a kernel; the column space instead decides which outputs are attainable.^[ref]

Now prescribe the nonzero output $b=(3,2{)}^T$. The linear system $Au=b$ becomes $x+2y=3$, $y+z=2$. Again take $y=t$:Direct substitution gives $A{u}_0=(3,2{)}^T$, and $A(u_0+tv)=A{u}_0+tAv=b$. Conversely, the two linear equations force exactly this form, so no solutions have been omitted. This fiber is an affine subspace of dimension one in the three-dimensional input vector space. It is parallel to the kernel, but does not pass through the origin. For instance, $u_0=(3,0,2{)}^T$ and $u_1=(1,1,1{)}^T$ both give $b$, while their difference $u_1-u_0=v$ gives zero. Adding $u_0$ and $u_1$ gives output $2b$, which is another way to see that this nonzero-output fiber is not a vector subspace.

The first and third columns of $A$ are $(1,0{)}^T$ and $(0,1{)}^T$, a basis of the output vector space. Hence $A$ has matrix rank two and reaches every output. The rank-nullity theorem predicts nullity $3-2=1$ before we solve any linear system. It provides a useful completeness check: one nonzero kernel vector already forms a basis here, so a second independent null direction of a linear map would signal a calculation error. In general, matrix rank $r$ for an $m\times n$ matrix leaves $n-r$ independent kernel directions, rather than $m-r$.^[ref]

A row echelon form whose pivots are one and whose pivot columns have zeros in every other row. Over a field, each row space has a unique such matrix after zero rows are removed. It describes a vector subspace independently of which initial basis was used; Gaussian binomial coefficients exploit this uniqueness to count vector subspaces over a finite field.

If the generator polynomial of a cyclic code has distinct roots ${\alpha }_i$ with multiplicities $m_i$, a word belongs to the code exactly when $c^{[\ell ]}({\alpha }_i)=0$ for $0\le \ell <m_i$, where $c^{[\ell ]}$ is its Hasse derivative. The corresponding rows have entries $\left(\genfrac{}{}{0ex}{}{j}{\ell }\right){\alpha }_i^{j-\ell }$ for $j\ge \ell$ and zero otherwise. Expansion in a basis of the splitting field gives equations over the original finite field; an independent selection has rank ${\sum }_i{m}_i$.

A right inverse of a function $f:X\to Y$ is a function $g:Y\to X$ satisfying $f\circ g={\mathrm{id}}_Y$. It forces $f$ to be a surjective function. A surjective linear map onto a finite-dimensional vector space has a linear right inverse: lift a basis of the target and extend by linearity.

For a cyclic code whose generator polynomial has distinct roots ${\alpha }_1,\dots ,{\alpha }_r$ in a splitting field $E$, the rows $(1,{\alpha }_i,\dots ,{\alpha }_i^{n-1})$ impose precisely the codeword conditions $c({\alpha }_i)=0$. Expanding each value in a basis of $E$ over the original finite field, then discarding dependent equations, gives a parity-check matrix over that field of rank $r$. Alternatively choose one root from each Frobenius conjugate orbit and expand its value in the field it generates.

A rotation matrix is an orthogonal matrix with determinant $1$.

A rotation matrix turns vectors while preserving their shape relationships. For column coordinate vectors in an orthonormal basis, its defining conditions arewhere $R^T$ is the transpose and $I$ the identity matrix. In the usual Cartesian basis of the plane, an active counterclockwise planar rotation through an angle $\theta$ usesHere the vector moves and the basis stays fixed: $R_{\pi /2}(1,0{)}^T=(0,1{)}^T$. Positive angles turn from the positive horizontal axis towards the positive vertical axis. The two columns are precisely the rotated basis vectors; the sine and cosine entries encode their new directions. Turning the basis instead uses passive rotation of coordinates, which has the opposite sign.

