Vector Spaces: Definition and Axioms

What do arrows in the plane, polynomials, and lists of numbers have in common? Each can be added together and stretched or shrunk by a scalar. These two operations always obey the same handful of arithmetic rules. Vector space theory isolates exactly those rules. It discards everything else about arrows, polynomials, or lists. This lets a single body of theorems apply to all of them at once. This is the founding abstraction of linear algebra. Instead of studying each concrete family of objects separately, we study any set. That set is equipped with an addition and a scalar multiplication satisfying a short list of axioms. The payoff is enormous economy of thought. Once a theorem is proved from the axioms alone, it holds everywhere. For instance, the zero vector is unique, and 0⋅v=00 \cdot v = 0 for every vector vv. That theorem is simultaneously true for Rn\mathbb{R}^n, for spaces of matrices, and for spaces of continuous functions. It also holds for spaces of polynomials, and for every other structure anyone will ever check against the axioms. This is why the definition, though it looks like a bureaucratic checklist at first glance, is one of the most productive definitions in all of mathematics. The scalars are drawn from a field F\mathbb{F}, most commonly R\mathbb{R} or C\mathbb{C}. This choice matters: the same set of vectors can be a real vector space or a complex vector space. Their behavior can genuinely differ. The vector space axioms say nothing about dimension, coordinates, or bases. Those are later refinements built on top of this bare-bones structure. What the axioms do guarantee is a coherent, cancellation-friendly arithmetic of vectors. It is robust enough to support everything from linear maps to eigenvalues to inner products. This article gives the precise definition of a vector space over a field. It develops the immediate logical consequences of the axioms. These include uniqueness of zero and additive inverses, and the laws 0v=00v = 0 and (−1)v=−v(-1)v = -v. Each fact is proved directly from the axioms, with no appeal to coordinates or examples. It sets up the vocabulary needed for Examples of Vector Spaces and for the notion of a Subspace. These are the two natural next steps in building up the theory.

Key takeaways
  • A vector space is a set VV with an addition and a scalar multiplication (scalars from a field F\mathbb{F}) satisfying ten axioms.
  • The axioms split into two groups: (V,+)(V, +) is an abelian group, and scalar multiplication interacts with ++ via distributivity and compatibility laws.
  • Uniqueness of the zero vector and of additive inverses follows purely from the group axioms, not extra assumptions.
  • Two "obvious" facts, 0⋅v=0V0 \cdot v = 0_V and (−1)v=−v(-1)v = -v, are theorems, not axioms — they must be derived.
  • The choice of scalar field is part of the data: Cn\mathbb{C}^n over R\mathbb{R} and Cn\mathbb{C}^n over C\mathbb{C} are different vector spaces on the same underlying set.
  • Key idea: a vector space is any set where "add vectors" and "scale by a number" behave exactly as they do for arrows in the plane — nothing more is assumed, and nothing less will do.

Where this lives today

The vector space axioms are the load-bearing foundation beneath essentially all of linear algebra and much of applied mathematics. In physics, quantum states live in complex vector spaces (Hilbert spaces), and the entire formalism of superposition rests on the additivity and scalar-multiplication axioms proved here. In computer science, data structures such as feature vectors in machine learning, images represented as pixel arrays, and word embeddings in natural language processing are all treated as elements of a vector space so that linear algebra algorithms (projections, least squares, principal component analysis) apply directly. In cryptography, vector spaces over finite fields Fp\mathbb{F}_p underlie coding theory and lattice-based cryptographic schemes, where the "no zero divisors" property (part (f)) and field arithmetic are essential to correctness proofs. In engineering, state-space representations of linear systems and signal spaces in Fourier analysis are vector spaces, often infinite-dimensional, where the same eight axioms — verified once, abstractly — justify manipulations performed routinely in control theory and signal processing. In pure mathematics, the axioms generalize directly to modules over a ring (as the counterexample above illustrates), linking linear algebra to abstract algebra, algebraic geometry, and representation theory.

Ask the tutor: Why does the proof that (−1)v=−v(-1)v = -v need the uniqueness of additive inverses, and would the theorem still make sense without it?

