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For a real quadratic form on an $n$-dimensional vector space, with rank $r$ and signature $s$ defined as positive minus negative index, the maximum dimension of a totally isotropic subspace is $n-(r+|s|)/2$. Quotienting by the radical of a bilinear form leaves a nondegenerate form of indices $p,q$. Projection of an isotropic subspace to each sign-coordinate space is injective, bounding its dimension by $\min (p,q)$. The radical and vectors pairing one positive with one negative unit coordinate attain the bound $n-r+\min (p,q)$.

Checking only midpoints is not enough for an arbitrary subset of a real vector space. The rational numbers contain the midpoint of every pair of rational numbers, but the line segment from $0$ to $1$ also contains irrational numbers. For instance $\sqrt{2}/2$ lies on that line segment and is not rational. This shows why the definition quantifies over every real $t\in [0,1]$.

There is a useful additional hypothesis: a closed subset $C\subseteq {\mathbb{R}}^d$ containing the midpoint of every pair of its points is a convex set. Repeated bisection includes all points $(1-t)x+ty$ with dyadic rational $t\in [0,1]$. These coefficients approach every real $t\in [0,1]$; being a closed subset then includes each limit. This is a short proof outline, and the rational numbers example identifies precisely why a closed subset was required.

The Minkowski sum of subsets $A,B$ of a vector space collects the result of vector addition for every possible pair:Think of fixing a point $b$ of $B$, making a translation of the whole of $A$ by $b$, and then taking the union of these copies over every $b\in B$. The points of $B$ specify displacement vectors; they need not be the centre of a shape. Minkowski addition applies to nonconvex sets too. It is different from set union: for $A=B=\{0,1\}$ on the real line, $A\cup B=\{0,1\}$ but $A+B=\{0,1,2\}$.

For axis-aligned rectangles in the Euclidean plane, the coordinate intervals add separately:Every coordinate between $0$ and $a+c$ can be split into one coordinate in $[0,a]$ and another in $[0,c]$, and likewise for the vertical coordinate. Thus a $2\times 1$ rectangle plus a $1\times 2$ rectangle gives a $3\times 3$ square of area $9$, although the two input areas total only $4$. The Minkowski sum adds possible displacements, so its area need not be the sum of input areas.

For closed Euclidean disks $D_r=\{x\in {\mathbb{R}}^2:\Vert x\Vert \le r\}$ centred at the origin, the triangle inequality gives $D_r+D_s\subseteq D_{r+s}$. Conversely, if $r+s>0$, split any $x\in D_{r+s}$ as $x=rx/(r+s)+sx/(r+s)$. Therefore $D_r+D_s=D_{r+s}$, including the case $r=s=0$. Disks of radii $1$ and $2$ give a disk of radius $3$, with areas $\pi ,4\pi ,9\pi$. For disks with centres $u,v$, the resulting centre is $u+v$. These two examples lead to the root-of-area comparison in the Brunn-Minkowski theorem.

Minkowski addition is commutative and associative because vector addition is. Its identity is the singleton $\{0\}$, and $A+\varnothing =\varnothing$. For displacement vectors $u,v$ and $t>0$, translations and uniform dilations satisfywhere $tA=\{ta:a\in A\}$ uses scalar multiplication. A Minkowski sum of convex sets is convex: a line segment between $a_1+b_1$ and $a_2+b_2$ is obtained by adding the corresponding points of the line segments in $A$ and $B$. Closure requires care; see sum of two closed sets need not be closed and sum of a compact set and a closed set.

A module isomorphism is a bijection of modules over the same ring preserving addition and scalar multiplication. Its inverse preserves these operations too. When a linear operator $T$ gives a vector space an $F[t]$-module structure by $t\cdot v=T(v)$, a module isomorphism is exactly an invertible linear map intertwining the two operators. This translates classification of modules into matrix similarity and rational canonical form.

