Vector Spaces and Subspaces

The abstract subspace is quietly running much of modern computation, and often the whole point is that a set of interest turns out to be a subspace. In coding theory, a binary linear error-correcting code is literally a subspace of $\mathbb{F}_2^n$ over the two-element field: closure under addition means the sum of any two codewords is again a codeword, and that is exactly what lets a receiver detect and correct bit errors — the Hamming, Reed–Muller, and low-density parity-check codes protecting every hard drive, QR code, and 5G link are all subspaces. In machine learning, principal-component analysis finds the best-fit low-dimensional subspace through a cloud of data, and "subspace methods" like MUSIC split a space into orthogonal signal and noise subspaces to resolve directions of arrival in radar and antenna arrays. In large-scale numerical linear algebra, Krylov-subspace solvers — conjugate gradient, GMRES, Lanczos — attack linear systems and eigenvalue problems with billions of unknowns by searching a cleverly grown sequence of nested subspaces rather than the whole space, the engines behind every large physics simulation. And in quantum information, the code space of a quantum error-correcting code hides fragile logical qubits inside a carefully chosen subspace of a much larger Hilbert space.

Think of the eight axioms as the rules of a self-consistent arithmetic of directions. Two displacements combine tip-to-tail into a third; any displacement can be stretched, shrunk, or reversed by a number. The axioms only demand that this combining never contradict itself — the order in which you add, the way scalars cascade through a product, must always agree. A subspace is then a closed neighbourhood of the origin: a region you can never escape by adding two of its members or by scaling one of them. In $\mathbb{R}^3$ that constraint is astonishingly rigid. The only subspaces are the origin alone, every line through the origin, every plane through the origin, and the whole of $\mathbb{R}^3$ — dimensions $0,1,2,3$. Nothing offset from the origin can qualify, because closure forces $\mathbf{0}$ to belong. The first figure lays out how these nest into a single ordered structure, the subspace lattice.

The lattice hides a warning that the second figure makes vivid. Intersections of subspaces are always subspaces, and so are sums — but unions almost never are. Take the two coordinate axes in $\mathbb{R}^2$. Each axis, on its own, is a perfectly good one-dimensional subspace: closed under addition, closed under scaling, containing the origin. Their union looks harmless, yet it fails the very first test of closure. Pick the vector $(1,0)$ on the $x$-axis and $(0,1)$ on the $y$-axis; both lie in the union, but their sum $(1,1)$ lies on neither axis and so escapes the union entirely. Closure under addition is broken, and a union of subspaces is a subspace only in the degenerate case where one already contains the other. This single picture is worth memorising: it is the fastest way to see why "closed under linear combination" is a genuinely restrictive demand and not mere bookkeeping.

A vector space is defined by eight axioms, and it pays to see them grouped by what they do. The first four make addition into an abelian group: it is commutative and associative, it has an identity $\mathbf{0}$, and every vector has an additive inverse. Written together they are the entire content of "you can add vectors and undo the addition".

where: $\mathbf{u},\mathbf{v},\mathbf{w}\in V$ — arbitrary vectors; $+:\,V\times V\to V$ — vector addition; $\mathbf{0}$ — the zero vector (additive identity); $-\mathbf{v}$ — the additive inverse of $\mathbf{v}$. Reading: $(V,+)$ is an abelian group, so sums may be reordered, regrouped, and cancelled at will. The remaining four axioms tie the field of scalars to the vectors. Scalar multiplication distributes over both vector addition and scalar addition, it is associative with the field's own multiplication, and the field's multiplicative identity $1$ acts trivially. These are the compatibility laws that make scalars and vectors cooperate.

where: $\alpha,\beta\in\mathbb{F}$ — scalars in the field; $\alpha\beta$ — the product taken in $\mathbb{F}$; $1\in\mathbb{F}$ — the multiplicative identity. Reading: the two distributive laws are what let you expand and factor linear combinations exactly as in ordinary algebra. Several facts that look like axioms are actually theorems forced by the eight above. The zero scalar annihilates any vector, any scalar annihilates the zero vector, and multiplying by $-1$ produces the additive inverse. Derivation B proves the first and third from scratch.