DefinitionVector space over a field

Let F\mathbb{F} be a field (e.g., R\mathbb{R} or C\mathbb{C}). A vector space over F\mathbb{F} is a set VV together with two operations:

  • addition: a map +:V×V→V+ : V \times V \to V, written (u,v)↦u+v(u,v) \mapsto u+v;
  • scalar multiplication: a map ⋅:F×V→V\cdot : \mathbb{F} \times V \to V, written (λ,v)↦λv(\lambda, v) \mapsto \lambda v;

such that the following ten axioms hold for all u,v,w∈Vu, v, w \in V and all λ,μ∈F\lambda, \mu \in \mathbb{F}: 1. (Commutativity of addition) u+v=v+uu + v = v + u. 2. (Associativity of addition) (u+v)+w=u+(v+w)(u+v)+w = u+(v+w). 3. (Additive identity) There exists an element 0V∈V0_V \in V such that v+0V=vv + 0_V = v for all v∈Vv \in V. 4. (Additive inverses) For every v∈Vv \in V there exists w∈Vw \in V such that v+w=0Vv + w = 0_V. 5. (Associativity of scalar multiplication) (λμ)v=λ(μv)(\lambda\mu)v = \lambda(\mu v). 6. (Multiplicative identity) 1v=v1v = v, where 11 is the multiplicative identity of F\mathbb{F}. 7. (Distributivity over vector addition) λ(u+v)=λu+λv\lambda(u+v) = \lambda u + \lambda v. 8. (Distributivity over scalar addition) (λ+μ)v=λv+μv(\lambda + \mu)v = \lambda v + \mu v. (Axioms are traditionally numbered up to 8 with commutativity/associativity of ++ and distributivity counted as above; some texts list closure of ++ and ⋅\cdot as separate axioms, giving a count of ten. Both conventions describe the same structure.) Elements of VV are called vectors; elements of F\mathbb{F} are called scalars. If F=R\mathbb{F} = \mathbb{R}, VV is a real vector space; if F=C\mathbb{F} = \mathbb{C}, VV is a complex vector space.

Intuition

Think first of arrows in the plane based at the origin. You can add two arrows tip-to-tail (the parallelogram rule), and you can stretch or reverse an arrow by multiplying by a real number. Every algebraic property you would ever want from "adding arrows" and "scaling arrows" — that order doesn't matter, that scaling twice is the same as scaling by the product, that scaling a sum distributes — is exactly one of the axioms above. The abstract definition is nothing but a faithful transcription of what your hand already does when you draw vectors on paper. The genius of the abstraction is that the same transcription applies to objects that don't look like arrows at all. A polynomial like 3x2−x+53x^2 - x + 5 can be added to another polynomial and scaled by a real number, and the resulting arithmetic satisfies precisely the same eight rules. So can an m×nm \times n matrix. So can a sequence (a1,a2,a3,… )(a_1, a_2, a_3, \dots) of real numbers. So can a continuous function f:[0,1]→Rf: [0,1] \to \mathbb{R}. Once you prove a fact from the axioms alone — say, that there is only one zero vector — you get that fact for free in every one of these settings simultaneously, without re-deriving it each time. It helps to keep two pictures in mind side by side: the concrete geometric picture of R2\mathbb{R}^2, where addition is the parallelogram law and scalar multiplication is stretching, and the abstract algebraic picture, where VV is just a set with two symbols ++ and ⋅\cdot satisfying a list of equations. The theorems proved from the axioms are statements about the second picture, but their content is calibrated so that they match your intuition from the first.

Figure
The parallelogram law for vector addition in R2\mathbb{R}^2, illustrating axioms 1–4
x y 0 u v u + v
TheoremElementary consequences of the vector space axioms

Let VV be a vector space over a field F\mathbb{F}. Then: (a) (Uniqueness of the additive identity) There is exactly one element 0V∈V0_V \in V satisfying axiom (3); i.e., if 0V0_V and 0V′0_V' both satisfy v+0V=vv + 0_V = v and v+0V′=vv + 0_V' = v for all v∈Vv \in V, then 0V=0V′0_V = 0_V'. (b) (Uniqueness of additive inverses) For each v∈Vv \in V there is exactly one w∈Vw \in V with v+w=0Vv + w = 0_V. This unique ww is denoted −v-v. (c) (Zero scalar law) 0F v=0V0_{\mathbb{F}} \, v = 0_V for every v∈Vv \in V, where 0F0_{\mathbb{F}} is the additive identity of F\mathbb{F}. (d) (Zero vector law) λ 0V=0V\lambda \, 0_V = 0_V for every λ∈F\lambda \in \mathbb{F}. (e) (Sign law) (−1)v=−v(-1) v = -v for every v∈Vv \in V, where −1-1 is the additive inverse of 11 in F\mathbb{F}. (f) (No zero divisors) If λ∈F\lambda \in \mathbb{F}, v∈Vv \in V, and λv=0V\lambda v = 0_V, then λ=0F\lambda = 0_\mathbb{F} or v=0Vv = 0_V. (g) (Cancellation) If u,v,w∈Vu, v, w \in V and u+w=v+wu + w = v + w, then u=vu = v.

