Dimension — basis cardinality: Worked examples — Hinglish deep dive

4.5.18 · D3 · Maths › Linear Algebra (Full) › Dimension — basis cardinality

Yeh page ek shooting gallery hai. Parent note ne woh ek theorem prove ki thi: har basis mein vectors ki sankhya same hoti hai, aur woh sankhya hi dimension hai. Yahan hum har tarah ki situation ko us theorem par throw karte hain — chote spaces, bade spaces, empty basis, infinite-dimensional spaces, redundant piles, word problems, aur ek exam trap — taaki koi bhi case aisa na ho jo tumne pehle dekha na ho.

Symbols aane se pehle, teen simple-words ke reminders (har ek parent mein define hai, yahan re-anchor kiya hai taaki pehli line padhne layak ho):

Definition Teen words jo hum baar baar use karte hain

Linearly independent set — vectors ka ek pile jahan koi bhi vector doosron se build nahi ho sakta. Socho aisi arrows jo sach mein nayi directions mein point karti hain; koi bhi baaki ki chhaya nahi hai. Linear Independence mein banaya gaya hai.
Spanning set — vectors ka ek pile jo har vector ko space mein scale aur add karke build kar sakta hai. Socho arrows ka ek set jinka combinations poora space paint kar dete hain. Spanning Sets mein banaya gaya hai.
Basis — ek pile jo dono ek saath hai: independent bhi aur spanning bhi. Exactly-fit wala pile. Basis of a Vector Space mein banaya gaya hai.

Aaj ke saare kaam ka engine Steinitz Exchange Lemma hai: $(size of any independent set) \leq (size of any spanning set) .$ Ise aise padho: IN ≤ SPAN. Tum kabhi bhi itne independent arrows pack nahi kar sakte jitne slots ek spanning pile mein hain.

Scenario matrix

Neeche har example ek cell ke saath tagged hai is table se. Saath milake yeh har cell ko hit karte hain.

Cell	Scenario class	Kya galat ho sakta hai / kya test karna hai
A	Ordinary finite space, do honest bases	kya counts sach mein match karte hain?
B	Degenerate: zero space ${0}$	"sirf origin" ki basis kya hoti hai?
C	Too-big spanning set (dependent pile)	Steinitz phir bhi kaam karta hai — kya yeh over-large independent sets ko forbid karta hai?
D	Too-small independent set	independence akele dimension pin nahi karta
E	Non-arrow space (polynomials / matrices)	un cheezon ka dimension jo $R^{n}$ nahi hain
F	Limiting / infinite case	kya hota hai jab koi finite spanning set exist nahi karta?
G	Real-world word problem	dimension as "free dials ki sankhya"
H	Exam twist — bade space ke andar subspace	basis list kiye bina count karo

Cell A — do honest bases, same count

Step 1. Independence test karo: $a (2, 1) + b (1, 3) = (0, 0)$ set karo. Isse milta hai $2 a + b = 0$ aur $a + 3 b = 0$ . Yeh step kyun? Independence matlab hai ki zero arrow banane ka sirf ek tarika hai aur woh hai $a = b = 0$ . Agar koi non-zero $a, b$ kaam karta, toh ek arrow doosre ki chhaya hota.

Step 2. Solve karo. Pehle se, $b = - 2 a$ ; substitute karo: $a + 3 (- 2 a) = a - 6 a = - 5 a = 0 \Rightarrow a = 0 \Rightarrow b = 0$ . Yeh step kyun? Sirf trivial solution milna independence prove karta hai — do arrows sach mein alag directions mein point karte hain (figure dekho).

Step 3. $R^{2}$ mein do independent vectors automatically $R^{2}$ ko span karte hain (independent count $=$ dimension $\Rightarrow$ basis). Toh $B_{2}$ ek basis hai aur $∣ B_{2} ∣ = 2$ . Yeh step kyun? Yeh theorem ka fayda hai: jab count dimension se match kare, toh "independent" free mein "basis" mein upgrade ho jaata hai.

Verify: Columns ka determinant hai $2 \cdot 3 - 1 \cdot 1 = 5 \neq = 0$ , toh vectors independent hain aur span karte hain. Count $= 2$ ✓ — standard basis jaisa hi, exactly jaisa Dimension Theorem promise karta hai.

Cell B — zero space, degenerate input

Step 1. Ek basis independent honi chahiye. Lekin set ${0}$ dependent hai: $1 \cdot 0 = 0$ ek non-trivial combination hai jo zero ke barabar hai. Yeh step kyun? Zero vector kabhi independent set mein nahi baith sakta — tum hamesha usse non-zero number se scale kar sakte ho aur phir bhi $0$ milta hai.

Step 2. Toh basis mein $0$ nahi aa sakta. Sirf remaining candidate hai empty set $\emptyset$ . Yeh step kyun? Step 1 ki wajah se hum mazboor hain ki jedh available tha woh bhi throw out kar dein.

