Dimension — basis cardinality — Hinglish notes

4.5.18 · Maths › Linear Algebra (Full)

Hum claim kya kar rahe hain?

Hum kyun care karte hain? Is theorem ke bina, "dimension" ambiguous hoti — tum nahi keh sakte ki $R^{3}$ 3-dimensional hai, kyunki shayad kisi weird basis mein 5 vectors hon. Yeh theorem guarantee karta hai ki answer unique hoga.

Hum isse prove kaise karte hain — Scratch se Derivation

Poora result ek workhorse lemma par tikaa hai. Use derive karo aur sab kuch derive ho jaata hai.

Exchange Lemma Derive Karna

Maano $S = {w_{1}, \dots, w_{n}}$ $V$ ko span karta hai, aur $L = {v_{1}, \dots, v_{m}}$ linearly independent hai. Goal: dikhao ki $m \leq n$ hai.

Idea: $v$ 's ko spanning set mein ek ek karke swap karo, har baar ek $w$ ko bahar nikaalte hue, aur poori process mein $n$ size ka ek spanning set banaye rakho. Agar $w$ 's kabhi khatam na hon, toh $m \leq n$ .

Step 1. Kyunki $S$ , $V$ ko span karta hai, likho $v_{1} = a_{1} w_{1} + a_{2} w_{2} + \dots + a_{n} w_{n} .$ Yeh step kyun? Spanning ka matlab hai har vector, including $v_{1}$ , $w$ 's ka combination hai.

Step 2. Koi $a_{i} \neq = 0$ hoga (warna $v_{1} = 0$ , jo independent set ke liye impossible hai). Reorder karo taaki $a_{1} \neq = 0$ ho. Solve karo: $w_{1} = \frac{1}{a _{1}} (v_{1} - a_{2} w_{2} - \dots - a_{n} w_{n}) .$ Yeh step kyun? Isse pata chalta hai ki $w_{1}$ redundant hai — hum use $v_{1}$ aur baaki $w$ 's se express kar sakte hain. Toh ${v_{1}, w_{2}, \dots, w_{n}}$ abhi bhi $V$ ko span karta hai.

Step 3 (induction). Maano $k$ swaps ke baad hamare paas spanning set hai ${v_{1}, \dots, v_{k}, w_{k + 1}, \dots, w_{n}} .$ $v_{k + 1}$ ko iske terms mein likho: $v_{k + 1} = old v ’s b_{1} v_{1} + \dots + b_{k} v_{k} + remaining w ’s c_{k + 1} w_{k + 1} + \dots + c_{n} w_{n} .$

Step 4 (key). Koi $c_{j} \neq = 0$ hoga. Kyun? Agar saare $c_{j} = 0$ hon, toh $v_{k + 1}$ $v_{1}, \dots, v_{k}$ ka combination hota — jo $L$ ke independent hone ko contradict karta hai. Toh koi $w$ abhi bhi hona chahiye jise bahar nikala ja sake. Yahi jagah hai jahan $L$ ki independence use hoti hai. Reorder karo, $w_{k + 1}$ ko bahar nikalo, $v_{k + 1}$ ko swap in karo. $n$ size ka spanning set bachaa rehta hai.

Step 5 (conclude). Har swap ek $v$ aur ek $w$ consume karta hai. Hamare paas $v$ 's se kam se kam utne hi $w$ 's hone chahiye, yaani $m \leq n$ . $■$

Lemma se Dimension Theorem tak

Maano $B_{1}$ (size $p$ ) aur $B_{2}$ (size $q$ ) dono $V$ ke bases hain.

$B_{1}$ independent hai, $B_{2}$ spans karta hai $\Rightarrow$ lemma apply karo: $p \leq q$ .
$B_{2}$ independent hai, $B_{1}$ spans karta hai $\Rightarrow$ lemma apply karo: $q \leq p$ .

Isliye $p = q$ . $■$ Woh number hi dimension hai.

Yeh step kyun? Ek basis simultaneously independent aur spanning hota hai, toh har ek lemma mein dono roles play karta hai — dono directions mein inequalities deta hai.

Worked Examples

Steel-man the Mistakes

Forecast-then-Verify

Flashcards

Vector space ki dimension kya hoti hai?

Kisi bhi basis ki cardinality (vectors ki sankhya) — well-defined isliye hai kyunki saare bases ka size equal hota hai.

State karo Dimension Theorem (invariance of basis cardinality).

Ek hi finite-dimensional space ke kisi bhi do bases mein vectors ki sankhya same hoti hai.

Steinitz Exchange Lemma kya kehta hai?

Agar

V

n

vectors span karte hain, toh har linearly independent set mein at most

n

vectors hote hain (independent ≤ spanning).

Exchange lemma ke proof mein

L

ki independence kahan use hoti hai?

Yeh guarantee karne ke liye ki kisi remaining

w

ka coefficient nonzero hai, taaki ek

w

hamesha swap out kiya ja sake.

Kya Steinitz mein spanning set ko linearly independent hona chahiye?

Nahi — use sirf span karna hai; woh dependent bhi ho sakta hai.

Exchange lemma se Dimension Theorem kaise derive karte hain?

Lemma dono taraf apply karo:

B_{1}

indep ≤

B_{2}

spanning deta hai

p \leq q

; roles reverse karo toh

q \leq p

; toh

p = q

4 vectors

R^{3}

ka basis kyun nahi ban sakte?

R^{3}

ko 3 vectors span karte hain, toh koi bhi independent set mein ≤3 vectors hote hain; 4 vectors dependent honge, isliye basis nahi.

dim P_{n}

(degree ≤ n ke polynomials) =

n + 1

(basis

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

Recall Feynman: 12-saal ke bachche ko samjhao

Socho tum LEGO ka ek kila banaa rahe ho. Tum laal bricks use kar sakte ho ya neeli bricks, lekin wahi exact kila banane ke liye tumhe hamesha utni hi bricks chahiye hongi. Bricks tumhare basis vectors hain, kila woh space hai. Chahe tum kaunsa color (kaunsa basis) chuno, bricks ki count fixed rehti hai — woh count hi dimension hai. Isse prove karne ki trick: apni independent bricks lo aur unhe ek ek karke ek aisi pile mein swap karo jo pehle se kila banaa sakti hai; tumhe kabhi jagah khatam nahi hogi, jo prove karta hai ki tumhare paas pile se zyada independent bricks nahi ho sakti. Dono directions mein karo aur dono counts match karni chahiye.

Connections

Linear Independence — "some coefficient ≠ 0" step provide karta hai.
Spanning Sets — woh slots provide karta hai jo lemma fill karta hai.
Basis of a Vector Space — woh dono-at-once object jise theorem measure karta hai.
Steinitz Exchange Lemma — is proof ka engine.
Rank-Nullity Theorem — linear maps ke liye dimension counting action mein.
Coordinates and Change of Basis — kyun ek fixed dimension hume coordinate vectors use karne deta hai.

Dimension — basis cardinality