Rank-nullity theorem — proof: Question bank — Hinglish deep dive

4.5.14 · D5 · Maths › Linear Algebra (Full) › Rank-nullity theorem — proof

Recall Basis Extension Theorem actually kya kehta hai (neeche zaroorat padegi)

Statement. Agar ${u_{1}, \dots, u_{k}}$ ek linearly independent set hai (jaise koi subspace ka basis) ek finite-dimensional space $V$ ke andar, toh aise vectors $v_{1}, \dots, v_{n - k} \in V$ exist karte hain jaise ki ${u_{1}, \dots, u_{k}, v_{1}, \dots, v_{n - k}}$ $V$ ka basis ho. Woh hypothesis jo matter karti hai: starting set independent honi chahiye, aur $V$ finite-dimensional hona chahiye. Added $v_{j}$ specifically isliye choose kiye jaate hain taaki independence preserve ho — tum bas koi bhi vectors nahi grab kar sakte. Yahi woh tool hai jo hume kernel basis ko $V$ ke full basis tak extend karne deta hai.

Recall Proof ke woh steps jinka is page mein reference hai (parent se)

Proof dikhata hai ki $B = {T (v_{1}), \dots, T (v_{n - k})}$ do moves mein $im T$ ka basis hai. Spanning: koi bhi $v$ ek kernel part (jo $0$ pe map hota hai) plus ek survivor part mein split hota hai, toh har output $T (v_{j})$ ka combination hai. Independence (woh step jo asli kaam karta hai): agar $\sum c_{j} T (v_{j}) = 0$ toh $\sum c_{j} v_{j} \in ker T$ ; usse $u_{i}$ ke zariye rewrite karo aur $V$ ke full basis ki independence invoke karo toh har $c_{j} = 0$ force hota hai. Jahan bhi is page pe "the independence step" likha ho, wahan exactly yahi argument hai.

True or false — justify

Rank + nullity equals

dim W

, the codomain.

False. Yeh

dim V = n

, yaani domain ke barabar hota hai. Image ek bahut bade

W

ke andar baith sakta hai jisme bahut saari unused room ho — woh spare room is counting se irrelevant hai ki

V

ki directions kaise sort hoti hain.

T : R^{5} \to R^{2}

, its rank can be at most

2

True. Image

R^{2}

ka subspace hai, isliye

rank \leq dim W = 2

; phir nullity

= 5 - rank \geq 3

, toh domain ka bahut saara hissa crush hona chahiye.

A linear map

T : R^{2} \to R^{5}

can have rank

5

False. Rank

\leq dim V = 2

bhi (image

V

ke basis ke images se span hota hai, jisme sirf

2

vectors hain), toh rank

\leq min (dim V, dim W) = 2

ker T = {0}

then

T

is injective (one-to-one).

True. Nullity

0

ka matlab sirf

0

0

pe map hota hai; kyunki

T

linear hai,

T (a) = T (b) \Leftrightarrow T (a - b) = 0 \Leftrightarrow a = b

. Isliye kernel exactly injectivity ki failure measure karta hai.

For

T : V \to V

with

dim V = n

, injective implies surjective.

True. Injective se nullity

0

milta hai, toh rank

= n = dim V

, yaani image pura

V

fill kar leta hai. Yeh equivalence equal finite dimensions ke liye special hai — dekho Invertible Linear Maps and Isomorphisms.

T : R^{3} \to R^{4}

then

T

can be surjective.

False. Rank

\leq dim V = 3 < 4 = dim W

, toh image kabhi

R^{4}

fill nahi kar sakta. Tum jitni independent directions feed ki hain usse zyada ship nahi kar sakte.

Nullity is always strictly less than

dim V

False. Zero map mein

ker T = V

hota hai, toh nullity

= dim V

(yaani

k = n

) aur rank

= 0

. Har input direction crush ho jaati hai.

Rank

0

forces

T

to be the zero map.

True. Rank

0

ka matlab

im T = {0}

hai, toh

T (v) = 0

har

v

ke liye. Yahi zero map ki definition hai.

If two maps

T, S : V \to W

have the same rank, they have the same kernel.

False. Same rank se same nullity force hota hai (dono

dim V - rank

ke barabar hain), lekin kernel ki equal dimension ka matlab same kernel subspace nahi — origin se alag alag lines ek dimension share kar sakti hain.

The independence of the survivors

{v_{1}, \dots, v_{n - k}}

V

guarantees

{T (v_{1}), \dots, T (v_{n - k})}

is independent.

False in general — ek map independent vectors ko collapse kar sakta hai. Yeh yahan isliye hold karta hai kyunki woh

v_{j}

Basis Extension Theorem ke zariye kernel ke bahar choose kiye gaye the; proof ka independence step (upar recall kiya hua) isse earn karne ke liye asli kaam karta hai.

Rank–nullity applies to

T = \frac{d}{d x}

on all polynomials

R [x]

(unbounded degree).

False as stated. Woh space infinite-dimensional hai, aur counting proof ko

dim V < \infty

chahiye (dekho Dimension of a Vector Space). Finite-degree version

P_{n}

theek hai.