Proof

Throughout, we use only the eight axioms and previously established parts of this theorem; no coordinates, bases, or examples are used. This is a direct algebraic proof from the axioms.

(a) Uniqueness of 0V0_V. Suppose 0V0_V and 0V′0_V' are both additive identities, so that for every v∈Vv \in V,

v+0V=vandv+0V′=v.v + 0_V = v \quad \text{and} \quad v + 0_V' = v.

Apply the first identity with v=0V′v = 0_V': 0V′+0V=0V′0_V' + 0_V = 0_V'. Apply the second identity with v=0Vv = 0_V: 0V+0V′=0V0_V + 0_V' = 0_V. By commutativity (axiom 1), 0V′+0V=0V+0V′0_V' + 0_V = 0_V + 0_V'. Chaining these three equalities,

0V=0V+0V′=0V′+0V=0V′.0_V = 0_V + 0_V' = 0_V' + 0_V = 0_V'.

Hence 0V=0V′0_V = 0_V', proving uniqueness.

(b) Uniqueness of additive inverses. Fix v∈Vv \in V and suppose w,w′∈Vw, w' \in V both satisfy v+w=0Vv + w = 0_V and v+w′=0Vv + w' = 0_V. Using associativity (axiom 2), commutativity (axiom 1), and the identity axiom (axiom 3):

w=w+0V=w+(v+w′)=(w+v)+w′=(v+w)+w′=0V+w′=w′+0V=w′.w = w + 0_V = w + (v + w') = (w+v) + w' = (v + w) + w' = 0_V + w' = w' + 0_V = w'.

Every equality above uses only axioms 1–4 (commutativity, associativity, identity, and the hypotheses v+w=0Vv+w=0_V, v+w′=0Vv+w'=0_V). Hence w=w′w = w', so the additive inverse is unique; we denote it −v-v.

(c) 0Fv=0V0_{\mathbb F} v = 0_V. Since 0F=0F+0F0_\mathbb{F} = 0_\mathbb{F} + 0_\mathbb{F} in the field F\mathbb{F}, distributivity over scalar addition (axiom 8) gives

0Fv=(0F+0F)v=0Fv+0Fv.0_\mathbb{F} v = (0_\mathbb{F} + 0_\mathbb{F}) v = 0_\mathbb{F} v + 0_\mathbb{F} v.

Let x=0Fv∈Vx = 0_\mathbb{F} v \in V; the displayed equation reads x=x+xx = x + x. Add −x-x (which exists by axiom 4, and is unique by part (b)) to both sides:

x+(−x)=(x+x)+(−x).x + (-x) = (x+x) + (-x).

The left side is 0V0_V by axiom 4. On the right side, associativity (axiom 2) gives (x+x)+(−x)=x+(x+(−x))=x+0V=x(x+x)+(-x) = x + (x + (-x)) = x + 0_V = x by axiom 4 and axiom 3. Hence 0V=x=0Fv0_V = x = 0_\mathbb{F} v, as claimed.

(d) λ0V=0V\lambda 0_V = 0_V. Since 0V+0V=0V0_V + 0_V = 0_V by axiom 3 (applying the identity property to v=0Vv = 0_V), distributivity over vector addition (axiom 7) gives

λ0V=λ(0V+0V)=λ0V+λ0V.\lambda 0_V = \lambda(0_V + 0_V) = \lambda 0_V + \lambda 0_V.

Set y=λ0Vy = \lambda 0_V; then y=y+yy = y + y. Exactly as in part (c), adding −y-y to both sides and using axioms 2, 3, 4 yields 0V=y=λ0V0_V = y = \lambda 0_V.

