Abstract vector spaces — axioms, examples beyond ℝⁿ: Worked examples

Before anything, one word we use constantly: an operation is "closed" means when you do it to elements of the set, the result lands back inside the set. Picture a fenced field: closed = the ball never rolls out of the fence.

The scenario matrix

Every "is it a vector space?" question falls into one of these cells. Each worked example below is tagged with the cell it kills.

Cell	What goes wrong (or right)	Cheapest test
A. Passes cleanly	all 8 axioms + closure hold	verify closure, name $0$
B. No zero vector	the required $0$ is missing from the set	is there a $0$ ?
C. Not closed under $+$	sum leaves the set	pick two elements, add
D. Not closed under scaling	scaling leaves the set	scale by $- 1$ or $\frac{1}{2}$
E. Weird operations, still a space	$+$ or $\cdot$ redefined but axioms survive	find the hidden zero
F. Weird operations, an axiom breaks	redefined ops violate axiom 8 or 5	test distributivity / identity
G. Degenerate / limiting	the one-point space ${0}$ , or field changes	check trivial case
H. Word problem	a real system phrased in disguise (ODE, signals)	translate to axioms

Example 1 — Cell A: a clean pass (upper-triangular matrices)

Worked example Is

$V = {(a 0 b c) : a, b, c \in R}$ a vector space over $R$ ? Forecast: guess before reading — does the "zero below the diagonal" survive addition and scaling?

Step 1. Add two of them: $(a 0 b c) + (a^{'} 0 b^{'} c^{'}) = (a + a^{'} 0 b + b^{'} c + c^{'})$ . Why this step? Closure under $+$ (Axiom 0) is the cheapest disqualifier. The bottom-left entry is $0 + 0 = 0$ , so the result is still upper-triangular — closed.

Step 2. Scale: $r (a 0 b c) = (r a 0 r b r c)$ . Why? Test scaling closure (also Axiom 0). Bottom-left is $r \cdot 0 = 0$ — still inside.

Step 3. Name the zero: $0 = (0000)$ , which is upper-triangular, so $0 \in V$ (Axiom 3: zero vector ✓). The inverse of any element is its entrywise negation, also upper-triangular (Axiom 4: additive inverse ✓). Why? Axioms 3 and 4 are the next cheapest after closure.

Step 4. Axioms 1,2,5,6,7,8 (the two commutative/associative laws, scalar identity, compatibility, and both distributive laws) are all inherited from $M_{2 \times 2} (R)$ (which the parent note already certified) because our operations are exactly the matrix operations restricted to a subset. Why? If the big set obeys an algebra rule, any subset that stays inside obeys it too — this is the essence of the Subspaces and the subspace test.

Verify: $V$ is a 3-dimensional subspace with basis ${(1000), (0010), (0001)}$ . Dimension $3$ , matching the 3 free entries $a, b, c$ . ✓

Example 2 — Cell B: no zero vector (positive reals)

Worked example Is

$V = R_{> 0}$ (strictly positive reals) with ordinary $+$ and ordinary $\cdot$ a vector space? Forecast: you can add two positives and get a positive — feels closed. So where's the catch?

Step 1. Closure under $+$ : $2 + 3 = 5 > 0$ ✓. Looks fine. Why? Confirm the tempting-but-incomplete evidence, so you feel the trap.

Step 2. Hunt for $0$ : we need an element $z \in V$ with $v + z = v$ . Ordinary addition forces $z = 0$ . But $0 \in / R_{> 0}$ . Why? Axiom 3 (zero vector) demands the zero live inside the set. It doesn't.

Step 3. Even scaling breaks: $(- 1) \cdot 2 = - 2 \in / V$ — not closed under scaling either. Why? One failure is enough, but showing two makes the disqualification airtight.

Verify: Axioms 3 and 4 fail, closure under scaling fails. Not a vector space under these operations. (Sequel: Example 6 rescues the same set with cleverer operations — cell E.)

Example 3 — Cell C: not closed under addition (degree-exactly-2 polynomials)

Worked example Is

$V = {$ polynomials of degree exactly $2}$ a vector space? Forecast: $P_{2}$ was fine in the parent note — does insisting on degree exactly 2 change anything?

Step 1. Take $p (x) = x^{2} + x$ and $q (x) = - x^{2} + 3$ . Both have degree exactly 2. Why? To test closure you must actually pick two elements and add.

Step 2. $p (x) + q (x) = (x^{2} - x^{2}) + x + 3 = x + 3$ , which has degree $1$ , not $2$ . Why? The leading terms cancelled — the sum fell out of the set. Closure under $+$ (Axiom 0) fails.

Step 3. Also $0 \cdot p = 0$ , the zero polynomial, has degree $- \infty$ (by the convention above, not $2$ ) — so $0 \in / V$ too. Why? A second independent failure; scaling by $0$ escapes the set, killing Axiom 3.