Spot the error

"Rank

=

number of columns of the matrix, so an

m \times n

matrix has rank

n

Wrong: rank

=

independent columns ki sankhya (column space ki dimension), jo

n

se bahut kam ho sakti hai agar columns repeat karein ya ek doosre par depend karein.

"Since

ker T \subseteq V

and

im T \subseteq V

, and they only meet at

0

, we get

V = ker T \oplus im T

Wrong:

im T

codomain

W

mein rehta hai,

V

mein nahi, toh unka same space mein hona bhi zaruri nahi. Theorem dimensions add karta hai, yeh

V

ko in dono subspaces mein split nahi karta.

T (v_{j})

spans the image because there are

n - k

of them and

dim (im T) = n - k

Circular: tum conclusion (

dim = n - k

) ko spanning prove karne ke liye use nahi kar sakte. Proof pehle span karta hai (koi bhi

w = T (v)

T (v_{j})

par collapse hota hai kyunki kernel parts vanish ho jaate hain), tab dimension deduce karta hai.

"If

\sum c_{j} T (v_{j}) = 0

then each

c_{j} T (v_{j}) = 0

, so

c_{j} = 0

Wrong: vectors ki ek sum ka

0

hona har term ko

0

force nahi karta jab tak vectors pehle se known independent na hon — aur exactly yahi hum prove karne ki koshish kar rahe hain. Sahi route

\sum c_{j} v_{j}

ker T

mein pull karta hai aur

V

ki full-basis independence use karta hai.

"Extend the kernel basis by any

n - k

vectors of

V

; they'll form a basis."

Wrong: arbitrary added vectors kernel basis par dependent ho sakte hain. Basis Extension Theorem specifically aise vectors supply karta hai jo set ko independent rakhte hain.

"Row rank and column rank can differ, so 'rank' in the theorem is ambiguous."

Wrong: row rank

=

column rank hamesha (dekho Rank of a Matrix (row vs column space)). Rank–nullity mein rank

dim (im T) =

column rank hai, jo row rank ke barabar hai.

Why questions

Why does the sum land on

dim V

and never on

dim W

Kyunki hum

V

ka basis count karna shuru karte hain (

n

vectors) aur un

n

input directions ko "crushed" (

k

) aur "independently shipped" (

n - k

) mein sort karte hain. Codomain kabhi count mein nahi aata.

Why must we choose the kernel basis first, then extend?

Taaki crushed directions (

u_{1}, \dots, u_{k}

) ko cleanly isolate kar sakein aur remaining

v_{j}

kernel ke bahar guaranteed hon; yahi separation hai jo unke images ko independent banata hai.

Why does each

T (u_{i}) = 0

matter in the spanning step?

Kisi bhi input

v

ko full basis mein likhte waqt, kernel-coefficient part

0

pe map hota hai, toh har output sirf survivors

T (v_{j})

se banta hai — exactly isliye woh image ko span karte hain.

Why is finite-dimensionality of

V

essential?

Proof ek basis extend karta hai aur vectors ki finite ginti karta hai (

k + (n - k) = n

). Infinitely many basis vectors ke saath dimensions ka woh arithmetic toot jaata hai aur cardinality ki care chahiye.

Why does more nullity force less rank (for fixed domain)?

Dono

rank = dim V - nullity = n - k

se tied hain: har extra crushed direction ek surviving output direction kam karta hai — free variables pivots ke saath trade off karte hain.

Why is rank

=

number of pivots when we row-reduce?

Har pivot column space (image) ki ek independent surviving direction mark karta hai, toh pivots count karna rank count karna hai; non-pivot (free) columns nullity se correspond karte hain.

Edge cases

The zero map

T : V \to W

: what are its rank and nullity?

Rank

= 0

(image sirf

{0}

hai) aur nullity

= dim V

(sab kuch crush ho jaata hai,

k = n

). Unka sum

dim V

hai. ✓

The identity map

T : V \to V

: what are its rank and nullity?

Rank

= dim V

(kuch lost nahi) aur nullity

= 0

(sirf

0

0

pe map hota hai,

k = 0

). Sum

= dim V

. ✓

A map from the zero space

T : {0} \to W

: does the theorem hold?

Haan trivially:

dim V = 0

, aur rank aur nullity dono

0

hain, toh

0 + 0 = 0

. Empty basis case bhi law follow karta hai.

An isomorphism

T : V \to W

with

dim V = dim W

: what does rank–nullity say?

Nullity

= 0

aur rank

= dim V = dim W

, toh

T

injective aur surjective dono hai — exactly isomorphism condition.

Can nullity and rank both be zero at once?

Sirf tab agar

dim V = 0

ho, yaani

V = {0}

. Warna unka sum positive number

dim V

hai, toh kam se kam ek nonzero hona chahiye.

T

is injective but

dim V < dim W

, is it surjective?

Nahi. Injective se rank

= dim V < dim W

milta hai, toh image ek proper subspace hai —

W

ka bahut saara hissa kabhi reach nahi hota. Injective ≠ surjective jab dimensions alag hon.