(e) (−1)v=−v(-1)v = -v. We must show (−1)v(-1)v is the additive inverse of vv, i.e., that v+(−1)v=0Vv + (-1)v = 0_V; uniqueness from part (b) then forces (−1)v=−v(-1)v = -v. Compute, using axiom 6 (1v=v1v = v), axiom 8 (distributivity over scalar addition), the field fact 1+(−1)=0F1 + (-1) = 0_\mathbb{F}, and part (c):

v+(−1)v=1v+(−1)v=(1+(−1))v=0Fv=0V.v + (-1)v = 1v + (-1)v = (1 + (-1)) v = 0_\mathbb{F} v = 0_V.

Thus (−1)v(-1)v satisfies the defining property of −v-v, and by the uniqueness established in part (b), (−1)v=−v(-1)v = -v.

(f) No zero divisors. Suppose λv=0V\lambda v = 0_V and λ≠0F\lambda \neq 0_\mathbb{F}. Since F\mathbb{F} is a field, λ\lambda has a multiplicative inverse λ−1∈F\lambda^{-1} \in \mathbb{F}. Using associativity of scalar multiplication (axiom 5) and axiom 6:

v=1v=(λ−1λ)v=λ−1(λv)=λ−10V.v = 1v = (\lambda^{-1} \lambda) v = \lambda^{-1}(\lambda v) = \lambda^{-1} 0_V.

By part (d) (applied with scalar λ−1\lambda^{-1}), λ−10V=0V\lambda^{-1} 0_V = 0_V. Hence v=0Vv = 0_V. This shows: if λv=0V\lambda v = 0_V and λ≠0F\lambda \ne 0_\mathbb{F}, then v=0Vv = 0_V; equivalently, λ=0F\lambda = 0_\mathbb{F} or v=0Vv = 0_V.

(g) Cancellation. Suppose u+w=v+wu + w = v + w. Add −w-w (which exists by axiom 4) to both sides and use associativity (axiom 2):

u=u+0V=u+(w+(−w))=(u+w)+(−w)=(v+w)+(−w)=v+(w+(−w))=v+0V=v.u = u + 0_V = u + (w + (-w)) = (u+w) + (-w) = (v+w) + (-w) = v + (w + (-w)) = v + 0_V = v.

Each step uses only axioms 2, 3, 4. Hence u=vu = v. ■\blacksquare

∎

Worked examples

Example 1: A genuine vector space with a twist — upper-triangular 2×22\times 2 real matrices under nonstandard-looking but valid operations. Let

V={(ab0c):a,b,c∈R},V = \left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} : a, b, c \in \mathbb{R} \right\},

with ordinary matrix addition and ordinary scalar multiplication by real numbers. We verify VV is a real vector space directly from the axioms, paying attention to closure (implicit in the maps +:V×V→V+: V\times V \to V and ⋅:R×V→V\cdot: \mathbb{R}\times V \to V being well-defined). Closure of ++: if A=(ab0c)A = \begin{pmatrix} a & b \\ 0 & c\end{pmatrix} and A′=(a′b′0c′)A' = \begin{pmatrix} a' & b' \\ 0 & c'\end{pmatrix} are both upper triangular, then

A+A′=(a+a′b+b′0c+c′)A + A' = \begin{pmatrix} a+a' & b+b' \\ 0 & c+c' \end{pmatrix}

is again upper triangular (the (2,1)(2,1)-entry is 0+0=00+0=0), so A+A′∈VA + A' \in V. Closure of scalar multiplication: for λ∈R\lambda \in \mathbb{R}, λA=(λaλb0λc)∈V\lambda A = \begin{pmatrix} \lambda a & \lambda b \\ 0 & \lambda c\end{pmatrix} \in V since the (2,1)(2,1)-entry is λ⋅0=0\lambda \cdot 0 = 0. Axioms 1, 2, 5–8: these are inherited entrywise from the fact that R\mathbb{R} itself satisfies commutativity, associativity, and distributivity (each matrix entry is a real number, and matrix addition/scaling acts entrywise), so, for instance,