Verify: Not a vector space. Contrast: degree $\leq 2$ (the " $\leq$ ") is fine because cancellation still leaves you $\leq 2$ . The word "exactly" is the whole trap. ✓

Example 4 — Cell D: closed under $+$ but not scaling (integer vectors)

Example 5 — Cell G: the degenerate one-point space

Worked example Is

$V = {0}$ (a set containing only the zero vector) a vector space? Forecast: a space with a single element sounds too small to count. Does it?

Step 1. Define $0 + 0 = 0$ and $r \cdot 0 = 0$ for every scalar $r$ . Why? There's only one element, so both operations have exactly one possible output — automatically closed (Axiom 0).

Step 2. Check the axioms: commutativity (Axiom 1), associativity (Axiom 2) are trivial (only one thing to combine). The zero is $0$ itself (Axiom 3 ✓); its inverse is $0$ (Axiom 4 ✓). Identity $1 \cdot 0 = 0$ (Axiom 5 ✓); distributivity (Axioms 7, 8) holds since every side equals $0$ . Why? Degenerate cases still must be checked — but here every check collapses to " $0 = 0$ ".

Verify: It is a vector space, the trivial (or zero) vector space, with dimension $0$ and empty basis $\emptyset$ . This is the limiting case: the smallest legal space. ✓ (Compare Linear independence, basis and dimension: the empty set spans ${0}$ .)

Example 6 — Cell E: weird operations that STILL form a space

Worked example On

$V = R_{> 0}$ define new operations: $u \oplus v = u \cdot v (old multiplication), r ⊙ u = u^{r} (old exponentiation) .$ Is $(V, \oplus, ⊙)$ a vector space over $R$ ? Forecast: the same positive reals that FAILED in Example 2 — can new operations save them? Where is the hidden zero?

Step 1. Closure (Axiom 0). $u, v > 0 \Rightarrow uv > 0$ (closed under $\oplus$ ). $u > 0, r \in R \Rightarrow u^{r} > 0$ (closed under $⊙$ ). ✓ Why? Positives multiply to positives; a positive raised to any real power is positive — no escape.

Step 2. Find the hidden $0$ (Axiom 3). We need $z$ with $u \oplus z = u$ , i.e. $u \cdot z = u$ , so $z = 1$ . The zero vector is the number $1$ . Why? Axiom 3's $0$ is whatever element acts neutrally under the chosen $+$ ; here that's $1$ , not $0$ . This is exactly the mistake the parent note warns about.

Step 3. Inverses (Axiom 4). Need $u \oplus w = 1$ , i.e. $u \cdot w = 1$ , so $w = 1/ u$ , which is positive — inside $V$ . ✓ Why? The additive inverse of $u$ is its reciprocal under this disguised addition.

Step 4. Scalar identity (Axiom 5) $1 ⊙ u = u^{1} = u$ . ✓ Why? The cheapest scalar axiom — check it directly, don't assume it.

Step 5. Scalar compatibility (Axiom 6) $a ⊙ (b ⊙ u) = a ⊙ (u^{b}) = (u^{b})^{a} = u^{ab}$ , and $(ab) ⊙ u = u^{ab}$ . Equal ✓. Why? This mixes two $⊙$ 's; the power-of-a-power law $(u^{b})^{a} = u^{ab}$ is exactly what makes it hold.

Step 6. Distributive over vector sum (Axiom 7) $r ⊙ (u \oplus v) = r ⊙ (uv) = (uv)^{r} = u^{r} v^{r} = (r ⊙ u) \oplus (r ⊙ v)$ . Equal ✓. Why? This mixes $⊙$ with $\oplus$ ; the product rule $(uv)^{r} = u^{r} v^{r}$ is the load-bearing identity here.

Step 7. Distributive over scalar sum (Axiom 8) $(a + b) ⊙ u = u^{a + b}$ , while $(a ⊙ u) \oplus (b ⊙ u) = u^{a} \cdot u^{b} = u^{a + b}$ . Equal ✓. Why? This one uses the exponent-sum law $u^{a + b} = u^{a} u^{b}$ . Along with Axioms 1–2 (commutativity/associativity of $\oplus$ , inherited from ordinary multiplication of reals), all 8 now hold explicitly.

Verify: All 8 axioms hold. It is a vector space! In fact $lo g : R_{> 0} \to R$ turns $\oplus$ into $+$ and $⊙$ into ordinary scaling — an isomorphism to $R^{1}$ . Dimension $1$ , basis ${e}$ (since $r ⊙ e = e^{r}$ reaches every positive). ✓

Example 7 — Cell F: weird operations where an axiom FAILS

Worked example On

$V = R^{2}$ keep ordinary $+$ but define scaling by $r ⊙ (x, y) = (r x, 0)$ . Vector space? Forecast: it kills the second coordinate on scaling. Which single axiom does that violate?

