Rank-nullity theorem — proof: Worked examples — Hinglish deep dive

4.5.14 · D3 · Maths › Linear Algebra (Full) › Rank-nullity theorem — proof

Yeh rank–nullity theorem ka drill-ground hai. Theorem ek hi cheez kehta hai: $rank (T) + nullity (T) = dim V,$ jahan $rank (T) = dim (im T)$ (wo dimensions jo bach jaati hain) aur $nullity (T) = dim (ker T)$ (wo dimensions jo $0$ mein crush ho jaati hain). Agar koi bhi word unfamiliar lage toh Kernel and Image of a Linear Map dekho.

Humara goal yahan: har tarah ki situation ko cover karna jisme theorem aa sakta hai, taaki koi bhi exam problem surprise na kare.

The scenario matrix

Har rank–nullity problem actually yeh poochhti hai ki kitni input directions collapse hoti hain. Neeche diye gaye cases mein woh saari shapes cover hoti hain jo ho sakti hain.

Cell	Case class	Kya cheez ise distinct banati hai	Example
A	Injective, nullity $= 0$	kuch crush nahi; image utni badi jitna domain	Ex 1
B	Surjective, rank $= dim W$	image codomain ko fill karti hai	Ex 2
C	General middle case ( $0 <$ nullity $< dim V$ )	kuch crush, kuch survive	Ex 3
D	Zero map (fully degenerate)	sab kuch crush, rank $= 0$	Ex 4
E	Wide matrix ( $m < n$ , forced nullity)	outputs se zyada inputs	Ex 5
F	Tall matrix ( $m > n$ , forced non-surjective)	inputs se kam outputs	Ex 6
G	Abstract space (not $R^{n}$ )	polynomials / functions	Ex 7
H	Real-world word problem	ek scenario ko $T$ mein translate karo	Ex 8
I	Exam-style twist (solve for unknown dim)	rank/nullity diya hai, ek dhundo	Ex 9

Worked examples

Cell A — Injective map, kuch crush nahi

Worked example Example 1

$T : R^{2} \to R^{3}$ , $T (x, y) = (x, y, x + y)$ . Rank aur nullity dhundo.

Forecast: map karne ke baad dono inputs independent lagte hain — nullity guess karo, phir rank.

Kernel dhundo. $T (x, y) = (0, 0, 0)$ set karo: yeh force karta hai $x = 0$ , $y = 0$ . Toh $ker T = {(0, 0)}$ . Yeh step kyun? Kernel hamesha pehle compute karo — iska dimension nullity hai, aur rank theorem se seedha milti hai.
Nullity $= 0$ . Kernel sirf zero vector hai, ek $0$ -dimensional space. Yeh step kyun? Ek akela point (origin) ki koi independent direction nahi hoti, isliye $dim = 0$ . Dekho Dimension of a Vector Space.
Rank by the theorem. $rank = dim V - nullity = 2 - 0 = 2$ . Yeh step kyun? Image ko directly describe karne ki zaroorat nahi padi — rank–nullity ne seedha de diya.

Verify: $T (1, 0) = (1, 0, 1)$ aur $T (0, 1) = (0, 1, 1)$ $R^{3}$ mein independent hain (koi ek doosre ka scalar multiple nahi hai), toh image $2$ -dimensional hai. $2 + 0 = 2 = dim R^{2}$ . ✓ Nullity $0$ $\Leftrightarrow$ injective — dekho Invertible Linear Maps and Isomorphisms.

Cell B — Surjective map, image codomain fill karti hai

Cell C — General middle case

Cell D — The zero map (fully degenerate)

Cell E — Wide matrix (forced nonzero nullity)

Cell F — Tall matrix (forced non-surjective)

Worked example Example 6

$A = 101011$ as $T : R^{2} \to R^{3}$ . Rank, nullity dhundo aur decide karo ki $T$ surjective hai ya nahi.

Forecast: sirf $2$ input directions hain lekin $3$ output slots — kya image $R^{3}$ fill kar sakti hai?