A+A′=(a+a′b+b′0c+c′)=(a′+ab′+b0c′+c)=A′+AA + A' = \begin{pmatrix} a+a' & b+b'\\ 0& c+c'\end{pmatrix} = \begin{pmatrix} a'+a & b'+b \\ 0 & c'+c\end{pmatrix} = A' + A

using commutativity of real addition entry by entry. Analogous entrywise arguments verify axioms 2, 5, 6, 7, 8. Axiom 3 (identity): take 0V=(0000)∈V0_V = \begin{pmatrix} 0&0\\0&0\end{pmatrix} \in V; then A+0V=AA + 0_V = A for all A∈VA \in V. Axiom 4 (inverses): given A=(ab0c)A = \begin{pmatrix} a&b\\0&c\end{pmatrix}, define −A=(−a−b0−c)∈V-A = \begin{pmatrix}-a & -b \\ 0 & -c\end{pmatrix} \in V; then A+(−A)=0VA + (-A) = 0_V. All ten conditions hold, so VV is a real vector space. Note the specific numerical instance: take

A=(3−207),λ=−4.A = \begin{pmatrix} 3 & -2 \\ 0 & 7\end{pmatrix}, \qquad \lambda = -4.

By the theorem's part (e), (−1)A=−A=(−320−7)(-1)A = -A = \begin{pmatrix} -3 & 2\\0&-7\end{pmatrix}, which one checks directly satisfies A+(−1)A=(0000)=0VA + (-1)A = \begin{pmatrix}0&0\\0&0\end{pmatrix} = 0_V. Also λA=−4(3−207)=(−1280−28)\lambda A = -4\begin{pmatrix}3&-2\\0&7\end{pmatrix} = \begin{pmatrix}-12 & 8\\0&-28\end{pmatrix}, consistent with entrywise scalar multiplication.

Example 2: Applying the theorem's cancellation law to solve a vector equation numerically, in R3\mathbb{R}^3. Let V=R3V = \mathbb{R}^3 over F=R\mathbb{F} = \mathbb{R}, with coordinatewise addition and scaling — a standard example whose axiom verification is entrywise exactly as in Example 1, and which we take as already established. Suppose u,v,w∈R3u, v, w \in \mathbb{R}^3 satisfy

u+w=v+w,where w=(2,−5,9).u + w = v + w, \qquad \text{where } w = (2, -5, 9).

Concretely, say we are told

u+(2,−5,9)=(7,−1,3)andv+(2,−5,9)=(7,−1,3).u + (2,-5,9) = (7, -1, 3) \quad\text{and}\quad v + (2,-5,9) = (7,-1,3).

We want to conclude u=vu = v without solving for uu and vv individually first — using only the abstract cancellation law from part (g) of the theorem, which is valid because R3\mathbb{R}^3 satisfies all eight vector space axioms. By hypothesis u+w=v+wu + w = v+w with w=(2,−5,9)w = (2,-5,9). Part (g) gives immediately u=vu = v, with no further computation needed. As a numerical check, we may still solve directly: adding −w=(−2,5,−9)-w = (-2,5,-9) to both sides,

u=(7,−1,3)+(−2,5,−9)=(5,4,−6),u = (7,-1,3) + (-2,5,-9) = (5, 4, -6),

and identically v=(7,−1,3)+(−2,5,−9)=(5,4,−6)v = (7,-1,3)+(-2,5,-9) = (5,4,-6). Indeed u=v=(5,4,−6)u = v = (5,4,-6), confirming the abstract cancellation argument matches direct computation. This illustrates the value of the theorem: cancellation lets us reason about equality of vectors structurally, via axioms 2–4 alone, rather than by re-deriving it coordinate by coordinate every time.

CounterexampleWhy the field axioms on scalars cannot be dropped: replacing F\mathbb{F} by Z\mathbb{Z}

The theorem's proof of part (f), "no zero divisors," used that F\mathbb{F} is a field — specifically, that every nonzero scalar has a multiplicative inverse. Drop that hypothesis by letting the "scalars" come from Z\mathbb{Z} (the integers, which form only a commutative ring, not a field: 22 has no multiplicative inverse in Z\mathbb{Z}) instead of from a field, keeping V=ZV = \mathbb{Z} itself with ordinary addition and multiplication as "scalar multiplication." One can check axioms 1–8 verbatim still hold with F\mathbb{F} replaced by Z\mathbb{Z} (this structure is called a Z\mathbb{Z}-module, not a vector space). But now consider λ=2∈Z\lambda = 2 \in \mathbb{Z} and v=3∈Zv = 3 \in \mathbb{Z}. Working modulo nothing — just in the additive group V=ZV=\mathbb{Z} — take instead V=Z/6ZV = \mathbb{Z}/6\mathbb{Z} (integers mod 6) as the vector-like set, with scalars also from Z\mathbb{Z} acting by λ⋅v=(λv)  6\lambda \cdot v = (\lambda v) \bmod 6. Then with λ=2\lambda = 2 and v=3v = 3:

