Dimension — basis cardinality: Question bank — Hinglish deep dive

4.5.18 · D5 · Maths › Linear Algebra (Full) › Dimension — basis cardinality

Traps se pehle, har symbol ko clearly define karte hain, aur poori kahaani ke peeche ek picture draw karte hain.

Definition Is page par use hone wali Notation (scratch se bani)

$V$ — wo vector space jisme hum kaam kar rahe hain. Ek vector sirf ek arrow/numbers ki list hai jise tum add aur scale kar sakte ho.
Ek spanning set $S = {w_{1}, w_{2}, \dots, w_{n}}$ — $w_{i}$ ("double-u sub $i$ " padho) wo vectors hain jinke combinations $V$ ke har vector tak pahunchte hain. Socho "raw material ka dher."
Ek independent set $L = {v_{1}, v_{2}, \dots, v_{m}}$ — $v_{j}$ ("vee sub $j$ ") wo vectors hain jahan koi bhi doosron se nahi banta. Socho "vectors jo apni jagah earn karte hain."
Coefficient $c_{j}$ — ek plain number jo vector $w_{j}$ ko multiply karta hai jab hum ek vector ko combination ki tarah likhte hain jaise $c_{1} w_{1} + c_{2} w_{2} + \dots$ . " $c_{j} \neq = 0$ " ka matlab sirf yeh hai ki " $w_{j}$ us combination mein actually dikhta hai."
$v_{k + 1}$ — wo agla independent vector jo hum exchange-lemma proof ke dauran swap karne ki koshish karte hain (woh " $(k + 1)$ -waan" wala).
$∣ B ∣$ — set $B$ ki cardinality (vectors ki sankhya).

Yaad karo woh ek weapon jo lagbhag har jawab ke peeche hai, ab counts ke baare mein carefully bataya gaya:

True or false — justify

Ek basis of

V

dono independent aur spanning honi chahiye.

True — akela ek property kaafi nahi; ek independent set bahut choti ho sakti hai aur ek spanning set bahut badi, isliye sirf dono-ek-saath wali cheez ka exact dimension count hota hai.

V

ka har spanning set ek basis hota hai.

False — ek spanning set bahut bada ho sakta hai aur usme redundant (dependent) vectors ho sakte hain; pehle usse thin karke ek independent subset banana padta hai tabhi woh basis banta hai.

V

mein har linearly independent set ek basis hota hai.

False — woh span karne ke liye bahut chota ho sakta hai; e.g.

{(1, 0)}

R^{2}

mein independent hai lekin sirf

x

-axis tak pahunchta hai, isliye woh basis nahi hai.

Ek hi space ke do bases mein hamesha same actual vectors hote hain.

False — unka sirf same count hota hai;

{(1, 0), (0, 1)}

aur

{(1, 1), (1, - 1)}

dono

R^{2}

ke bases hain phir bhi koi vector share nahi karte.

Agar ek space

5

vectors se span hota hai, toh uski basis size

6

ki bhi ho sakti hai.

False — basis independent hoti hai, aur independent

\leq

spanning

= 5

, toh koi bhi basis (actually koi bhi independent set)

5

se zyaada nahi ho sakta.

Steinitz Exchange Lemma ke liye spanning set ka independent hona zaroori hai.

False — usse sirf span karna hai; spanning set dependent ho sakta hai, aur yahi generality lemma ko powerful banati hai.

Agar

V

ki har basis mein

3

vectors hain, toh

V

mein kuch independent set mein

4

vectors hain.

False — maximum independent size dimension

3

ke barabar hai; koi bhi

4

vectors exchange lemma se dependent honge.

Ek basis mein vector add karna use basis banaye rakhta hai.

False — woh spanning rehta hai lekin independence toot jaati hai (naya vector purane ka combination hai), isliye woh ab basis nahi raha.

R^{3}

ki basis se ek vector hataane ke baad

2

-dimensional subspace ki basis milti hai.

True — baaki do vectors ab bhi independent hain (ek independent set ka subset), isliye woh apne

2

-dimensional span ki basis banaate hain.

Zero vector ek basis mein ho sakta hai.