Module theory studies abelian groups with scalar multiplication by a ring. It extends linear algebra: a module over a field is a vector space, while ring scalars let the same language describe integer arithmetic and linear operators. The new phenomena are visible before any abstract classification: some nonzero scalars cannot be divided out, some modules have torsion elements of a module, and some have no basis of a module.

The distinction is useful because an integer relation such as $6m=0$ carries information which disappears if one assumes all nonzero scalars have multiplicative inverses. A finitely generated module over a principal ideal domain can be separated into a finite free module and a torsion module; the structure theorem for finitely generated modules over a principal ideal domain unifies the classification of finitely generated abelian groups with the study of linear operators through polynomial rings. See Keith Conrad's Modules over a PID, Theorem 4.1 and the operator applications.

A historical anchor is Emmy Noether's 1921 paper ^[ref]. Its first section studies rings and ideals under the condition that every ideal is a finitely generated module over its ring, and connects this to the ascending chain condition; later sections develop primary decomposition. This is an early source for the structural viewpoint underlying Noetherian rings. ^[ref] makes those statements accessible in English.

Integrate a polynomial times the conjugate of another over the product of unit circles, with normalized product angular measure. Fourier orthogonality makes the monomials $z_1^{{\alpha }_1}\cdots z_d^{{\alpha }_d}$ an orthonormal basis of the polynomial vector space. The Hermitian form is positive definite because the squared norm is the sum of squared absolute values of the finitely many coefficients. No completion or infinite series is needed to obtain this algebraic basis.

A multilinear map $F:V_1\times \cdots \times V_r\to W$ is a map between vector spaces that is linear in each argument when all other arguments are fixed. For example the evaluation $(\omega ,v)\mapsto \omega (v)$ is bilinear, and the determinant is multilinear in its columns. The universal property of a tensor product replaces this separate linearity by a single linear map from $V_1\otimes \cdots \otimes V_r$.

For a nilpotent endomorphism $N$ of a vector space, the commutator action $\mathrm{ad}N:T\mapsto NT-TN$ on its endomorphisms is nilpotent. The commuting left and right multiplication maps give $(\mathrm{ad}N{)}^r(T)={\sum }_{j=0}^r(-1{)}^j\left(\genfrac{}{}{0ex}{}{r}{j}\right)N^{r-j}T{N}^j$. If $N^q=0$ and $r\ge 2q-1$, every summand vanishes. This works in every characteristic of a field and is useful when applying Engel theorem to an Adjoint representation.

For endomorphisms of a finite-dimensional complex vector space, suppose repeated commutation with $T$ annihilates $S$. Write the blocks of $S$ using the generalized eigenspaces of $T$. On $\mathrm{Hom}(V_{\lambda },V_{\mu })$, the operator $\mathrm{ad}T$ is $(\mu -\lambda )I$ plus a nilpotent endomorphism. If $\lambda \ne \mu$, it is invertible, so $(\mathrm{ad}T{)}^{N}S=0$ forces that off-diagonal block of $S$ to vanish. Thus $S$ preserves each generalized eigenspace. This allows a nilpotent Lie algebra representation to be split by generalized eigenvalues even when its acting operators do not commute.

When a measurement model is a linear map $T$, its kernel records changes that its readout cannot distinguish. For example, the rectangular kernel and solution-fibre example measures only $x+2y$ and $y+z$. Even exact readouts $(3,2)$ leave the whole fiber $u_0+\ker A$ possible. Repeating either readout with greater precision cannot distinguish those inputs: their outputs agree exactly. This ambiguity follows from a nontrivial kernel, rather than from measurement error.

Additional information can remove a null direction of a linear map. In that example, adding a third linear functional that reads $y$ distinguishes $u_0+tv$, because its third readout is $t$. More generally, additional linear functionals $c_1,\dots ,c_k$ give unique recovery of all inputs exactly whenThis follows by applying the trivial kernel criterion to the combined linear map $x\mapsto (Tx,c_1(x),\dots ,c_k(x))$. Merely adding repeated linear equations does not meet this condition. On a finite-dimensional input vector space, each added scalar readout can increase the matrix rank by at most one, so at least $\dim \ker T$ extra independent readouts are needed for recovery of unrestricted inputs.

