Linear transformations — definition, kernel, image: Question bank

4.5.27 · D5 · Maths › Linear Algebra (Full) › Linear transformations — definition, kernel, image

True or false — justify

Har linear map

T (0_{V}) = 0_{W}

satisfy karta hai.

True. Likho

0_{V} = 0 \cdot v

; homogeneity deta hai

T (0_{V}) = 0 \cdot T (v) = 0_{W}

. Ye linearity ka ek free sanity check hai.

Agar

T (0) = 0

hai, toh

T

linear hai.

False.

T (0) = 0

necessary hai but sufficient nahi. Jaise

T (x, y) = (x^{2}, y)

0

0

par bhejta hai phir bhi additivity fail karta hai, isliye akele yeh ek test pass karna kuch prove nahi karta.

T : V \to W

ka kernel

W

ka ek subset hai.

False. Kernel un inputs ka set hai jo zero par map hote hain, isliye

ker T \subseteq V

. Sirf output value

0_{W}

W

mein rehti hai.

T : V \to W

ka image

V

ka ek subspace hai.

False. Outputs

W

mein rehte hain, isliye

im T \subseteq W

. Ye codomain ka subspace hai, domain ka nahi.

Kernel aur image dono mein hamesha zero vector hota hai (apne respective spaces ka).

True.

T (0_{V}) = 0_{W}

dikhata hai ki

0_{V} \in ker T

aur

0_{W} \in im T

; har subspace mein uska zero hona zaroori hai.

Ek linear map do alag nonzero vectors ko ek hi output par bhej sakta hai.

True, precisely tab jab kernel nontrivial ho: agar

T (u) = T (v)

toh

u - v \in ker T

, isliye nonzero kernel genuine collisions produce karta hai.

Agar

ker T = {0}

toh

T

surjective hai.

False. Trivial kernel injectivity deta hai, surjectivity nahi. Concretely

T (x, y) = (x, y, 0)

R^{2} \to R^{3}

hai, mein

ker T = {0}

hai (injective) lekin uska image sirf flat

x y

-plane hai, isliye woh

z \neq = 0

wale poore

W

ko miss kar deta hai.

T (x) = A x

ke liye, image

A

ka row space hai.

False. Kyunki

A x = \sum x_{i} (column_{i})

, image

A

ka column space hai, row space nahi.

Rank–nullity kehta hai

dim V = dim W

minus nullity.

False. Ye kehta hai

dim V = nullity + rank

, dono domain

V

ke relative measure kiye jaate hain (aur

V

finite-dimensional hona zaroori hai);

dim W

formula mein kabhi appear nahi karta.

Ek affine map

T (x) = A x + b

jahan

b \neq = 0

ho, linear hai.

False.

T (0) = b \neq = 0

T (0) = 0

ko violate karta hai, isliye ye affine hai (ek linear map ke baad ek shift), linear nahi.

Spot the error

ker T

ek subspace hai kyunki ismein

0

hai."

0

contain karna necessary hai lekin woh reason nahi hai. Ek subspace addition aur scaling ke under closed hona chahiye:

T (u) = T (v) = 0 \Rightarrow T (c u + d v) = 0

. Sirf

0

contain karna (jaise ek single point) kaafi nahi hai.

T

injective hai, isliye uske kernel mein ek nonzero vector ho sakta hai jab tak woh unique ho."

Galat. Koi bhi nonzero

v \in ker T

deta hai

T (v) = T (0) = 0

with

v \neq = 0

— do inputs, ek output — toh

T

injective nahi hai. Injective hone ke liye exactly

ker T = {0}

hona zaroori hai.

dim (im T)

dim V

se zyada ho sakta hai."

Impossible. Rank

= dim V - nullity \leq dim V

, kyunki nullity

\geq 0

hoti hai. Ek map kabhi bhi input dimensions se zyada independent outputs produce nahi kar sakti.

T (x, y) = (x + 1, y)

linear hai kyunki iska graph straight hai."

Straight graph linearity nahi hai. Origin plug in karo:

T (0, 0) = (1, 0) \neq = (0, 0)

, toh poori picture

+ 1

ki wajah se origin se shift ho gayi hai; ye affine hai, linear nahi.

"Kyunki

A

ke columns image ko span karte hain, columns ki sankhya rank ke barabar hai."

Nahi. Rank independent columns ki sankhya hai. Extra dependent columns count badhate hain bina span ko bade kiye, isliye rank

\leq

columns ki sankhya.

T

3

-dim space ko

4

-dim space par map karta hai, toh ye surjective ho sakta hai."