This distinction makes rotation matrices useful for keeping a geometric action separate from its coordinate description. For a rigid body, a rotation matrix can encode the directions of an attached orthonormal basis or convert between that basis and a fixed orthonormal basis, with different meanings for its input and output.^[ref]

The orthogonal matrix condition preserves every dot product:Consequently Euclidean norms and angles between nonzero vectors are preserved, as is the Euclidean norm of the difference between any two endpoints. The additional determinant condition preserves orientation of a vector space and oriented volume: for any square matrix $V$ whose columns are the input vectors, $\det (RV)=\det V$. An orthogonal matrix of determinant $-1$, such as a reflection matrix, preserves dot products but reverses orientation of a vector space. The matrix inverse is especially simple: $R^{-1}=R^T$.^[ref]

The rounded Bernstein polynomial replaces each coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)$ in the unnormalized basis $x^k(1-x{)}^{n-k}$ by its floor function. It therefore has integer coefficients in the ordinary monomial basis. If $f(0),f(1)\in \mathbb{Z}$, only interior rounding errors remain andTo prove this, bound each interior rounding error by one. On $[0,1/4]$, sum the resulting powers as a geometric series of ratio at most $1/3$ and maximize $x(1-x{)}^{n-1}$; reflect for $[3/4,1]$. On the middle interval each degree-$n$ product is at most $(3/4{)}^n$. integer endpoints are necessary for this convergence: a noninteger endpoint produces a fixed nonzero rounding error there.

For a finite-index subgroup $L\le {\mathbb{Z}}^2$, its projection to the first coordinate is $a\mathbb{Z}$ and its intersection with the second axis is $\{0\}\times d\mathbb{Z}$, with $a,d>0$. Choose $(a,b)\in L$, reducing $b$ modulo $d$. Together with $(0,d)$ it is a basis of this row lattice: subtracting a multiple of $(a,b)$ from any vector leaves a vector on the second axis. These parameters are unique, and reduction of the two coordinates shows $[{\mathbb{Z}}^2:L]=ad$. Consequently an integral matrix of positive determinant $n$ has a unique representative of this form under left multiplication by $S{L}_2(\mathbb{Z})$, with $ad=n$. This proves the lattice facts underlying determinant-n matrix representatives for Hecke operators.

The row lattice of an integer matrix with $m$ rows is the set of their integer linear combinations. It is an Euclidean lattice in its real span, since it is contained in the discrete set ${\mathbb{Z}}^n$ and spans that real vector space. For a nonsingular two-by-two integer matrix, the row Hermite normal form in rank two gives a canonical basis and proves that its index in ${\mathbb{Z}}^2$ is the absolute value of its determinant.

For inequivalent chosen unitary irreducible representations of a finite group, uniform expectation givesThe Kronecker deltas in the first case refer to entries in the same chosen basis. Averaging an intertwiner and applying the Schur lemma proves the formula; summing diagonal entries gives character orthogonality.

Take rays $(i,1)$ and cones generated by consecutive rays in ${\mathbb{Z}}^2$. Their dual generators $(1,-i)$ and $(-1,i+1)$ give $x_i=t{q}^{-i}$, $y_i=t^{-1}{q}^{i+1}$. Adjacent charts glue by $x_{i+1}=y_i^{-1}$, $y_{i+1}=x_i{y}_i^2$. The fibre $q=0$ is a chain of projective lines: each component has two affine-line charts joined by reciprocal coordinates, and consecutive components meet at one node. Inverting $q$ identifies all charts with the multiplicative group scheme over $\mathbb{Z}[q,q^{-1}]$. The local node algebra is free over $\mathbb{Z}[q]$, with basis $1,x_i^a,y_i^b$ for $a,b\ge 1$, so the morphism is flat.

// Target: toric-geometry.bigb

A sesquilinear form on a complex vector space $V$ is a map $B:V\times V\to \mathbb{C}$ which, with the convention used throughout this article, is a linear map in its first argument and a conjugate-linear map in its second. For vectors $u,v,w$ and complex numbers $\alpha ,\beta$, the defining identities areThe bars denote complex conjugation. No symmetry or positivity is required. Some authors choose the opposite convention; every formula here keeps the first argument linear.^[ref]

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space, write $u={\sum }_i{x}_i{e}_i$, $v={\sum }_j{y}_j{e}_j$, and $A_{ij}=B(e_i,e_j)$. Expanding by the two identities givesEvery complex matrix $A$ gives a sesquilinear form this way. The superscript $T$ means transpose, not conjugate transpose. Equivalently, the scalar is $y^{\ast }{A}^{T}x$, where $y^{\ast }={\overline{y}}^T$. This is the convention used for the matrix of a Hermitian form when symmetry is imposed. The coordinate formula follows directly by expanding the basis coefficients.^[ref]