Restrictions on the possible inputs can also make recovery unique. If the allowed inputs form a set $D$, two allowed inputs are indistinguishable exactly when their difference is in the kernel. Thus $T$ is injective on $D$ precisely when no difference of distinct points of $D$ is a kernel vector. Such a restriction is extra information about the input; the original readout alone determines only its fiber.

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$. The norm must be positive on nonzero vectors, scale by the absolute value of a scalar, and satisfy the triangle inequality. These conditions make distance compatible with addition and scaling.

On $C([0,1])$, both the supremum norm $\Vert f{\Vert }_{\infty }={\max }_{0\le x\le 1}|f(x)|$ and the L2 norm $\Vert f{\Vert }_2=({\int }_0^1|f(x){|}^{2}dx{)}^{1/2}$ are genuine norms. A nonzero continuous function stays nonzero on an interval of positive length, so its squared integral cannot vanish. For $f(x)=x$, they give $1$ and $1/\sqrt{3}$, respectively. Always $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$, but the shrinking peaks below show that no uniform reverse bound exists. Unlike the finite-dimensional equivalence of norms, infinite-dimensional vector spaces can support inequivalent norm topologies.

A number field is a finite field extension $K/\mathbb{Q}$ of the rational numbers. Its degree of a field extension $[K:\mathbb{Q}]$ is its dimension as a vector space over $\mathbb{Q}$. Every element is an algebraic number; the primitive element theorem also says that $K=\mathbb{Q}(\theta )$ for some single algebraic number $\theta$.

For example,The basis $1,\sqrt{2}$ makes arithmetic explicit: use $(\sqrt{2}{)}^2=2$ to reduce products. Every nonzero element has its inverse in the same field:The denominator cannot vanish unless $a=b=0$, since $\sqrt{2}$ is not a rational number. For instance, $(3+\sqrt{2}{)}^{-1}=(3-\sqrt{2})/7$ belongs to the number field, although it is not an algebraic integer.

A degree-$n$ number field has exactly $n$ distinct field embeddings into the complex numbers fixing $\mathbb{Q}$, because every finite field extension in characteristic zero is a separable field extension. For $\mathbb{Q}(\sqrt{2})$ these are the two real mapsBoth preserve addition and multiplication. Keeping both values, rather than just the positive choice of $\sqrt{2}$, is what makes the field norm and the Minkowski embedding of a number field useful.

For endomorphisms $S,T$ of the same finite-dimensional vector space, $ST=I$ makes $S$ surjective and $T$ injective. The rank-nullity theorem makes both invertible, hence $TS=I$. The finite-dimensional and equal-domain assumptions matter; unilateral shift operators on an infinite-dimensional sequence space give one-sided inverses that are not two-sided.

A group of exponent two is abelian. Give such a finite group the additive notation of a vector space over ${\mathbb{F}}_2$, with zero acting as zero and one as the identity scalar. The group axioms and $g+g=0$ verify the vector-space laws. A basis with $d$ vectors then gives exactly $2^d$ elements. Thus a finite group whose elements all have order at most two must have power-of-two order.

An orthogonal basis of an inner-product vector space consists of nonzero mutually orthogonal vectors that span the space. The orthogonal projection of $v$ onto a finite-dimensional span is ${\sum }_i⟨v,u_i⟩u_i/⟨u_i,u_i⟩$. Dividing each vector by its norm produces an orthonormal basis. For centered random variables, the inner product is covariance; an orthogonal basis therefore consists of uncorrelated variables, without requiring unit variance.