Step 3. Kya $\emptyset$ ek basis hai? Yeh independent hai (vacuously — koi vector nahi, toh koi buri combination nahi) aur ${0}$ ko span karta hai (empty sum define hota hai $0$ ke barabar, jo ek maatra vector present hai). Yeh step kyun? Basis ki dono defining properties hold karti hain, toh $\emptyset$ qualify karta hai, aur $∣ \emptyset ∣ = 0$ .

Verify: $dim {0} = 0$ . Theorem se sanity check: kisi bhi do bases ka size equal hona chahiye, aur yahan sirf ek basis hai $\emptyset$ , size $0$ — koi contradiction nahi ✓.

Cell C — ek too-big, dependent spanning pile

Step 1. Steinitz apply karo us spanning pile ko use karke jo hum pehle se jaante hain: $R^{3}$ ko 3 standard vectors span karte hain. Yeh step kyun? Lemma ki ceiling sabse choti spanning set se set hoti hai jo hum name kar sakein, na ki bloated 5-vector wali se. IN ≤ SPAN, aur sabse tight SPAN jo available hai woh $3$ hai.

Step 2. Isliye kisi bhi independent set ka size $\leq 3$ hai. Toh 4 independent vectors impossible hain. Yeh step kyun? Bhaale kisine size-5 spanning set advertise ki ho, ek choti spanning set (size 3) phir bhi 4-independent piles ko forbid karti hai. Lemma ko spanning pile ke independent hone ki zaroorat nahi — dependence allowed hai.

Step 3. Maximum independent count exactly $3$ hai, achieve hota hai jaise ${(1, 0, 0), (0, 1, 0), (0, 0, 1)}$ se. Yeh step kyun? Teen independent vectors exist karte hain (standard basis), aur Step 2 kehta hai tum 3 se zyada nahi kar sakte, toh 3 ceiling hai aur yeh reach ho jaati hai.

Verify: $dim R^{3} = 3$ ; size-5 spanning set legal hai (bas dependent hai) lekin independent ceiling $3$ par rehti hai. Largest independent set $= 3$ ; chaar independent vectors $\Rightarrow$ impossible ✓.

Cell D — ek too-small independent set

Step 1. Independence check karo: $a (1, 0, 0) = (0, 0, 0) \Rightarrow a = 0$ . Independent ✓ (ek single non-zero vector hamesha hota hai). Yeh step kyun? Confirm karta hai ki set ek legitimate independent pile hai — trap yeh hai ki yeh "enough" lagta hai.

Step 2. Spanning check karo: kya hum $(0, 1, 0)$ ko $a (1, 0, 0)$ ke roop mein build kar sakte hain? Uske liye chahiye $(a, 0, 0) = (0, 1, 0)$ — impossible. Yeh step kyun? Spanning maangti hai ki har vector reachable ho. $x$ -axis ke saath ek arrow kabhi off-axis vectors tak nahi pahunch sakta.

Step 3. Toh yeh sirf ek line span karta hai (ek 1-dimensional subspace), na ki poora $R^{3}$ . Yeh $R^{3}$ ka basis nahi hai; yeh us line ka basis zaroor hai. Yeh step kyun? Independence $IN \leq n \leq SPAN$ ka lower half deta hai; tumhe spanning bhi chahiye pehle count $dim = 3$ par pin ho.

Verify: Size $1 < 3 = dim R^{3}$ , toh yeh $R^{3}$ ka basis nahi ho sakta — theorem se consistent: ek basis ko dimension exactly hit karni chahiye, kam nahi ✓.

Cell E — ek aisa space jo arrows nahi hai: $2 \times 2$ matrices

Step 1. Pile propose karo $E_{11} = (1000), E_{12} = (0010), E_{21} = (0100), E_{22} = (0001) .$ Yeh step kyun? Har $E_{ij}$ "ek slot mein 1, baaki mein 0" hai. Koi bhi matrix $(a c b d) = a E_{11} + b E_{12} + c E_{21} + d E_{22}$ hai, toh yeh clearly span karte hain.

Step 2. Independence: agar $a E_{11} + b E_{12} + c E_{21} + d E_{22} = (0000)$ , toh entries match karne par $a = b = c = d = 0$ force hota hai. Yeh step kyun? Har basis matrix ek alag slot ka maalik hai, toh koi combination cancel nahi kar sakti jab tak har coefficient zero na ho — sach mein nayi "directions".

Step 3. Dono properties hold karti hain, toh yeh 4 elements wala basis hai: $dim V = 4$ . Yeh step kyun? Dimension Theorem se yeh $4$ forced hai — $2 \times 2$ matrices ka har doosra basis bhi exactly 4 members rakhega.

Verify: $dim R^{2 \times 2} = 2 \times 2 = 4$ ✓. General pattern $dim R^{m \times n} = mn$ ; yahan $mn = 4$ .

Cell F — limiting case: koi finite spanning set nahi

Step 1. Contradiction ke liye maano ki ek finite pile ${p_{1}, \dots, p_{n}}$ , $P$ ko span karti hai. Yeh step kyun? Hum limiting behaviour test karte hain yeh assume karke ki Steinitz ke finite tools apply ho sakte hain, phir koi crack dhundhte hain.

