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

4.5.43 · D3 · Maths › Linear Algebra (Full) › Abstract vector spaces — axioms, examples beyond ℝⁿ

Shuru karne se pehle, ek word jo hum constantly use karte hain: operation "closed" hai matlab jab tum use set ke elements par karo, to result set ke andar hi aata hai. Ek fenced field imagine karo: closed = ball kabhi fence ke bahar nahi jaati.

Scenario matrix

Har "is it a vector space?" question in cells mein se kisi ek mein aata hai. Neeche har worked example us cell ke saath tagged hai jise woh kill karta hai.

Cell	Kya galat jaata hai (ya sahi)	Sabse sasta test
A. Cleanly pass karta hai	saare 8 axioms + closure hold karte hain	closure verify karo, $0$ name karo
B. Zero vector nahi	required $0$ set mein missing hai	kya koi $0$ hai?
C. $+$ ke under closed nahi	sum set se bahar chala jaata hai	do elements pick karo, add karo
D. Scaling ke under closed nahi	scaling set se bahar chala jaata hai	$- 1$ ya $\frac{1}{2}$ se scale karo
E. Weird operations, phir bhi ek space	$+$ ya $\cdot$ redefine hua par axioms bachte hain	hidden zero dhundho
F. Weird operations, ek axiom toot jaata hai	redefined ops axiom 8 ya 5 violate karte hain	distributivity / identity test karo
G. Degenerate / limiting	one-point space ${0}$ , ya field changes	trivial case check karo
H. Word problem	disguise mein ek real system (ODE, signals)	axioms mein translate karo

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

Worked example Kya

$V = {(a 0 b c) : a, b, c \in R}$ ek vector space hai $R$ ke upar? Forecast: padhne se pehle guess karo — kya "diagonal ke neeche zero" addition aur scaling ke baad survive karta hai?

Step 1. Unhe add karo: $(a 0 b c) + (a^{'} 0 b^{'} c^{'}) = (a + a^{'} 0 b + b^{'} c + c^{'})$ . Yeh step kyun? $+$ ke under Closure (Axiom 0) sabse sasta disqualifier hai. Bottom-left entry $0 + 0 = 0$ hai, toh result abhi bhi upper-triangular hai — closed.

Step 2. Scale karo: $r (a 0 b c) = (r a 0 r b r c)$ . Kyun? Scaling closure test karo (also Axiom 0). Bottom-left $r \cdot 0 = 0$ hai — abhi bhi andar.

Step 3. Zero name karo: $0 = (0000)$ , jo hai upper-triangular, toh $0 \in V$ (Axiom 3: zero vector ✓). Kisi bhi element ka inverse uski entrywise negation hai, jo bhi upper-triangular hai (Axiom 4: additive inverse ✓). Kyun? Axioms 3 aur 4 closure ke baad sabse saste hain.

Step 4. Axioms 1,2,5,6,7,8 (do commutative/associative laws, scalar identity, compatibility, aur dono distributive laws) sab $M_{2 \times 2} (R)$ se inherited hain (jise parent note ne pehle se certify kar diya hai) kyunki hamare operations exactly wahi matrix operations hain jo ek subset par restricted hain. Kyun? Agar bada set ek algebra rule obey karta hai, to koi bhi subset jo andar rehta hai woh rule bhi obey karta hai — yeh Subspaces and the subspace test ka essence hai.

Verify: $V$ ek 3-dimensional subspace hai jiska basis ${(1000), (0010), (0001)}$ hai. Dimension $3$ , 3 free entries $a, b, c$ se match karta hai. ✓

Example 2 — Cell B: zero vector nahi (positive reals)

Worked example Kya

$V = R_{> 0}$ (strictly positive reals) ordinary $+$ aur ordinary $\cdot$ ke saath ek vector space hai? Forecast: tum do positives add kar sakte ho aur ek positive milta hai — feel closed hota hai. Toh catch kahan hai?

Step 1. Closure under $+$ : $2 + 3 = 5 > 0$ ✓. Theek lagta hai. Kyun? Tempting-but-incomplete evidence confirm karo, taaki tum trap feel karo.

Step 2. $0$ dhundho: humein ek element $z \in V$ chahiye jiske saath $v + z = v$ . Ordinary addition force karta hai $z = 0$ . Lekin $0 \in / R_{> 0}$ . Kyun? Axiom 3 (zero vector) demand karta hai ki zero set ke andar rahe. Yeh nahi rehta.

