Linear transformations — definition, kernel, image: Visual walkthrough

4.5.27 · D2 · Maths › Linear Algebra (Full) › Linear transformations — definition, kernel, image

Yeh parent note ka visual companion hai. Agar koi symbol naya lage, toh use picture milegi use use karne se pehle.

Step 0 — Woh words jo hum use kar sakte hain

Kisi bhi derivation se pehle, chaar plain-English ideas pakki kar lete hain. Inki hamen baar baar zaroorat padegi.

Aakhri wala yaad rakho: dimension = basis mein arrows ki count. Poora theorem ek basis ko do groups mein baantne aur unhe count karne ki kahaani hai.

Hamari running machine hogi $T : R^{3} \to R^{2}, T (x, y, z) = (x + y + z, x - z) .$ Yahan $V = R^{3}$ (input space, $3$ directions) aur $W = R^{2}$ (output space, $2$ directions) hai.

Step 1 — Machine ko ek direction crush karte dekhna

KYA. Machine ko $R^{3}$ mein har possible arrow dete hain aur dekhte hain woh $R^{2}$ mein kahan land karte hain.

KYUN. Theorem is baare mein hai ki kya bachta hai aur kya marta hai. Toh pehli cheez dekhni hai ek aisi direction jo mar jaati hai — ek aisa input jo machine zero output par flatten kar deta hai.

PICTURE. Neeche, left box 3-D input space $V$ hai; right box 2-D output space $W$ hai. Magenta arrow $(1, - 2, 1)$ dala jaata hai — aur woh $W$ mein exactly origin dot par land karta hai. Woh crush ho gaya.

Figure — Linear transformations — definition, kernel, image

Haath se check karo: $T (1, - 2, 1) = (x 1 + y (- 2) + z 1, x 1 - z 1) = (0, 0) .$

Pehla output slot hai $x + y + z = 1 - 2 + 1 = 0$ .
Doosra output slot hai $x - z = 1 - 1 = 0$ .

Toh yeh arrow zero arrow par map karta hai. Yeh kernel ka member hai.

Step 2 — Saare crush hue arrows collect karna: the kernel

KYA. Har woh input arrow dhundho jo zero par land ho, sirf ek nahi.

KYUN. Ek crush hua arrow ek coincidence hai; saare crush hue arrows ka set information carry karta hai. Woh set hai kernel, $ker T = {v : T (v) = 0}$ .

PICTURE. $V$ mein, saare crush hue arrows origin se guzarti magenta line banate hain. Us line par har arrow mar jaata hai; us se door koi nahi marta.

$T (x, y, z) = (0, 0)$ solve karo, yaani do equations: $x + y + z = 0 x - z = 0 (top output = 0) (bottom output = 0)$

Doosri equation se $x = z$ (yeh zyada simple hai, toh ise substitute karne ke liye use karo).
$x = z$ pehli mein daalo: $z + y + z = 0 \Rightarrow y = - 2 z$ .
Toh har kernel arrow $(z, - 2 z, z) = z (1, - 2, 1)$ hai.

$ker T = span {(1, - 2, 1)} .$ Yeh ek independent direction hai, toh $dim (ker T) = 1$ . Is number ko hum nullity kehte hain. Isko pakde rakho: nullity = 1.

Step 3 — Bachne wali directions ko bahar aate dekhna: the image

KYA. Dekho ki machine kahan bhej sakti hai — saare reachable outputs ka set.

KYUN. Jo bhi crush nahi hota use $W$ mein kahin toh jaana hi hai. Saare outputs ka collection hai image, $im T = {T (v) : v \in V}$ . Iska dimension hai rank.

PICTURE. $W$ mein, hum teen standard input directions $e_{1}, e_{2}, e_{3}$ ke outputs plot karte hain. Unmen se sirf do bhi poore plane ko cover karne ke liye fan out kar lete hain.

Kyunki $T (x) = A x$ jahan $A = (1110 1 - 1)$ hai, ek basis vector feed karna ek column pick karta hai: $T (e_{1}) = col 1 (1, 1), T (e_{2}) = col 2 (1, 0), T (e_{3}) = col 3 (1, - 1) .$

$(1, 1)$ aur $(1, 0)$ alag alag directions mein point karte hain → independent → milke saare $R^{2}$ ko span karte hain.
$(1, - 1)$ phir sirf un donon ka ek combination hai — koi nai direction nahi jodta.