False —

0

wala koi bhi set dependent hota hai, kyunki

1 \cdot 0 = 0

ek nontrivial combination hai jo zero ke barabar hai.

Error dhundo

{(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)}

mein

4

vectors hain, isliye

dim R^{3} = 4

Error yeh hai ki inhe basis keh diya;

R^{3}

3

vectors se span hota hai, isliye koi bhi independent set

\leq 3

ka hoga — yeh

4

dependent hain kyunki

(1, 1, 1) = (1, 0, 0) + (0, 1, 0) + (0, 0, 1)

, toh yeh count nahi karte.

"Dimension woh size hai jitne vectors tum

V

mein sabse zyaada fit kar sakte ho."

Galat extremal quantity — tum

V

mein infinitely many vectors fit kar sakte ho; dimension ek maximal independent set ki size hai (equivalently ek minimal spanning set). "Maximal independent" matlab: independent, aur tum koi bhi vector add nahi kar sakte independent rehte hue — woh maximal independent set exactly ek basis hai.

"Exchange lemma proof mein, humein kuch

c_{j} \neq = 0

chahiye tha; woh isliye aaya kyunki

w

's independent hain."

Error galat hypothesis name karta hai — yahan

c_{j}

w_{j}

par coefficient hai jab hum

v_{k + 1} = c_{k + 1} w_{k + 1} + \dots + c_{n} w_{n}

plus earlier

v

's likhte hain. Kuch

c_{j} \neq = 0

incoming set

L = {v_{1}, \dots, v_{m}}

ki independence se aata hai; agar sab

c_{j} = 0

hote toh

v_{k + 1}

earlier

v

's ka combination hota, jo $L$ ki independence ko contradict karta.

B_{1}

spans aur

B_{2}

spans, toh lemma dono taraf apply karne se

∣ B_{1} ∣ = ∣ B_{2} ∣

milta hai."

Error dono sides par spanning use karta hai; lemma independent-vs-spanning compare karta hai, isliye tumhe har basis ko dono roles play karne hote hain — indep(

B_{1}

)

\leq

span(

B_{2}

) aur indep(

B_{2}

)

\leq

span(

B_{1}

"Kyunki

{1, 1 + x, 1 + x + x^{2}}

aur

{1, x, x^{2}}

bilkul alag dikhte hain, inhe differently-sized space span karni chahiye."

Appearance count nahi hota — dono mein

3

elements hain aur dono

P_{2}

span karte hain, isliye dono certify karte hain

dim P_{2} = 3

; theorem size ko basis change ke saath badalne se rokta hai.

"Rank-nullity kehta hai

dim V = rank + nullity

, isliye dimension chosen map par depend karti hai."

Error

dim V

ko ek map se baandh deta hai — rank

= dim (image)

aur nullity

= dim (ker)

ke definitions se,

dim V

akele basis cardinality se fixed hai; rank-nullity us fixed number ko alag alag maps ke liye alag tarike se split karta hai, total kabhi nahi badalta.

Why questions

Exchange lemma inequality (indep

\leq

span) kyun deta hai, equality kyun nahi?

Kyunki ek spanning set mein redundancy (extra vectors) ho sakti hai, isliye uske members independent set se strictly zyaada ho sakte hain; equality tabhi aati hai jab dono sets bases hon.

Dimension Theorem paane ke liye lemma do baar kyun apply karna padta hai?

Ek application

∣ B_{1} ∣ \leq ∣ B_{2} ∣

deta hai; roles swap karke hi

∣ B_{2} ∣ \leq ∣ B_{1} ∣

milta hai, aur do opposite inequalities hi equality force karti hain.

Proof mein independence of

L

hi ek zyada jagah independence kyun use hoti hai?

L = {v_{1}, \dots, v_{m}}

ki independence guarantee karti hai ki koi bhi incoming vector already swapped-in walon se nahi bana, isliye ek

w

hamesha kick out hone ke liye bachta hai — yahi swap process ko saare

m

steps tak chalata rehta hai.

Dimension well-defined hona matter kyun karta hai?

