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A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

A linear map between ordered vector spaces is positive if it carries the positive cone of the domain into the positive cone of the codomain. For function spaces this means $f\ge 0⟹Tf\ge 0$. A Feller semigroup consists of such positive maps on continuous functions.

A tied largest modulus can prevent even direction convergence. With $D=\mathrm{diag}(2,-2)$ and $x_0=(1,1{)}^T/\sqrt{2}$,The two directions form a cycle; neither is an eigenvector. There is no dominant-modulus spectral gap. A tied modulus is not a universal failure: an initial eigenvector remains on its line, and $A=2I$ preserves every direction. Subspace iteration addresses a separated cluster by targeting a dominant invariant subspace instead of one eigenvector.

An orthogonal start can fail despite a strict spectral gap. For $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $x_0=(1,-1{)}^T/\sqrt{2}$, the dominant component is zero, so $A{x}_0=x_0$ and every iterate equals $x_0$. The Rayleigh quotient is exactly $1$ and the eigenpair residual is zero, although the dominant eigenvalue is $3$. Exact arithmetic never creates a missing component.^[ref]

The third panel contrasts these failures with $N=\mathrm{diag}(-3,1)$, $x_0=(1,1{)}^T/\sqrt{2}$, for whichHere the horizontal eigenvector direction converges with a strict spectral gap, while the normalized vectors alternate signs. This is the successful direction convergence described under sign and phase in power iteration.

For a nonorthogonal eigenbasis, even the word “orthogonal” must be used carefully. The diagonalizable matrix $C=\left(\begin{array}{cc}3& 2\\ 0& 1\end{array}\right)$ has right eigenvectors $v_1=(1,0{)}^T$, $v_2=(-1,1{)}^T$, and dominant left eigenvector $w_1=(1,1{)}^T$. For $x_0=(a,b{)}^T$, the dominant coefficient is $c_1=w_1^T{x}_0=a+b$. The starting vector $v_2$ is nonorthogonal to $v_1$ but stays at the smaller eigenvalue $1$; $(0,1{)}^T$ is orthogonal to $v_1$ but has $c_1=1$ and converges to the dominant direction. The left eigenvector test, not the right-eigenvector inner product, controls the outcome.

A random vector drawn from a continuous probability distribution with a probability density function on the full real or complex vector space has $c_1\ne 0$ with probability one. This avoids the exact missing-component hyperplane, but does not guarantee a large component or fast convergence. Floating-point arithmetic may seed a missing component, yet a structurally preserved zero can remain zero, so accidental perturbations are not a convergence guarantee.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

A projective representation is a group homomorphism into the projective linear group of a vector space. Choosing invertible linear representatives makes multiplication hold up to nonzero scalar factors. A projective unitary representation chooses unitary operators as representatives, whose scalar factors have modulus one.

A proper vector subspace of a vector space is one strictly smaller than the ambient space. In a finite-dimensional vector space, this is equivalent to having smaller dimension: a basis of the subspace can be extended to a basis of the ambient space, and strict inclusion requires at least one additional vector.

For even $d=2n$, the roots of $X^{q^d}+X^{q^2}+TX$ form an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$. Ratios $(c\alpha )/\alpha$ for a nonzero root $\alpha$ show that the splitting field contains every $c\in {\mathbb{F}}_{q^2}$. The subgroup fixing that quadratic constant extension is normal of index two and lies in $GL(n,q^2)$; applying projective transitivity over ${\mathbb{F}}_{q^2}(T)$ makes it contain $SL(n,q^2)$. It therefore normalizes that special linear group, while the full group acts semilinearly.

A quadratic form on a vector space over a field $F$ is a function $Q:V\to F$ satisfying $Q(av)=a^{2}Q(v)$ whose polar bilinear form of a quadratic form $B_Q(v,w)=Q(v+w)-Q(v)-Q(w)$ is a bilinear form. In finite dimension, it is represented in coordinates by a homogeneous degree-two polynomial, or by $Q(v)=v^{T}Av$ with a suitable matrix $A$.