where: $0\in\mathbb{F}$ — the zero scalar; $\mathbf{0}\in V$ — the zero vector. Reading: nothing here is assumed — each identity is squeezed out of distributivity plus the uniqueness of inverses, which is why they hold in every vector space without exception. For a subset the eight axioms collapse into one. A nonempty $W\subseteq V$ is a subspace exactly when it is closed under linear combinations; this single condition is the workhorse of the entire subject, tested far more often than the full definition.

where: $W$ — the candidate subset; $\iff$ — logical equivalence. Reading: closure under one two-parameter combination simultaneously secures $\mathbf{0}\in W$ (take $\alpha=\beta=0$), closure under addition ($\alpha=\beta=1$), scalar closure ($\beta=0$), and inverses ($\alpha=-1$) — Derivation A spells this out. The subspace generated by a set is its span: all finite linear combinations of its members. The span is always the smallest subspace containing the set, which is why it is the bridge from "some vectors" to "a whole subspace".

where: $S\subseteq V$ — any set of vectors; $k$ — a finite number of terms; $\operatorname{span}(S)$ — the smallest subspace containing $S$. Reading: adding a vector already in $\operatorname{span}(S)$ changes nothing, which is precisely the redundancy that linear independence will later measure. Subspaces combine into new subspaces by intersection and sum, and their dimensions are linked by an exact accounting rule. The union is deliberately absent — it fails, as the second figure showed.

where: $U,W$ — subspaces of a common $V$; $U+W$ — the smallest subspace containing both; the dimension identity (Grassmann's formula) generalises inclusion–exclusion to subspaces. Reading: two planes in $\mathbb{R}^3$ ($2+2$) sharing only a line ($\dim 1$) sum to all of $\mathbb{R}^3$ since $2+2-1=3$.

The abstract definition specialises to a zoo of concrete spaces; recognising the same eight-axiom skeleton inside each is the skill being built here.

1 · A plane through the origin versus an affine plane (fully worked)

In $\mathbb{R}^3$, decide whether $W=\{(x,y,z):x+2y-z=0\}$ is a subspace, find its dimension, and contrast it with the shifted set $W'=\{(x,y,z):x+2y-z=1\}$.

Step 1. Check the origin: $0+2(0)-0=0$, so $\mathbf{0}\in W$. For $W'$, $0+2(0)-0=0\neq 1$, so $\mathbf{0}\notin W'$ and $W'$ is disqualified at once — it is an affine plane, not a subspace.

Step 2. Test closure in $W$ with concrete vectors. Take $\mathbf{u}=(1,0,1)$ (since $1+0-1=0$) and $\mathbf{v}=(2,1,4)$ (since $2+2-4=0$). Their sum $\mathbf{u}+\mathbf{v}=(3,1,5)$ gives $3+2-5=0$, so $\mathbf{u}+\mathbf{v}\in W$. Scaling, $3\mathbf{u}=(3,0,3)$ gives $3+0-3=0$, so $3\mathbf{u}\in W$. Both operations stay inside $W$.

Step 3. Prove closure in general: if $x_1+2y_1-z_1=0$ and $x_2+2y_2-z_2=0$, then for any $\alpha,\beta$ the combination $\alpha\mathbf{u}+\beta\mathbf{v}$ has coordinate sum $\alpha(x_1+2y_1-z_1)+\beta(x_2+2y_2-z_2)=\alpha\cdot 0+\beta\cdot 0=0$. The single linear constraint is preserved by every linear combination, so $W$ is a subspace.

Step 4. Dimension: solve the constraint as $x=z-2y$ with $y,z$ free. Then $(x,y,z)=y(-2,1,0)+z(1,0,1)$, so $W=\operatorname{span}\{(-2,1,0),(1,0,1)\}$ with two free parameters. $W$ is a plane through the origin of dimension $2$ — one homogeneous equation cuts $\mathbb{R}^3$ down by exactly one dimension, from $3$ to $2$.

2 · A subspace of polynomials (fully worked)

Let $P_3$ be the polynomials of degree $\le 3$ (so $\dim P_3=4$). Is $W=\{p\in P_3:p(1)=0\}$ a subspace? Find its dimension and a basis.