Step 3. Yahan tak ki scaling bhi toot jaati hai: $(- 1) \cdot 2 = - 2 \in / V$ — scaling ke under bhi closed nahi. Kyun? Ek failure kaafi hai, lekin do dikhana disqualification ko airtight banata hai.

Verify: Axioms 3 aur 4 fail karte hain, scaling ke under closure fail karta hai. In operations ke under vector space nahi hai. (Sequel: Example 6 same set ko zyada clever operations ke saath rescue karta hai — cell E.)

Example 3 — Cell C: addition ke under closed nahi (degree-exactly-2 polynomials)

Worked example Kya

$V = {$ polynomials of degree exactly $2}$ ek vector space hai? Forecast: $P_{2}$ parent note mein theek tha — kya degree par exactly 2 insist karna kuch badalta hai?

Step 1. $p (x) = x^{2} + x$ aur $q (x) = - x^{2} + 3$ lo. Dono ki degree exactly 2 hai. Kyun? Closure test karne ke liye tumhe actually do elements pick karke add karne honge.

Step 2. $p (x) + q (x) = (x^{2} - x^{2}) + x + 3 = x + 3$ , jis ki degree $1$ hai, 2 nahi. Kyun? Leading terms cancel ho gaye — sum set se bahar gir gaya. Closure under $+$ (Axiom 0) fail karta hai.

Step 3. Also $0 \cdot p = 0$ , zero polynomial, ki degree $- \infty$ hai (upar diye convention ke hisaab se, $2$ nahi) — toh $0 \in / V$ bhi. Kyun? Ek doosri independent failure; $0$ se scaling set se bahar jaati hai, Axiom 3 kill karta hai.

Verify: Vector space nahi hai. Contrast: degree $\leq 2$ (" $\leq$ " wala) theek hai kyunki cancellation phir bhi tumhe $\leq 2$ par chhod deti hai. Word "exactly" poora trap hai. ✓

Example 4 — Cell D: $+$ ke under closed hai lekin scaling nahi (integer vectors)

Worked example Kya

$V = Z^{2}$ (integers ke pairs) $R$ par ek vector space hai? Forecast: integer + integer = integer, toh addition safe hai. Real number se scaling ke baare mein kya?

Step 1. Addition: $(1, 2) + (3, 4) = (4, 6)$ , integers ✓. Zero $(0, 0) \in Z^{2}$ ✓ (Axiom 3), inverses $(- a, - b)$ ✓ (Axiom 4). Kyun? Confirm karo ki additive group bilkul theek hai — failure kahin aur hai.

Step 2. $r = \frac{1}{2} \in R$ se scale karo: $\frac{1}{2} (1, 3) = (\frac{1}{2}, \frac{3}{2}) \in / Z^{2}$ . Kyun? Scalars $R$ se aate hain, lekin result ko $V$ mein land karna chahiye. Ek non-integer scalar scaling ke under closure (Axiom 0) tod deta hai.

Verify: Real vector space nahi hai. (Yeh ek abelian group ke taur par theek hai, aur yahan tak ki ek "module over $Z$ " ke taur par bhi — lekin parent note ki definition ko scalars ka ek field chahiye; Fields and scalars dekho. $Z$ ek field nahi hai.) ✓

Example 5 — Cell G: degenerate one-point space

Worked example Kya

$V = {0}$ (ek set jisme sirf zero vector hai) ek vector space hai? Forecast: ek hi element wala space count karne ke liye bahut chhota lagta hai. Kya yeh hai?

Step 1. Define karo $0 + 0 = 0$ aur $r \cdot 0 = 0$ har scalar $r$ ke liye. Kyun? Sirf ek hi element hai, toh dono operations ka exactly ek possible output hai — automatically closed (Axiom 0).

Step 2. Axioms check karo: commutativity (Axiom 1), associativity (Axiom 2) trivial hain (combine karne ke liye sirf ek cheez hai). Zero $0$ khud hai (Axiom 3 ✓); uska inverse $0$ hai (Axiom 4 ✓). Identity $1 \cdot 0 = 0$ (Axiom 5 ✓); distributivity (Axioms 7, 8) hold karta hai kyunki har side $0$ ke barabar hai. Kyun? Degenerate cases ko phir bhi check karna hota hai — lekin yahan har check " $0 = 0$ " par collapse ho jaata hai.

Verify: Yeh ek vector space hai, trivial (ya zero) vector space, dimension $0$ ke saath aur empty basis $\emptyset$ ke saath. Yeh limiting case hai: sabse chhota legal space. ✓ (Linear independence, basis and dimension se compare karo: empty set ${0}$ ko span karta hai.)