Step 2. Maano $d$ , $p_{1}, \dots, p_{n}$ mein sabse zyada degree hai. Unke har combination ki degree $\leq d$ hogi. Lekin $x^{d + 1} \in P$ ki degree $d + 1$ hai. Yeh step kyun? Polynomials ko combine karne se top degree kabhi nahi badhti, toh ek fixed finite pile ki ek degree ceiling hoti hai — ek genuine limiting obstruction.

Step 3. Toh $x^{d + 1}$ span nahi hota. Koi finite pile $P$ ko span nahi karti; $P$ infinite-dimensional hai. Yeh step kyun? Set ${1, x, x^{2}, x^{3}, \dots}$ ek infinite basis hai. Dimension Theorem spirit mein phir bhi hold karta hai — saari bases ki same infinite cardinality hoti hai — lekin Steinitz ki finite ceiling kabhi kick in nahi karti.

Verify: Har finite $n$ ke liye, $dim P_{n} = n + 1$ (parent se match). Jab $n$ bina bound ke badhta hai, $dim P = \infty$ ✓ — honest limiting value.

Cell G — word problem: dimension as free dials

Step 1. "Koi reproduce nahi ho sakta doosron se" = 5 recipe-vectors $R^{3}$ mein linearly independent hain. Yeh step kyun? Word "irreproducible" ko "independent" mein translate karna algebra ko decide karne deta hai.

Step 2. $R^{3}$ , 3 vectors se span hota hai, toh IN ≤ SPAN deta hai independent count $\leq 3$ . Yeh step kyun? Teen fruits = teen dials = ek 3-slot spanning pile. Lemma independence ko wahan cap kar deta hai.

Step 3. $5 > 3$ , toh 5 recipes sab independent nahi ho sakti — kam se kam ek doosron ka mix hai. Yeh step kyun? Claim ceiling violate karta hai; machine ki marketing mathematically impossible hai.

Verify: Max independent recipes $= dim R^{3} = 3$ ; 5 independent claim karna impossible hai kyunki $5 > 3$ ✓. (Tum zyada se zyada 3 truly independent signature blends offer kar sakte ho.)

Cell H — exam twist: ek bade space ke andar subspace ka dimension, koi basis listed nahi

Step 1. Constraint ko ek variable ke liye solve karo: $z = - x - y$ , jahan $x, y$ free hain. Yeh step kyun? Ek linear equation ek degree of freedom remove karti hai; free variables dimension count karte hain (yeh Rank-Nullity Theorem miniature mein hai).

Step 2. General element likho: $(x, y, - x - y) = x (1, 0, - 1) + y (0, 1, - 1)$ . Yeh step kyun? Har solution $(1, 0, - 1)$ aur $(0, 1, - 1)$ ka combination hai — toh yeh dono $W$ ko span karte hain.

Step 3. Independence: $x (1, 0, - 1) + y (0, 1, - 1) = (0, 0, 0)$ force karta hai $x = 0$ (pehla slot) aur $y = 0$ (doosra slot). Independent ✓. Toh yeh size 2 ka basis banate hain. Yeh step kyun? Dono basis properties hold karti hain, aur Dimension Theorem se count $2$ honest dimension hai — $W$ ka koi doosra basis alag nahi ho sakta.

Verify: $R^{3}$ par ek constraint se $dim W = 3 - 1 = 2$ ✓. Rank-nullity check: map $(x, y, z) \mapsto x + y + z$ ka rank 1 hai, toh nullity $= 3 - 1 = 2 = dim W$ ✓.

Roundup

Recall Kis cell ne kya sikhaya?

Matrix mein har scenario match kiya? Cell A — tilted basis phir bhi 2 count karta hai ::: determinant $\neq = 0 \Rightarrow$ independent, count $= dim = 2$ . Cell B — ${0}$ ka dimension ::: $0$ ; basis hai empty set $\emptyset$ . Cell C — 5-vector dependent spanning set ::: independent ceiling $3$ par rehti hai; 4 independent impossible hai. Cell D — $R^{3}$ mein single independent vector ::: basis nahi; sirf ek line span karta hai, too small. Cell E — $2 \times 2$ matrices ::: $dim = 4$ ; basis ${E_{11}, E_{12}, E_{21}, E_{22}}$ . Cell F — saare polynomials $R [x]$ ::: infinite-dimensional; koi finite pile span nahi karti. Cell G — 3 fruits par 5 smoothie recipes ::: zyada se zyada 3 independent ho sakti hain, toh claim fail karta hai. Cell H — plane $x + y + z = 0$ ::: $dim = 2$ ek constraint / rank-nullity se.

Connections

Steinitz Exchange Lemma — Cells A, C, G mein use ki gayi ceiling.
Linear Independence — har cell mein "koi vector shadow nahi hai" wala test.
Spanning Sets — woh slots jo fill ho rahe hain ya kam pad rahe hain.
Basis of a Vector Space — both-at-once object jo har example build karta hai.
Rank-Nullity Theorem — Cell H mein counting shortcut.
Coordinates and Change of Basis — kyun same-count bases itni alag dikh sakti hain (Cell A).