Toh $im T = R^{2}$ , jis se rank = 2 milta hai.

Step 4 — Key idea: $V$ ka basis "crushed" + "extra" mein split karna

KYA. Poori input space $V$ ka basis banao kernel ke basis se shuru karke aur sirf utne extra arrows jodte hue jitne se $V$ fill ho jaye.

KYUN. Yeh proof ka engine hai. Agar hum input directions ko cleanly crush hui onon aur extra onon mein group kar sakein, toh har group alag count karna hamein nullity + (kuch) dega, aur hum dikhayenge ki woh "kuch" rank hai.

PICTURE. $V$ mein: magenta arrow $k_{1} = (1, - 2, 1)$ kernel ko span karta hai. Hum do violet arrows $w_{1}, w_{2}$ add karte hain jo is tarah choose kiye gaye hain ki teeno milke independent hon aur saare $R^{3}$ ko span karein.

$basis of k e r T, p = 1 {k_{1}} \cup extra vectors, q = 2 {w_{1}, w_{2}} = p + q = 3 basis of V .$

$p = dim (ker T) = 1$ — crush hua group.
$q$ = jitne extras ki zaroorat padi — survive karne wala group.
$p + q = dim V = 3$ kyunki $R^{3}$ ke basis mein exactly $3$ arrows hote hain.

Humne theorem abhi prove nahi ki — humne sirf $q$ ko "bacha hua count" ke roop mein define kiya hai. Agle steps prove karte hain ki $q$ rank ke barabar hai.

Step 5 — Extras poori image produce karte hain (woh use span karte hain)

KYA. Dikhao ki $T$ ka har output sirf $T (w_{1}), T (w_{2})$ se bana hai — extras ke images se.

KYUN. Agar woh do vectors already har output generate kar lete hain, toh woh image ko span karte hain. Crush hue arrows kuch contribute nahi karte, kyunki woh zero par map hote hain.

PICTURE. Left side ek general input feed karte hain, kernel part gayab ho jaata hai aur sirf violet extras ke images right side par bachte hain.

Koi bhi $v \in V$ hamare basis mein aise likha jaata hai: $v = kernel part a_{1} k_{1} + extra part b_{1} w_{1} + b_{2} w_{2} .$ $T$ apply karo aur linearity use karo ( $T$ sums par split hota hai aur scalars bahar kheenchta hai): $T (v) = a_{1} = 0 T (k_{1}) + b_{1} T (w_{1}) + b_{2} T (w_{2}) = b_{1} T (w_{1}) + b_{2} T (w_{2}) .$

$T (k_{1}) = 0$ kyunki $k_{1}$ kernel mein hai — yahi toh kernel ka matlab hai.
Toh har output sirf $T (w_{1}), T (w_{2})$ ka combination hai.

Isliye ${T (w_{1}), T (w_{2})}$ $im T$ ko span karta hai. Yeh "it's a basis" ka aadha hissa hai.

Step 6 — Extras ke images collapse nahi hote (woh independent hain)

KYA. Dikhao ki $T (w_{1})$ aur $T (w_{2})$ genuinely alag alag directions mein point karte hain — koi accidental overlap nahi.

KYUN. Image ka basis hone ke liye, span karna kaafi nahi; unhe independent bhi hona chahiye. Agar woh secretly collapse ho jaate, toh rank $q$ se chhota hota aur count toot jaata.

PICTURE. Hum suppose karte hain ki woh collapse hote hain ( $W$ mein ek hypothetical dashed collapse) aur contradiction ko $V$ ke andar trace karte hain: iska matlab ek bachne wali direction secretly kernel mein rehti.

Maano koi combination zero deta hai: $b_{1} T (w_{1}) + b_{2} T (w_{2}) = 0.$ $T$ ko bahar karo (linearity, ulti direction mein): $T (ise u kaho b_{1} w_{1} + b_{2} w_{2}) = 0 \Rightarrow u \in ker T .$

$u$ sirf extra arrows se bana hai.
Lekin $ker T$ sirf $k_{1}$ se span hota hai, toh $u = c k_{1}$ kisi scalar $c$ ke liye.