When the characteristic is not two, one may choose $A$ to be a symmetric matrix and recover it from $B_Q/2$. In characteristic two, the symmetric-matrix restriction would lose examples: $Q(x,y)=xy$ has an upper-triangular representing matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while $v^{T}Av$ for a symmetric matrix has no $xy$ cross term. Its polar form is $B_Q((x,y),(u,v))=xv+yu$, an alternating bilinear form. The distinction matters when studying quadratic forms over finite fields.^[ref]

A quasi-norm on a vector space is nonnegative, vanishes only at zero, is absolutely homogeneous, and satisfies a triangle inequality with one constant factor $C\ge 1$. For $0<p<1$, $(\int |f{|}^p{)}^{1/p}$ on an Lp space is a quasi-norm: $\Vert f+g{\Vert }_p^p\le \Vert f{\Vert }_p^p+\Vert g{\Vert }_p^p$ gives a quasi-triangle inequality. Measure-preserving composition preserves this quantity even though the Minkowski inequality is unavailable below exponent one.

For a Lie algebra ideal $I$, the quotient vector space $\mathfrak{g}/I$ has Lie bracket $[x+I,y+I]=[x,y]+I$. The ideal condition makes this independent of representatives. Its Lie algebra ideals correspond to the ideals of $\mathfrak{g}$ containing $I$, by inverse image under the quotient Lie algebra homomorphism.

A subset $C$ of a real vector space is radially open when, for each $x\in C$ and each vector $v$, the points $x+tv$ remain in $C$ for all sufficiently small real $t$. No ambient topology is required. If $C$ is also convex and contains zero, it is absorbing and its Minkowski functional $p_C(x)=\inf \{s>0:x\in sC\}$ is finite and sublinear, with $C=\{p_C<1\}$. Convexity gives subadditivity; radial openness permits a slight dilation of any point of $C$, giving the strict inequality.

Let $U,V$ be nonzero finite-dimensional complex vector spaces, and let $\varphi :U\otimes V\to W$ have injective restrictions when either nonzero factor is fixed. It induces a map from ${\mathbb{CP}}^{a-1}\times {\mathbb{CP}}^{b-1}$ to ${\mathbb{CP}}^{r-1}$, where $a=\dim U$, $b=\dim V$, and $r=\mathrm{rank}\varphi$. Pullback of the degree-two generator is $x+y$, since both factor restrictions are projective linear embeddings. Its $(a+b-2)$th power is $\left(\genfrac{}{}{0ex}{}{a+b-2}{a-1}\right)x^{a-1}{y}^{b-1}\ne 0$ in integral cohomology. The target must therefore have complex dimension at least $a+b-2$. Positivity of both factor dimensions is necessary; a zero factor can make the claimed inequality false.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

The rank of a vector bundle at a point is the dimension of its fiber as a vector space. A vector bundle trivialization identifies nearby fibers with the same vector space, so this rank is locally constant. A rank-$r$ bundle has fibers of dimension $r$ everywhere.

In the simply connected homotopy category, a rational space is a simply connected CW complex whose positive homotopy groups are vector spaces over $\mathbb{Q}$. Multiplication by every nonzero integer on these groups is invertible. More generally one uses nilpotent spaces with localized fundamental group and higher homotopy groups; the simply connected version avoids local-coefficient twisting.

If a real linear operator $A$ on an $n$-dimensional vector space has $n$ distinct real eigenvalues, any real square root $B$ commutes with $A$ and preserves each one-dimensional eigenspace. There $B$ is multiplication by a real $b_i$, hence ${\lambda }_i=b_i^2\ge 0$. Conversely, if all the eigenvalues are nonnegative, choosing $b_i=\sqrt{{\lambda }_i}$ on an eigenbasis supplies a real square root. Without simple spectrum the necessity fails: a real quarter-turn squares to $-I$.

A real vector space is a vector space over the field $\mathbb{R}$ of real numbers.