For a bilinear form on a finite-dimensional vector space, the displayed right bilinear orthogonal complement is a subspace. Nondegeneracy identifies the ambient space with its dual and gives $\dim W^{\perp }=\dim V-\dim W$. If the form is symmetric or alternating, orthogonality is reciprocal and $(W^{\perp }{)}^{\perp }=W$. For a general nonsymmetric form, left and right complements must be distinguished. A restricted symmetric or alternating form has radical of a bilinear form $W\cap W^{\perp }$; quotienting by that radical of a bilinear form gives a nondegenerate induced form.

Work in the finite-dimensional vector space $V$ implicit in the requested matrix representation and $n=\dim V$. Put $K_j=\ker {\alpha }^j$, with $K_0=0$. These form an increasing sequence ending in $K_m=V$, and $\alpha (K_j)\subseteq K_{j-1}$. Choose a basis of $K_1$, extend it to a basis of $K_2$, and continue until a basis of $V$ is obtained. Each basis vector introduced at stage $j$ maps into the span of vectors chosen at earlier stages. Its matrix column can therefore have nonzero entries only in earlier rows. This proves that the matrix is strictly upper triangular.

For a strictly upper-triangular $n\times n$ matrix $N$, a nonzero contribution to $(N^r{)}_{ij}$ requires an index chain $i<i_1<\cdots <i_{r-1}<j$. No such chain has $n$ links among $n$ indices. Consequently $\boxed{{\alpha }^n=0}$. An example with exact index four is the Nilpotent Jordan blockHere $M^3$ has entry one in position $(1,4)$ and all others zero, while $M^4=0$.

In the adapted basis, $I+A$ is upper triangular with diagonal entries one, so all its eigenvalues are one. This does not persist for the product of two such matrices with unrelated adapted bases. TakeBoth square to zero, buthas characteristic polynomial $t^2-3t+1$ and eigenvalues $(3\pm \sqrt{5})/2$, neither equal to one. The obstruction is that the two nilpotent matrices need not admit a common triangularizing basis.

A norm on a real vector space $V$ is a map $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly when $x=0$, $p(tx)=|t|p(x)$ for real $t$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d_p(x,y)=p(x-y)$. The three axioms respectively give definiteness, compatibility with scaling, and the triangle inequality.

Identify a subset $A\subseteq \Omega$ with its indicator vector $a=(a_1,\dots ,a_n)\in {\mathbb{F}}_2^n$. Symmetric difference becomes coordinatewise addition modulo two, so the subsets form a vector space of dimension $n$, with the singleton subsets as a basis. The proposed form becomes the dot productDistributivity proves bilinearity, and commutativity proves symmetry. Its Gram matrix in the singleton basis is $I_n$, so it is nondegenerate. A permutation merely permutes coordinates and preserves the dot product, proving invariance under $S_n$.

Write $u=(1,\dots ,1)$ for the full subset. Its bilinear orthogonal complement $E=u^{\perp }$ consists exactly of even-cardinality subsets. If $n\ge 2$ is even, $B(u,u)=n=0$ in ${\mathbb{F}}_2$, so $⟨u⟩\le E$. For any $v\in E$, $B(v,v)={\sum }_i{v}_i^2={\sum }_i{v}_i=0$, so the restricted form is alternating.

Nondegeneracy of the full form gives $E^{\perp }=⟨u⟩$; one can also see this directly by testing against all vectors $e_i+e_j$, which forces all coordinates of an element of $E^{\perp }$ to be equal. Hence the radical of a bilinear form of the restricted form isDefine the quotient form by $\overline{B}(v+⟨u⟩,w+⟨u⟩)=B(v,w)$. Adding $u$ to either representative does not change the pairing with $E$, so it is well-defined. It is alternating, and a class pairing to zero with all classes has a representative in $\mathrm{rad}(B{|}_E)=⟨u⟩$, hence is zero. ThereforeBoth spaces are $S_n$-invariant, so the coordinate action descends and preserves the quotient form. This is the symplectic quotient of the binary subset module. The evenness assumption is essential because otherwise $u$ does not lie in its own bilinear orthogonal complement.