Ab hamare paas hai $b_{1} w_{1} + b_{2} w_{2} - c k_{1} = 0.$ Lekin ${k_{1}, w_{1}, w_{2}}$ ko basis — independent! hone ke liye choose kiya gaya tha. Independent vectors ka combination zero tab hi hota hai jab har coefficient zero ho: $b_{1} = 0, b_{2} = 0, c = 0.$ Toh $b_{1} T (w_{1}) + b_{2} T (w_{2}) = 0$ ka ek hi tarika hai — trivial wala → $T (w_{1}), T (w_{2})$ independent hain. Step 5 ke saath milke, woh image ka ek basis banate hain, toh $rank = dim (im T) = q .$

Step 7 — Counts ko milana

KYA. Do groups jodo.

KYUN. Ab hum jaante hain har group ki size: crush hua group $= p =$ nullity, extra group $= q =$ rank, aur milke woh $V$ ko tile karte hain.

PICTURE. Ek single balance scale: $V$ ki $3$ input directions exactly $1$ crushed (magenta) $+$ $2$ surviving (violet) mein split hoti hain.

$dim V = Step 4: kernel basis size p + Step 4: extras q = nullity dim (ker T) + rank dim (im T) .$ Haari machine ke liye: $3 = 1 + 2$ . ✓ Theorem hold karta hai. $■$

Abstract statement ke liye Rank–Nullity Theorem dekho, aur "column space = image, null space = kernel" language ke liye Column space and null space dekho.

Step 8 — Edge aur degenerate cases (taaki kuch surprise na kare)

KYA. Counting ko uske extremes par push karo. KYUN. Ek visual proof har corner mein survive karna chahiye, sirf tidy example mein nahi.

PICTURE. Teeno mini-machines same axes par: zero map, injective map, aur bijective map.

Ek-picture summary

Upar sab kuch ek single flow mein compress: $V$ ki $3$ input directions split hoti hain, ek kernel mein mar jaati hai, do image mein survive karti hain, aur $1 + 2 = 3$ .

Recall Feynman retelling (plain words, koi symbols nahi)

Ek machine socho jisme teen input dials hain (matlab $V$ 3-dimensional hai) aur ek do-light output panel ( $W$ hai). Tum dials hilaate ho aur lights dekhte ho. Dials ka ek special combination lights par kuch nahi karta — yeh ek aisi direction hai jo machine bilkul ignore karti hai; woh "crush" ho jaati hai. Yeh ek dead direction hai (nullity = 1). Baaki do directions actually lights hilaate hain, aur milke woh panel ko koi bhi pattern dikhane de sakte hain — lights poori tarah controllable hain. Yeh do live directions hain (rank = 2). Ab punchline: teen dials total = ek dead + do live. Directions kahin se bhi gayab ya zahir nahi hoti — har input direction ya toh crush hoti hai ya survive karti hai, kabhi dono nahi, kabhi koi nahi. Woh accounting, jo hamesha balance rehne par majboor hai, Rank–Nullity Theorem hai: $dim V = nullity + rank$ .

Recall Self-test

Kaunsa step prove karta hai ki extras image ko span karte hain? ::: Step 5 (kernel terms gayab ho jaate hain, toh har output $T (w_{i})$ ka combo hai). Kaunsa step prove karta hai ki woh independent hain? ::: Step 6 (ek collapse basis combination ko zero hone par majboor karta hai, impossible hai). Hamare $T$ ke liye nullity $= 1$ kyun hai? ::: Kernel ek single line $span {(1, - 2, 1)}$ hai. Rank $= 2$ kyun hai? ::: Image $R^{2}$ fill karta hai; do columns already use span karte hain. Zero map mein, rank aur nullity kya hain? ::: rank $= 0$ , nullity $= dim V$ .

Connections

Rank–Nullity Theorem — yeh page uska visual derivation hai.
Column space and null space — image = column space (Step 3), kernel = null space (Step 2).
Matrix multiplication — Step 3 columns read off karta hai kyunki $T (e_{i})$ = column $i$ hai.
Injective surjective bijective — Step 8 ke Cases B aur C.
Change of basis — Step 8 mistake box mein invariance.
Eigenvalues and eigenvectors — bijective (Case C) ka matlab koi zero eigenvalue nahi.