Rank. Columns $(1, 0, 1)$ aur $(0, 1, 1)$ independent hain (koi ek doosre ka multiple nahi). Rank $= 2$ . Yeh step kyun? Rank $=$ column space ka dimension $=$ independent columns ki sankhya.
Nullity. $nullity = dim V - rank = 2 - 2 = 0$ . Yeh step kyun? Koi input direction collapse nahi hoti, toh $T$ injective hai.
Surjective? Image $3$ -dimensional $R^{3}$ ke andar $2$ -dimensional hai, toh $im T \neq = R^{3}$ . Surjective nahi hai. Yeh step kyun? Rank $2 < dim W = 3$ ; ek tall full-rank map kabhi bhi bade codomain ke har point tak nahi pahunch sakta.

Verify: $2 + 0 = 2 = dim R^{2}$ . ✓ Rank $2 < 3$ non-surjectivity confirm karta hai — theorem domain $R^{2}$ ke against balance karta hai, $R^{3}$ ke nahi.

Cell G — Abstract space (polynomials)

Cell H — Real-world word problem

Worked example Example 8

Ek factory $4$ raw ingredients (input vector in $R^{4}$ ) ko blend karke $2$ measured outputs banati hai — total mass aur total cost: $T (a, b, c, d) = (a + b + c + d, 2 a + 2 b + 2 c + 2 d) .$ Kitne independent "recipe adjustments" hain jo dono outputs unchanged rakhte hain?

Forecast: doosra output pehle ka proportional lagta hai — guess karo outputs mein kitni actually information hai.

Matrix ki tarah model karo. $A = (12121212)$ , $T : R^{4} \to R^{2}$ . Yeh step kyun? "Recipe adjustments jo kuch nahi badlate" exactly kernel hai.
Rank. Row $2 = 2 \times$ Row $1$ , toh ek pivot. Rank $= 1$ . Yeh step kyun? Mass aur cost mein sirf ek independent piece of information hai.
Nullity = answer. $nullity = dim V - rank = 4 - 1 = 3$ . Yeh step kyun? Teen independent tarike hain jisse ingredients tweak kar sako jab dono outputs fixed rahen.

Verify: $A x = 0$ ek equation hai $a + b + c + d = 0$ jisme $4$ unknowns hain $\Rightarrow 3$ free variables. $1 + 3 = 4 = dim R^{4}$ . ✓ Concretely, $(1, - 1, 0, 0)$ , $(1, 0, - 1, 0)$ , $(1, 0, 0, - 1)$ teen independent kernel vectors hain.

Cell I — Exam-style twist (unknown dimension solve karo)

Worked example Example 9

Ek linear map $T : V \to W$ hai jisme $dim V = 9$ . Bataya gaya hai ki $rank (T) = 6$ aur codomain ka $dim W = 12$ hai. Nullity dhundo, aur batao ki $T$ injective ya surjective ho sakta hai ya nahi.

Forecast: rank diya hua hai; theorem ek hi line mein nullity deta hai — lekin dhyan rakho ki yeh kis dimension use karta hai.

Nullity. $nullity = dim V - rank = 9 - 6 = 3$ . Yeh step kyun? Theorem $dim V = 9$ ke against balance karta hai, $dim W = 12$ ke nahi. Woh $12$ ek distractor hai.
Injective? Injective ke liye nullity $= 0$ chahiye. Yahan nullity $= 3 \neq = 0$ , toh injective nahi hai. Yeh step kyun? $3$ independent input directions $0$ par collapse ho rahi hain.
Surjective? Surjective ke liye $rank = dim W = 12$ chahiye. Yahan rank $= 6 < 12$ , toh surjective nahi hai. Yeh step kyun? Image ek $12$ -dimensional codomain ke andar ek $6$ -dimensional sliver hai.

Verify: $rank + nullity = 6 + 3 = 9 = dim V$ . ✓ $dim W = 12$ kabhi sum mein nahi aata — sirf surjectivity verdict pe affect karta hai.

Recall Quick self-test across the matrix

Injective ka matlab nullity kitne ke barabar hoti hai? ::: $0$ Surjective ka matlab rank kitne ke barabar hoti hai? ::: $dim W$ (codomain ka dimension) Zero map $R^{4} \to R^{5}$ ka rank aur nullity kya hoga? ::: rank $0$ , nullity $4$ Ek wide $2 \times 3$ rank-1 matrix ki nullity kya hogi? ::: $2$ Agar $dim V = 9$ aur rank $= 6$ , toh nullity kya hai? ::: $3$ Rank–nullity $dim V$ tak sum karta hai ya $dim W$ tak? ::: $dim V$ (the domain)