λv=2⋅3=6≡0(mod6),\lambda v = 2 \cdot 3 = 6 \equiv 0 \pmod 6,

so λv=0V\lambda v = 0_V, yet λ=2≠0\lambda = 2 \neq 0 and v=3≠0Vv = 3 \neq 0_V in Z/6Z\mathbb{Z}/6\mathbb{Z}. This directly violates conclusion (f) of the theorem: we have found λ≠0\lambda \ne 0 and v≠0Vv \ne 0_V with λv=0V\lambda v = 0_V. The proof of (f) breaks exactly where expected: it invoked "λ≠0F\lambda \ne 0_\mathbb{F} implies λ\lambda has a multiplicative inverse," which is true in a field but false for λ=2\lambda = 2 in Z\mathbb{Z} (there is no integer kk with 2k=12k=1), and false for λ=2\lambda = 2 acting on Z/6Z\mathbb{Z}/6\mathbb{Z} in the relevant sense. This shows the field hypothesis on the scalars is not a cosmetic convenience: without multiplicative inverses for nonzero scalars, zero divisors can and do appear, and part (f) of the theorem genuinely fails.

Scope, edge cases & common traps

SituationWhat the axioms sayCommon mistake
V={0V}V = \{0_V\}, the zero spaceThis trivially satisfies all eight axioms; it is a legitimate (if minimal) vector spaceAssuming a vector space must contain a nonzero vector
Scalars from Q\mathbb{Q}, R\mathbb{R}, C\mathbb{C}, or Fp\mathbb{F}_pAny field works; the same VV can be a vector space over different fields with different behaviorTreating "vector space" as automatically meaning real or complex
0Fv=0V0_\mathbb{F} v = 0_V vs. λ⋅0V=0V\lambda \cdot 0_V = 0_VTwo different theorems (parts (c) and (d)), both proved, neither an axiomConflating the two, or citing one as "the" zero law
(−1)v=−v(-1)v = -vA proved theorem (part (e)), not definitionalTreating −v-v as defined by (−1)v(-1)v rather than by axiom 4, then "proving" axiom 4 circularly
Scalar multiplication notation λv\lambda vAlways means the field element λ\lambda acting on the vector vv; never vector-times-vectorConfusing scalar multiplication with an undefined "vector times vector" product

The mathematician behind it

Origins · 1 minGiuseppe Peano and how this was discovered
GP
Giuseppe Peano
1858–1932 · Italian mathematician, foundational logician, and axiomatic pioneer

Giuseppe Peano is best remembered for the Peano axioms characterizing the natural numbers, but in 1888 he also gave, in Calcolo Geometrico, what is generally recognized as the first fully explicit axiomatic definition of an abstract vector space (which he called a "linear system"). Working in an era when mathematicians were newly obsessed with axiomatizing every corner of mathematics — arithmetic, geometry, set theory — Peano isolated the algebraic rules obeyed by geometric vectors and stated them as free-standing postulates, independent of any geometric picture. His formulation anticipated by decades the modern treatment found in every linear algebra course.

History
- 1844: Hermann Grassmann publishes Die lineale Ausdehnungslehre, developing a general theory of extended magnitudes that implicitly contains vector space ideas, decades ahead of its reception.
- 1888: Giuseppe Peano gives the first explicit axiomatic definition of a real vector space ("linear system") in Calcolo Geometrico.
- 1920s: Stefan Banach and Hans Hahn develop infinite-dimensional vector spaces with norms, feeding the axioms into functional analysis.
- 1930s: The Bourbaki school and contemporaries standardize the modern field-based axiomatic definition used in essentially all textbooks today.
- Present: The same ten axioms underlie linear algebra courses worldwide and generalize further to modules over rings.
Why do vector space axioms not mention "vectors times vectors"?