Step 1. The zero polynomial has $p(1)=0$, so $\mathbf{0}\in W$. If $p(1)=0$ and $q(1)=0$, then $(\alpha p+\beta q)(1)=\alpha p(1)+\beta q(1)=\alpha\cdot 0+\beta\cdot 0=0$, so $\alpha p+\beta q\in W$. $W$ is a subspace — evaluation at a point is a linear condition, and one linear condition always cuts out a subspace.

Step 2. Characterise the members: $p(1)=0$ means $(x-1)$ divides $p$, so every $p\in W$ has the form $p(x)=(x-1)q(x)$ with $\deg q\le 2$. The factor $q$ ranges over the $3$-dimensional space $P_2$.

Step 3. Read off a basis. The elements $x-1$, $x^2-1$, and $x^3-1$ all vanish at $x=1$ and are linearly independent (distinct degrees). Any $p\in W$ is a combination of them, so $W=\operatorname{span}\{x-1,\ x^2-1,\ x^3-1\}$. $\dim W=3$.

Step 4. Sanity check by counting: imposing the single independent linear constraint $p(1)=0$ on the $4$-dimensional $P_3$ removes exactly one dimension, giving $4-1=3$ — consistent with the basis found. One scalar constraint, one dimension lost.

3 · Two planes meet in a line (dimension bookkeeping)

In $\mathbb{R}^3$ let $U=\{z=0\}$ and $W=\{x=0\}$, both planes through the origin. Find $U\cap W$, $U+W$, and verify Grassmann's formula.

Step 1. $U\cap W$ requires $z=0$ and $x=0$ simultaneously, leaving the $y$-axis: $U\cap W=\operatorname{span}\{(0,1,0)\}$, a line of dimension $1$.

Step 2. $U+W$ contains $(1,0,0)$ and $(0,1,0)$ from $U$ and $(0,1,0),(0,0,1)$ from $W$, hence all three axis directions: $U+W=\mathbb{R}^3$, dimension $3$.

Step 3. Check the identity $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$: the right-hand side is $2+2-1=3$, matching $\dim(U+W)=3$. The books balance: $2+2-1=3$. This is the linear-algebra echo of two homogeneous constraints intersecting — and the same accounting governs the solution spaces of coupled homogeneous differential equations.

The vector-space axioms feel timeless, but they were written down astonishingly late — long after physicists were already computing with vectors. The person who first set them out in essentially their modern form was a logician obsessed with rigour, working from an even more obscure predecessor's ideas.

Texts. Axler, Linear Algebra Done Right, 4th ed., Ch. 1–2 (axiomatic vector spaces and subspaces) · Halmos, Finite-Dimensional Vector Spaces, §1–§10 · Stone & Goldbart, Mathematics for Physics, Ch. 1 · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 3 · Hassani, Mathematical Physics, Ch. 2 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 8 · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 3.

Provenance. The eight axioms (4 additive + 4 scalar), the derived identities, and the subspace test are cross-checked against Axler Ch. 1–2, Halmos §1–§10, and Stone & Goldbart Ch. 1. A verifier should confirm: (1) the four abelian-group axioms and four scalar-compatibility axioms; (2) the derivations $0\mathbf{v}=\mathbf{0}$ and $(-1)\mathbf{v}=-\mathbf{v}$; (3) the single-condition subspace test and that it forces $\mathbf{0}\in W$; (4) the $\mathbb{R}^2$ counterexample $(1,0)+(0,1)=(1,1)$ for unions; (5) the worked plane $x+2y-z=0$ has dimension $2$; (6) $W=\{p\in P_3:p(1)=0\}$ has dimension $3$ with basis $\{x-1,x^2-1,x^3-1\}$; (7) Grassmann's formula $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$ giving $2+2-1=3$. Dimensions quoted ($\dim P_n=n+1$, $\dim$ of $m\times n$ matrices $=mn$) are standard.

Prerequisites. Fields and basic algebra · Set notation and functions. Leads to. Linear Independence, Basis, and Dimension · Linear Operators and Matrix Representations · Inner-Product Spaces and Norms · Rank, Nullity, and the Fundamental Theorem of Linear Maps.

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