Example 6 — Cell E: weird operations jo PHIR BHI ek space form karte hain

Worked example

$V = R_{> 0}$ par nayi operations define karo: $u \oplus v = u \cdot v (old multiplication), r ⊙ u = u^{r} (old exponentiation) .$ Kya $(V, \oplus, ⊙)$ $R$ par ek vector space hai? Forecast: wahi positive reals jo Example 2 mein FAIL hue the — kya nayi operations unhe bacha sakti hain? Hidden zero kahan hai?

Step 1. Closure (Axiom 0). $u, v > 0 \Rightarrow uv > 0$ ( $\oplus$ ke under closed). $u > 0, r \in R \Rightarrow u^{r} > 0$ ( $⊙$ ke under closed). ✓ Kyun? Positives multiply hokar positives dete hain; ek positive ko kisi bhi real power par uthaya jaaye toh positive hi rehta hai — koi escape nahi.

Step 2. Hidden $0$ dhundho (Axiom 3). Humein $z$ chahiye jiske saath $u \oplus z = u$ , yani $u \cdot z = u$ , toh $z = 1$ . Zero vector number $1$ hai. Kyun? Axiom 3 ka $0$ woh element hai jo chosen $+$ ke under neutrally act karta hai; yahan woh $1$ hai, $0$ nahi. Yeh exactly wahi mistake hai jiske baare mein parent note warn karta hai.

Step 3. Inverses (Axiom 4). $u \oplus w = 1$ chahiye, yani $u \cdot w = 1$ , toh $w = 1/ u$ , jo positive hai — $V$ ke andar. ✓ Kyun? $u$ ka additive inverse is disguised addition ke under uska reciprocal hai.

Step 4. Scalar identity (Axiom 5) $1 ⊙ u = u^{1} = u$ . ✓ Kyun? Sabse sasta scalar axiom — ise directly check karo, assume mat karo.

Step 5. Scalar compatibility (Axiom 6) $a ⊙ (b ⊙ u) = a ⊙ (u^{b}) = (u^{b})^{a} = u^{ab}$ , aur $(ab) ⊙ u = u^{ab}$ . Equal ✓. Kyun? Yeh do $⊙$ 's mix karta hai; power-of-a-power law $(u^{b})^{a} = u^{ab}$ exactly wahi hai jo ise hold karaata hai.

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 ✓. Kyun? Yeh $⊙$ ko $\oplus$ ke saath mix karta hai; product rule $(uv)^{r} = u^{r} v^{r}$ yahan load-bearing identity hai.

Step 7. Distributive over scalar sum (Axiom 8) $(a + b) ⊙ u = u^{a + b}$ , jabki $(a ⊙ u) \oplus (b ⊙ u) = u^{a} \cdot u^{b} = u^{a + b}$ . Equal ✓. Kyun? Yeh wala exponent-sum law $u^{a + b} = u^{a} u^{b}$ use karta hai. Axioms 1–2 (commutativity/associativity of $\oplus$ , ordinary multiplication of reals se inherited) ke saath, saare 8 ab explicitly hold karte hain.

Verify: Saare 8 axioms hold karte hain. Yeh ek vector space hai! Actually $lo g : R_{> 0} \to R$ $\oplus$ ko $+$ mein aur $⊙$ ko ordinary scaling mein turn karta hai — $R^{1}$ ke saath ek isomorphism. Dimension $1$ , basis ${e}$ (kyunki $r ⊙ e = e^{r}$ har positive tak pahunchta hai). ✓

Example 7 — Cell F: weird operations jahan ek axiom FAIL karta hai

Worked example

$V = R^{2}$ par ordinary $+$ rakho lekin scaling $r ⊙ (x, y) = (r x, 0)$ se define karo. Vector space? Forecast: yeh scaling par doosra coordinate kill karta hai. Yeh kaun sa akela axiom violate karta hai?

Step 1. Closure & addition ordinary hain, toh theek. Zero $(0, 0)$ theek (Axiom 3). Kyun? Suspect isolate karo: sirf scaling change hua tha.

Step 2. Axiom 5 test karo (scalar identity $1 ⊙ v = v$ ): $1 ⊙ (x, y) = (1 \cdot x, 0) = (x, 0)$ . Lekin humein $(x, y)$ chahiye. Kyun? Axiom 5 sabse sasta scalar axiom hai aur yeh directly mangled coordinate ko touch karta hai.

Step 3. Kisi bhi $v = (x, y)$ ke liye jahan $y \neq = 0$ , jaise $v = (3, 7)$ : $1 ⊙ (3, 7) = (3, 0) \neq = (3, 7)$ . Kyun? Ek concrete counterexample exam mein general claim se behtar hota hai.

