Question bank — Abstract vector spaces — axioms, examples beyond ℝⁿ

True. Addition aur scalar multiplication entrywise hain, closure hold karta hai, zero matrix $0$ hai, aur saare 8 axioms $\mathbb{R}$ se inherit hote hain; iska dimension $4$ hai.

True. Axiom 3 ek zero vector $0$ ka existence force karta hai, isliye koi bhi vector space empty nahi ho sakta — sabse chhota possible wala $\{0\}$ hai.

False. Yeh closed nahi hai: $(x^3+x)+(-x^3)=x$ ka degree $1$ hai, $3$ nahi, aur zero polynomial (degree undefined) missing hai, isliye Axioms 3 aur closure fail karte hain.

False. Closure necessary hai par sufficient nahi; tumhe phir bhi zero vector, inverses, aur distributive/compatibility laws chahiye — jaise weird custom operations closed ho sakti hain par phir bhi Axiom 8 break kar sakti hain.

False. Dimension field par depend karta hai: $\mathbb{C}$ ka dimension $\mathbb{R}$ ke upar $2$ hai (basis $\{1,i\}$) par $\mathbb{C}$ ke upar $1$ hai (basis $\{1\}$) — same set, alag scalars. Dekho Fields and scalars.

True. Homogeneous equation linear combinations ke neeche closed hai aur zero function contain karta hai; inhomogeneous wala closure fail karta hai ($y_1+y_2$ solve karta hai $y^{\prime \prime }+y=6\ne 3$) aur $0$ nahi hai.

False. Infinite-dimensional hona ek property hai, disqualifier nahi; $C[0,1]$ saare 8 axioms satisfy karta hai aur ek bilkul valid vector space hai. Dekho Function spaces and Fourier series.

True. Yeh apne aap ke upar ek $1$-dimensional vector space hai jiska basis $\{1\}$ hai; "vectors" aur "scalars" same objects hote hain, jo ki legal hai.

False. Iska koi zero vector nahi hai ($0\notin {\mathbb{R}}_{>0}$) aur koi additive inverses nahi hain ($-v<0$), isliye Axioms 3 aur 4 directly fail karte hain.

Error: yeh scalar $0\in \mathbb{F}$ ko vector $0\in V$ se confuse karta hai. Theorem kehta hai ki kisi bhi vector ko scalar $0$ se scale karne par $0$ milta hai; yeh dono objects ko equal nahi karta, jo bilkul alag worlds mein ho sakte hain (jaise functions vs. numbers).

Error: "cancelling" ek axiom nahi hai. Tumhe isse legitimise karna hoga dono sides mein additive inverse $-(0v)$ (Axiom 4) add karke aur associativity use karke; tabhi wo term vanish hoti hai.

Error: "flips its sign" number-arithmetic assume karta hai, par $v$ ek function ya matrix ho sakta hai. Asli proof yeh use karta hai: $v+(-1)v=(1+(-1))v=0v=0$, isliye $(-1)v$ $v$ ka additive inverse hai, jo ki $-v$ denote karta hai.

Error: ek subspace mein $0=(0,0)$ hona zaroori hai, par $(0,0)$ $y=x+1$ satisfy nahi karta. Yeh scaling mein bhi fail karta hai: $2\cdot (1,2)=(2,4)$ line par nahi hai. Sirf origin se guzarne wali lines qualify karti hain. Dekho Subspaces and the subspace test.

Error: dimension basis vectors count karta hai, roots nahi. Basis $\{1,x,x^2\}$ mein teen elements hain, isliye $\dim P_2=3$; polynomial ke roots polynomial space ke dimension se irrelevant hain.

Error: ek basis linearly independent aur spanning honi chahiye. Infinitely many functions phir bhi dependent ho sakti hain (jaise $\{1,x,2x,x^2,\dots \}$), isliye infinite hona kaafi nahi hai. Dekho Linear independence, basis and dimension.

Kyunki closure operations $+:V\times V\to V$ aur $\cdot :\mathbb{F}\times V\to V$ ke type mein baked in hai; outputs $V$ mein wapas aane chahiye. Kyunki yeh operation signatures mein rehta hai, ise aksar count nahi kiya jaata, par yeh check karne ki sabse aasaan cheez hai aur wahi hai jo beginners bhool jaate hain.

Kyunki yeh sirf abstract axioms use karte hain, jo har vector space obey karta hai chahe uske elements "kaisa bhi dikhein". Yahi abstraction ka poora fayda hai: ek proof arrows, functions, matrices, aur ODE solutions ke liye simultaneously kaam karta hai.

Kyunki scalar multiplication tabhi sense deta hai jab tum fix karo ki kaunse scalars allowed hain. Same set alag alag fields ke upar alag dimensions ke saath vector space ho sakta hai, isliye "vector space" $\mathbb{F}$ ka naam liye bina incomplete hai.

Kyunki iske elements ordinary functions hain, aur functions pehle se hi pointwise operations ke through saare 8 axioms obey karte hain. Closure hi ek cheez hai jo fail ho sakti hai, isliye yeh check karna kaafi hai ki solutions ke linear combinations abhi bhi solutions hain.

Kyunki ek vector space ko $+$ aur ek compatible scalar action chahiye jo ise ek field se link kare, saath mein ek zero, inverses, aur distributive laws. Akela $+$ zyada se zyada ek group ya monoid deta hai, poora linear-algebra structure nahi.

Complex scalars ke saath, ek basis vector $\{1\}$ $z\cdot 1$ ke zariye har complex number tak pahunch jaata hai, isliye dimension $1$. Sirf real scalars ke saath tumhe do chahiye, $\{1,i\}$, kyunki $1$ ka koi bhi real multiple $i$ produce nahi karta — isliye dimension $2$.

Haan — trivial/zero vector space. Yeh saare 8 axioms satisfy karta hai ($0+0=0$, $a0=0$) aur iska dimension $0$ hai jiska basis empty set hai.

Haan. $0$ wahi element hai jo $v+0=v$ satisfy karta hai: yeh $C[0,1]$ mein zero function hai, $M_{2\times 2}$ mein zero matrix hai, aur $P_n$ mein zero polynomial hai — inme se koi bhi number $0$ nahi hai.

Nahi. Yeh real scalar multiplication ke neeche closure mein fail karta hai: $\sqrt{2}\cdot 1=\sqrt{2}\notin \mathbb{Q}$. Yeh $\mathbb{Q}$ apne upar ek vector space hai, jo phir se dikhata hai ki field choice decisive hai.

Haan, agar tum set theory ke standard axioms accept karo, par infinite-dimensional spaces jaise "all polynomials" ke liye basis infinite hoti hai aur $C[0,1]$ ke liye basis kuch aisi nahi hai jo tum explicitly likh sako. Existence aur constructibility alag cheezein hain.

Woh element $0$ hona chahiye (Axiom 3), aur har axiom trivially true statement $0=0$ mein collapse ho jaata hai. Yeh confirm karta hai ki $\{0\}$ sabse chhota legal vector space hai aur ek vector space kabhi bhi empty nahi hota.