Step 1. Closure & addition are ordinary, so fine. Zero $(0, 0)$ fine (Axiom 3). Why? Isolate the suspect: only scaling was changed.

Step 2. Test Axiom 5 (scalar identity $1 ⊙ v = v$ ): $1 ⊙ (x, y) = (1 \cdot x, 0) = (x, 0)$ . But we need $(x, y)$ . Why? Axiom 5 is the cheapest scalar axiom and it directly touches the mangled coordinate.

Step 3. For any $v = (x, y)$ with $y \neq = 0$ , e.g. $v = (3, 7)$ : $1 ⊙ (3, 7) = (3, 0) \neq = (3, 7)$ . Why? A concrete counterexample beats a general claim in an exam.

Verify: Not a vector space — Axiom 5 fails. Lesson: passing most axioms (closure, addition, distributivity here) is not enough; a single axiom failure disqualifies. ✓

Example 8 — Cell H: a word problem in disguise (RC circuit / decay signals)

Worked example A sensor logs signals of the form

$f (t) = C e^{- t}$ for a constant $C \in R$ (decaying voltages). Is the set $V = {C e^{- t} : C \in R}$ of all such signals a vector space (over $R$ )? Forecast: these are curves, not arrows — but the parent note showed function spaces work. Does this subset close up?

Step 1. Add two signals. $C_{1} e^{- t} + C_{2} e^{- t} = (C_{1} + C_{2}) e^{- t}$ , still of the form $C e^{- t}$ with $C = C_{1} + C_{2}$ . Closed under $+$ (Axiom 0). ✓ Why? The $e^{- t}$ factors out — the "shape" is preserved, only the amplitude adds.

Step 2. Scale. $r \cdot (C e^{- t}) = (r C) e^{- t}$ , still the same form. Closed under scaling (Axiom 0). ✓ Why? Scaling multiplies amplitude, keeps the decay shape.

Step 3. Zero & inverse. $C = 0$ gives $0 =$ the zero signal $0 \cdot e^{- t} = 0$ , in $V$ (Axiom 3 ✓). Inverse of $C e^{- t}$ is $(- C) e^{- t} \in V$ (Axiom 4 ✓). Axioms 1,2,5,6,7,8 inherited from the function space $C [0, \infty)$ . Why? We're inside a known function space, so algebra axioms come free.

Verify: It is a vector space — a 1-dimensional subspace of all signals, basis ${e^{- t}}$ . Physically: any decaying-at-unit-rate voltage is a scalar multiple of one prototype, so the whole "space of these signals" is a single dial (amplitude $C$ ). This is the same logic as the ODE solution space in the parent note; deep-linked to Function spaces and Fourier series. ✓

The decision flowchart

Read the flowchart as a sieve you run top to bottom, bailing out at the first failure:

Closure first. Add two elements and scale one — do both land back in $V$ ? If no, stop: not a vector space (Examples 2, 3, 4 died here).
Zero next. Is there an element $0 \in V$ with $v + 0 = v$ ? Remember it may be disguised (the number $1$ in Example 6). If no, stop.
Inverses. Does every $v$ have a $- v$ inside $V$ ? If no, stop (Example 2 also failed here).
The rest. Only now grind Axioms 1, 2, 5, 6, 7, 8 (commutativity, associativity, scalar identity, compatibility, both distributive laws). A single failure — like Axiom 5 in Example 7 — still disqualifies.
If all survive, it is a vector space (Examples 1, 5, 6, 8).

Recall Rapid self-test

Positive reals with ordinary $+$ : vector space? ::: No — no zero (0 is missing) and no inverses. Positive reals with $u \oplus v = uv$ , $r ⊙ u = u^{r}$ : vector space? ::: Yes — hidden zero is $1$ , isomorphic to $R$ , dimension $1$ . Polynomials of degree exactly 2: vector space? ::: No — leading terms can cancel, not closed under $+$ ; and $0$ (degree $- \infty$ ) missing. $Z^{2}$ over $R$ : vector space? ::: No — scaling by $\frac{1}{2}$ leaves the set; $Z$ isn't a field anyway. The set ${0}$ : vector space? ::: Yes — the trivial space, dimension 0. Which axiom fails if $r ⊙ (x, y) = (r x, 0)$ ? ::: Axiom 5 (scalar identity), since $1 ⊙ v \neq = v$ .

Connections

Abstract vector spaces — axioms, examples beyond ℝⁿ
Subspaces and the subspace test
Linear independence, basis and dimension
Linear maps and matrix representations
Fields and scalars
Function spaces and Fourier series