Iske bina "

dim R^{3} = 3

" meaningless hota, kyunki kisi odd basis mein

5

vectors ho sakte the; theorem guarantee karta hai ki answer wahi rehta hai chahe tum koi bhi basis chuno.

Lemma mein spanning set dependent hone se argument kyun nahi tooti?

Proof sirf incoming vectors ko current set ke terms mein express karta hai aur swap karta hai — usse

w

's ka uniquely likha hona kabhi nahi chahiye, isliye sirf spanning kaafi hai.

Swapping har step par spanning property kyun preserve karti hai?

Har kicked-out

w_{j}

ko dikhaaya jaata hai ki woh baaki vectors plus incoming

v

ka combination hai, isliye pehle jo bhi reachable tha woh unreachable nahi hota — set ab bhi spans karta hai.

Edge cases

Zero vector space

{0}

ki dimension kya hai?

Woh

0

hai — empty set (vacuously) independent hai aur

{0}

span karta hai, isliye woh basis hai, jis ki cardinality

0

hai.

Kya empty set ek valid basis hai, aur kis cheez ka?

Haan — woh independent hai (koi bhi nontrivial relation exist nahi karta) aur uska span

{0}

hai, isliye woh zero space ki unique basis hai.

Infinite-dimensional space ke liye, kya "independent

\leq

spanning" ab bhi kuch kehta hai?

Haan — agar

V

mein har finite size ka independent set hota hai (e.g.

1, x, x^{2}, x^{3}, \dots

saare polynomials mein), toh koi bhi finite set usse span nahi kar sakta, kyunki size

n

ka ek finite spanning set independent sets ko

n

par cap kar deta. Concretely, saare polynomials ki space ka infinite basis

{1, x, x^{2}, \dots}

hai aur koi finite basis bilkul nahi hai.

Jab hum kehte hain ek infinite-dimensional space ki "dimension ek infinite cardinal hai," toh concretely iska kya matlab hai?

Iska matlab sirf yeh hai ki har basis mein infinitely many vectors hain aur koi bhi do bases ko one-to-one match kiya ja sakta hai; e.g. saare polynomials ka basis

{1, x, x^{2}, \dots}

hai, jo counting numbers

0, 1, 2, \dots

se pair off hota hai — isliye uski dimension "countably infinite" hai, us list ki size.

Infinite space ke liye, kya ek maximal independent set exist karna guaranteed hai jo basis ka kaam kare?

Kyunki tum independent vectors add karte rahe sakte ho, aur ek standard set-theory principle (Zorn's Lemma) guarantee karta hai ki yeh growing chain ek maximal tak pahunchti hai; woh maximal independent set kuch bhi add nahi kar sakta dependent hue bina, isliye woh span karta hai aur basis hai.

Agar

V

n

vectors se span kiya jaata hai aur tum

V

mein

n

independent vectors paate ho, toh kya woh necessarily basis banaate hain?

Haan — woh maximum independent size

n

tak pahunch gaye, aur ek

n

-dimensional space mein

n

independent vectors automatically span karte hain, isliye woh basis hain.

Kya ek akela vector

{v}

jahan

v \neq = 0

ek basis ho sakta hai?

Sirf agar

dim V = 1

; tab

v \neq = 0

independent hai aur uske scalar multiples poori line fill karte hain, isliye woh span karta hai aur basis hai.

Agar do spaces ki same dimension hai, toh kya woh "same" space hain?

Literally nahi, lekin woh isomorphic hain — equal dimension ka matlab hai ek coordinate map unke bases ko one-to-one match karta hai, isliye woh vector spaces ke roop mein identically behave karte hain.

Connections

Steinitz Exchange Lemma — yahan lagbhag har jawab ke peeche ek hi tool.
Linear Independence — "kuch coefficient ≠ 0" step aur too-small traps ka source.
Spanning Sets — slots aur too-big traps ka source.
Basis of a Vector Space — woh both-at-once object jiske around har trap ghoomta hai.
Rank-Nullity Theorem — dimension splitting, dimension changing nahi.
Coordinates and Change of Basis — kyun alag dikhne wale bases phir bhi ek hi count share karte hain.