Because a vector space, by definition, only equips VV with addition of vectors and multiplication of a vector by a scalar from the field F\mathbb{F} — there is no operation multiplying two vectors together in the axioms. Structures with an extra vector-times-vector product (like the dot product or cross product) are additional structure layered on top of a vector space, not part of the base definition. The dot product, for instance, gives Rn\mathbb{R}^n the structure of an inner product space, which is a vector space plus an inner product. Keeping the base definition free of any vector-vector multiplication is precisely what makes it so widely applicable, since spaces like polynomials or matrices support addition and scalar multiplication but no single natural "vector times vector" operation in general.

Is {0V}\{0_V\}, the space containing only the zero vector, really a vector space?

Yes. Define 0V+0V=0V0_V + 0_V = 0_V and λ⋅0V=0V\lambda \cdot 0_V = 0_V for every scalar λ∈F\lambda \in \mathbb{F}. Every axiom is satisfied: commutativity and associativity of addition hold trivially since there is only one element to work with, 0V0_V itself serves as the additive identity, 0V0_V is its own additive inverse, and both distributive laws and the scalar-multiplication axioms hold because both sides of every equation reduce to 0V0_V. This "zero vector space" or "trivial vector space" is the smallest possible vector space over any field, and it plays the role of a base case in many inductive arguments about dimension.

Can the same set VV be a vector space over two different fields, and does that matter?

Yes, and it matters a great deal. The set C\mathbb{C} (complex numbers) can be viewed as a vector space over C\mathbb{C} itself, where it behaves like a one-dimensional space, or as a vector space over R\mathbb{R}, where the same set behaves like a two-dimensional space (spanned by 11 and ii). The underlying set and the addition operation are identical in both cases, but the scalar multiplication — which scalars are allowed to multiply vectors — differs, and this changes structural properties like dimension that are studied once bases are introduced. This is exactly why the definition insists on specifying the field F\mathbb{F} as part of the data of a vector space, not as an afterthought.

Why must scalars come from a field rather than just any set with addition and multiplication?

The proofs of several basic consequences of the axioms genuinely require field properties. The proof that "λv=0V\lambda v = 0_V implies λ=0\lambda = 0 or v=0Vv=0_V" (part (f) of the theorem) used that every nonzero scalar has a multiplicative inverse — a defining property of fields, absent in general rings like Z\mathbb{Z}. If scalars only form a ring, the resulting structure is called a module, and several familiar vector space facts (existence of a basis with a well-defined size, no zero divisors) can fail, as the counterexample with Z/6Z\mathbb{Z}/6\mathbb{Z} demonstrates. Requiring a field is what makes vector space theory, especially the later theory of dimension and bases, as clean as it is.

Does every vector space need infinitely many elements?

No. Over a finite field such as F2={0,1}\mathbb{F}_2 = \{0,1\} (arithmetic mod 2), the set F2n\mathbb{F}_2^n of length-nn binary strings, with coordinatewise addition and scalar multiplication mod 2, is a perfectly valid vector space with exactly 2n2^n elements — all axioms are verified entrywise exactly as for Rn\mathbb{R}^n. Such finite vector spaces are the backbone of coding theory and much of cryptography. The axioms themselves say nothing about cardinality; they only constrain the algebraic behavior of whatever elements VV happens to have.

How is "additive inverse" different from just writing v−wv-w for subtraction?

Subtraction is not a primitive operation in the axioms; it is defined using addition and additive inverses, via v−w:=v+(−w)v - w := v + (-w), where −w-w is the unique additive inverse of ww guaranteed to exist by axiom 4 and shown unique in part (b) of the theorem. Every vector space therefore automatically has a well-defined subtraction, but this is a derived convenience, not an extra axiom. This mirrors how subtraction is defined in terms of addition and negation in the integers or in any field.

If I can verify the ten axioms for some strange set of objects, am I guaranteed all of linear algebra applies to it?

Yes, in the sense that every theorem proved purely from the vector space axioms — including everything about linear independence, bases, dimension, linear maps, and eigenvalues developed later in the theory — will automatically hold for your strange set once you have honestly verified the axioms, with no further case-by-case checking needed. This transferability is the entire point of the axiomatic method: the axioms are the complete list of assumptions used everywhere downstream, so verifying them once licenses the full machinery of linear algebra, from Subspaces onward, on your particular set.

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