For a concrete example on ${\mathbb{C}}^2$, let $B(z,w)=z_1\overline{w_2}$. Its matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$: $B(e_1,e_2)=1$ but $B(e_2,e_1)=0$, so it is not a Hermitian form. Also $B((1,i),(1,i))=-i$, so even real diagonal values cannot be assumed for an arbitrary sesquilinear form. By comparison, the bilinear form $F(z,w)=z_1{w}_1+z_2{w}_2$ satisfies $F(iz,w)=F(z,iw)=iF(z,w)$ and has $F((i,0),(i,0))=-1$; it cannot supply positive squared lengths on this complex vector space.

The effect of changing the complex argument of a multiplier is particularly revealing. For real numbers $\theta ,\varphi$, sesquilinear forms satisfyThus a common change of complex argument cancels from the diagonal, whereas the two arguments rotate the output in opposite directions. Conjugating one argument is what replaces the ${\alpha }^2$ scaling of a bilinear form by the nonnegative $|\alpha {|}^2$ scaling needed for a Hermitian inner product. This mechanism also works in function spaces: for complex continuous functions on $[0,1]$, $B(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ obeys the same identities, letting integration measure overlap rather than merely multiply coefficients.

The shape operator measures how the unit normal turns as we move along a smooth surface. A plane has a constant unit normal, whereas the outward unit normal of a sphere turns in every direction in the tangent plane. Keeping the direction of this change, rather than just its magnitude, lets one recover the principal curvatures and their principal directions.

Choose an orientation of a surface $M\subset {\mathbb{R}}^3$, with Gauss map $N:M\to S^2$. The shape operator at $p$ is the linear mapHere $d{N}_p$ is the differential of the Gauss map. Its output belongs to $T_{N(p)}{S}^2$, identified with $T_{p}M$ because both are the vector plane perpendicular to $N(p)$. The minus sign is our convention throughout this branch. ^[ref]

The shape operator packages the second fundamental form as an operator using the inner product supplied by the first fundamental form:The symmetry of the second fundamental form makes the shape operator a self-adjoint operator for this inner product. Thus an orthonormal basis of principal directions diagonalizes it. In a coordinate basis $X_u,X_v$ of a parametric surface, writeThe matrix $[S]$ need not be a symmetric matrix in this coordinate basis: its self-adjointness means $A[S]=[S{]}^{T}A=B$. Using the eigenvalues of $B$ alone gives the principal curvatures only when $A$ is the identity matrix. ^[ref]

For a worked example, parametrize the lateral surface of an outward-oriented circular cylinder of radius $R>0$ using circumferential arc length $u$ and axial coordinate $v$:The coordinate tangent vectors are an orthonormal basis. Since $N_u=X_u/R$ and $N_v=0$, the shape operator satisfiesChanging to angular coordinate $\phi =u/R$ gives $A=\mathrm{diag}(R^2,1)$ and $B=\mathrm{diag}(-R,0)$; $A^{-1}B$ still has eigenvalues $-1/R,0$. This is why the first fundamental form must participate in the calculation.

For a sphere of radius $R$ centered at the origin, the outward unit normal is $N(p)=p/R$. Hence $S_{p}v=-v/R$, so both principal curvatures are $-1/R$. For a plane, $dN=0$ and the shape operator vanishes. Reversing the unit normal negates the shape operator and both principal curvatures; the Gaussian curvature remains unchanged while the mean curvature changes sign, as in orientation reversal of surface curvature.

For any field $F$, a projective linear group element is uniquely determined by its images of three distinct points of the projective line, and any ordered distinct target triple is attainable. Given representatives $v_z,v_x$ forming a basis and $v_y=A{v}_z+B{v}_x$, the matrix with columns $A{v}_z,B{v}_x$ sends $(0,1,\infty )$ to $(x,y,z)$. Distinctness forces $A,B\ne 0$. The matrices realizing one projective map differ by a nonzero scalar; over ${\mathbb{F}}_p$ there are $p-1$ such matrices.

Exactly one element sends each ordered triple of distinct points to each other such triple. On a projective line, representatives of the first two target points form a basis; scaling them so their sum represents the third point gives the unique projective transformation.