Impossible. Rank

\leq dim V = 3 < 4

, isliye image ki dimension zyada se zyada

3

ho sakti hai aur woh

4

-dimensional codomain ko fill nahi kar sakti.

Why questions

T

apne basis par values se completely determined kyun hota hai?

Koi bhi

x = \sum x_{i} e_{i}

, aur linearity deta hai

T (x) = \sum x_{i} T (e_{i})

T (e_{i})

fix karne se har output fix ho jaata hai, isliye unhe columns ke roop mein stack karne se matrix ban jaati hai.

Nonzero kernel ka matlab input recover nahi ho sakta, kyun?

Kernel element se different do inputs ka ek hi output hota hai (

T (u) = T (u + k)

jab

k \in ker T

), isliye ek diye gaye output se aap nahi bata sakte ki aap kahan se shuru hue the — map un directions par invertible nahi hai.

Additivity aur homogeneity dono check karne kyun zaroori hain, sirf ek nahi?

Ye dono alag-alag operations (add karna, scale karna) control karte hain, isliye ek hold ho sakta hai jabki doosra fail ho. Concrete example:

T (x, y) = (∣ x ∣, y)

sirf

c \geq 0

ke liye homogeneity obey karta hai aur additivity break karta hai kyunki

∣ - 1 + 1∣ = 0 \neq = ∣ - 1∣ + ∣1∣ = 2

— ek property check karna use galat se "pass" kar deta.

Matrix maps ke liye image ko column space kyun kehte hain?

Kyunki

A x

A

ke columns ka weighted sum hai, isliye har reachable output columns ka linear combination hai — exactly columns ka span.

Rank–nullity koi bhi basis pick karne par hold kyun karta hai?

Proof

ker T

ka ek basis extend karke

V

ka basis banata hai; counts

p

(nullity) aur

q

(rank) dimensions hain, jo basis-independent hain, isliye identity

dim V = p + q

intrinsic hai.

Image origin se kabhi bend away kyun nahi kar sakta?

Image ek subspace hai: ye scaling aur addition ke under closed hai aur

0_{W}

contain karta hai, isliye ye ek flat set hona zaroori hai jo origin se guzre — ek line, plane, ya higher flat, kabhi shifted ya curved region nahi.

Edge cases

Zero map

T (v) = 0

(har

v

ke liye) ka kernel kya hai?

Poora domain:

ker T = V

, nullity

= dim V

, aur rank

= 0

kyunki image sirf

{0}

hai. Rank–nullity phir bhi hold karta hai:

dim V = dim V + 0

Identity map

T (v) = v

ka kernel kya hai?

Sirf

{0}

, kyunki

v = 0

T (v) = 0

ka akela solution hai. Nullity

= 0

, rank

= dim V

, aur map bijective hai.

Agar

dim V < dim W

ho, toh kya

T

kabhi surjective ho sakta hai?

Nahi. Rank

\leq dim V < dim W

, isliye image

W

ko fill nahi kar sakti; aisi map injective ho sakti hai lekin onto kabhi nahi.

Agar

dim V > dim W

ho, toh kya

T

ka nontrivial kernel hona zaroori hai?

Haan. Nullity

= dim V - rank \geq dim V - dim W > 0

, isliye kam se kam ek nonzero vector

0

tak crush hoga;

T

injective nahi ho sakta.

Finite-dimensional space par ek linear

T : V \to V

ke liye, kya injective aur surjective equivalent hain?

Haan. Rank–nullity force karta hai ki nullity

= 0 ⟺

rank

= dim V

, isliye trivial kernel aur full image saath-saath hote hain — injective, surjective aur bijective sab coincide karte hain.

Kya

R^{n}

R^{m}

tak ka ek linear map empty image rakh sakta hai?

Nahi. Image mein hamesha

0_{W} = T (0_{V})

hota hai, isliye woh minimum

{0}

hai aur kabhi empty nahi; "smallest possible" image zero subspace hai.

Dimension

0

wala subspace valid kernel ya image ho sakta hai?

Haan. Zero subspace

{0}

ki dimension

0

hai aur ye ek legitimate kernel (injective maps) ya image (zero map) hai. Dimension zero allowed hai, degenerate-invalid nahi.

Connections

Injective surjective bijective — yahan kernel/image tests exactly in teeno properties decide karte hain.
Rank–Nullity Theorem — upar ke har "edge case" ke peeche counting law.
Column space and null space — image aur kernel ke matrix names.
Change of basis — isliye rank aur nullity basis-independent hain.