Verify: Vector space nahi — Axiom 5 fail karta hai. Lesson: adhiktar axioms pass karna (closure, addition, distributivity yahan) kaafi nahi hai; ek akela axiom failure disqualify karta hai. ✓

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

Worked example Ek sensor signals of the form

$f (t) = C e^{- t}$ log karta hai ek constant $C \in R$ ke liye (decaying voltages). Kya $V = {C e^{- t} : C \in R}$ aise saare signals ka set ek vector space hai ( $R$ par)? Forecast: yeh curves hain, arrows nahi — lekin parent note ne dikhaya ki function spaces kaam karte hain. Kya yeh subset close up hota hai?

Step 1. Do signals add karo. $C_{1} e^{- t} + C_{2} e^{- t} = (C_{1} + C_{2}) e^{- t}$ , abhi bhi $C e^{- t}$ form ka hai jahan $C = C_{1} + C_{2}$ . $+$ ke under Closed (Axiom 0). ✓ Kyun? $e^{- t}$ factor out ho jaata hai — "shape" preserve hoti hai, sirf amplitude add hoti hai.

Step 2. Scale karo. $r \cdot (C e^{- t}) = (r C) e^{- t}$ , abhi bhi same form. Scaling ke under Closed (Axiom 0). ✓ Kyun? Scaling amplitude multiply karta hai, decay shape rakhe rehta hai.

Step 3. Zero & inverse. $C = 0$ deta hai $0 =$ zero signal $0 \cdot e^{- t} = 0$ , $V$ mein (Axiom 3 ✓). $C e^{- t}$ ka inverse $(- C) e^{- t} \in V$ hai (Axiom 4 ✓). Axioms 1,2,5,6,7,8 function space $C [0, \infty)$ se inherited hain. Kyun? Hum ek known function space ke andar hain, toh algebra axioms free mein aate hain.

Verify: Yeh ek vector space hai — saare signals ka ek 1-dimensional subspace, basis ${e^{- t}}$ ke saath. Physically: koi bhi unit-rate-par-decay karne wala voltage ek prototype ka scalar multiple hai, toh poora "inhe signals ka space" ek akela dial hai (amplitude $C$ ). Yeh parent note mein ODE solution space ki same logic hai; Function spaces and Fourier series se deep-linked. ✓

Decision flowchart

Flowchart ko ek sieve ki tarah padhho jo tum top to bottom run karte ho, pehli failure par bail out karte ho:

Pehle Closure. Do elements add karo aur ek ko scale karo — kya dono $V$ mein wapas land karte hain? Agar nahi, ruko: vector space nahi (Examples 2, 3, 4 yahan mare).
Aage Zero. Kya koi element $0 \in V$ hai jiske saath $v + 0 = v$ ? Yaad raho yeh disguised ho sakta hai (Example 6 mein number $1$ ). Agar nahi, ruko.
Inverses. Kya har $v$ ka ek $- v$ $V$ ke andar hai? Agar nahi, ruko (Example 2 yahan bhi fail hua).
Baaki. Sirf ab Axioms 1, 2, 5, 6, 7, 8 grind karo (commutativity, associativity, scalar identity, compatibility, dono distributive laws). Ek akeli failure — jaise Example 7 mein Axiom 5 — abhi bhi disqualify karta hai.
Agar sab survive karein, toh yeh ek vector space hai (Examples 1, 5, 6, 8).

Recall Rapid self-test

Ordinary $+$ ke saath positive reals: vector space? ::: Nahi — koi zero nahi ( $0$ missing hai) aur koi inverses nahi. $u \oplus v = uv$ , $r ⊙ u = u^{r}$ ke saath positive reals: vector space? ::: Haan — hidden zero $1$ hai, $R$ ke saath isomorphic, dimension $1$ . Degree exactly 2 ke polynomials: vector space? ::: Nahi — leading terms cancel ho sakte hain, $+$ ke under closed nahi; aur $0$ (degree $- \infty$ ) missing hai. $Z^{2}$ over $R$ : vector space? ::: Nahi — $\frac{1}{2}$ se scaling set chhod deta hai; $Z$ waise bhi ek field nahi hai. Set ${0}$ : vector space? ::: Haan — trivial space, dimension $0$ . Kaun sa axiom fail karta hai agar $r ⊙ (x, y) = (r x, 0)$ ? ::: Axiom 5 (scalar identity), kyunki $1 ⊙ v \neq = v$ .

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