Null space (kernel) and column space (image) — basis, dimension: Visual walkthrough

4.5.13 · D2 · Maths › Linear Algebra (Full) › Null space (kernel) and column space (image) — basis, dimens

Step 1 — Vector kya hai, aur ek matrix uske saath kya kar rahi hai?

KYA HAI. $R^{n}$ mein ek vector bas $n$ numbers ki ek list hai — origin (point $0$ ) se kisi jagah tak ek arrow. Input ko hum $x$ likhte hain. Ek matrix $A$ ek machine hai: isko ek arrow $x$ do, aur yeh ek naya arrow $A x$ return karta hai, jo shayad ek alag size ki space $R^{m}$ mein ho.

YAHAN SE KYUN SHURU KAREIN. Theorem ka har symbol — "input space", "output", "zero pe crush hona" — bekar hai jab tak aap arrow ko andar jaate aur bahar aate nahi dekh sakte. Toh pehle hum ek arrow ko poora safar karte hue dekhte hain.

PICTURE. Left mein, input plane $R^{2}$ ek blue arrow $x$ ke saath. Right mein, output space, jisme machine dwara banaya gaya yellow arrow $A x$ hai. Unke beech ka wavy arrow matrix $A$ ka action hai.

Step 2 — Output hamesha sirf columns ka ek blend ho sakta hai

KYA HAI. $x = (x_{1}, x_{2}, \dots, x_{n})$ ko split karo. Tab $A x = x_{1} col 1 c_{1} + x_{2} col 2 c_{2} + \dots + x_{n} c_{n} .$ Yahan $c_{j}$ $A$ ka $j$ -wa column hai (jo khud $R^{m}$ mein ek arrow hai), aur $x_{j}$ sirf ek number hai — us column ko kitna "on" karte hain.

YEH REWRITE KYUN. Yeh poore topic ki sabse gehri baat reveal karta hai: output hamesha columns ka ek weighted sum hota hai. $A$ jo bhi produce kar sakta hai woh columns ke blends ke set se bahar nahi ja sakta. Us set ka ek naam hai — column space $Col (A)$ . Toh "kaunse outputs reachable hain?" ka jawab sirf columns se milta hai.

PICTURE. Teen columns $c_{1}, c_{2}, c_{3}$ origin se coloured arrows ke roop mein drawn hain; output $A x$ unke scaled copies ko tip-to-tail stack karke bana hai. Sliders $x_{j}$ move karo aur $A x$ shaded region mein ghoomta hai — woh shaded region hi $Col (A)$ hai.

Step 3 — Kuch columns fake hain: woh koi nayi direction nahi add karte

KYA HAI. Maano column $c_{3}$ already $c_{1}$ aur $c_{2}$ ka blend hai, jaise $c_{3} = 2 c_{1} - c_{2}$ . Tab $c_{3}$ on karna kahin nahi pahunchata jahan pehle do pehle se nahi pahunche. Hum $c_{3}$ ko dependent kehte hain. Jo column kahin naya pahunchta hai woh independent hai — ek pivot column.

SIZE KE LIYE KYUN MAAYANE RAKHTA HAI. Column space ki dimension (kitni independent directions span karta hai) sirf pivot columns count karta hai. Woh count rank $r$ hai. Fakes free riders hain.

PICTURE. Do solid pivot arrows $c_{1}, c_{2}$ ek flat plane (shaded) span karte hain. Dependent arrow $c_{3}$ usi plane ke andar flat hai — dashed, koi nayi dimension contribute nahi karta.

Step 4 — Ek fake column aapko ek free direction deta hai jo crush ho jaati hai

KYA HAI. Wahi relation lo $c_{3} = 2 c_{1} - c_{2}$ , yani $2 c_{1} - c_{2} - c_{3} = 0$ . Isse ek aise input ki recipe ki tarah padho jo kuch bhi output nahi karta: $x = (2, - 1, - 1)$ choose karo. Tab $A x = 2 c_{1} - c_{2} - c_{3} = 0.$ Yeh input $x$ non-zero hai, phir bhi $A$ isse zero arrow pe bhej deta hai. Yeh null space $Null (A) = {x : A x = 0}$ ka member hai.

KYUN. Har dependent (free) column ek aisi recipe deta hai — input space mein ek genuine direction jo $A$ ek single point pe collapse kar deta hai. Toh free columns aur null-space directions ek hi sikke ke do pehlu hain, input side se dekhe gaye.

PICTURE. Left: input arrow $x = (2, - 1, - 1)$ $R^{3}$ mein, null line (red) pe baitha hua. Right: iska image single point $0$ hai — left mein poori red line right mein us ek dot pe squash ho gayi.

Step 5 — Har column ya toh pivot hai YA free. Kabhi dono nahi, kabhi koi nahi.

KYA HAI. $n$ columns mein se ek-ek karke guzro. Har column, jab row reduction wahan pahunchti hai, ya toh ek brand-new leading 1 introduce karta hai (pivot → nayi output direction, rank mein counted) ya nahi karta (free → dependent, ek crushed input deta hai, nullity mein counted). Koi teesra option nahi hai, aur koi column dono nahi ho sakta.

KYUN YAHI POORA THEOREM HAI. Hum ek set ko partition kar rahe hain — $n$ columns — bilkul do labelled buckets mein. Partitioning ka matlab hai bucket sizes total mein add hoti hain: $sab columns n = r (# pivots) + n - r (# free) .$

PICTURE. $n$ column-tokens ki ek row; har ek green (pivot) ya red (free) coloured hai — kabhi grey (unassigned) nahi, kabhi striped (dono) nahi. Green gino, red gino, woh poori strip tile karte hain.

Step 6 — Edge case A: kuch bhi crush nahi hota (injective)

KYA HAI. Agar har column pivot hai, $r = n$ , toh nullity $= n - n = 0$ . Sirf ek input $0$ pe crush hota hai, woh hai $x = 0$ khud. Null space sirf origin hai, ek single point jiska dimension $0$ hai.

KYUN. Koi free column nahi matlab koi free dial nahi, matlab zero ke liye koi non-trivial recipe nahi. Alag inputs alag outputs dete hain — map injective hai.

PICTURE. Input plane output plane pe map hota hai bina kisi squashing ke — ek full-dimensional shadow, red null "line" ek single origin dot tak simat jaati hai.

Step 7 — Edge case B: sab kuch crush ho jaata hai (zero matrix)

KYA HAI. Agar $A = 0$ (all-zeros $m \times n$ matrix) hai, toh koi bhi column kahin nahi pahunchta: $r = 0$ . Tab nullity $= n - 0 = n$ : poori input space ek single output point $0$ pe crush ho jaati hai. Column space ${0}$ hai, dimension $0$ .

KYUN DIKHAO. Yeh Step 6 ka mirror image hai aur confirm karta hai ki theorem extreme pe bhi balance rehti hai: $0 + n = n$ . Reader ko kabhi aisa case nahi milna chahiye jise hum ne skip kiya — yahan kuch survive nahi karta, sab kuch crush ho jaata hai, aur sum phir bhi check karta hai.

PICTURE. Poora blue input plane ek dot pe collapse ho jaata hai; output "space" bas woh origin dot hai. Rank bucket khaali, null bucket bhara hua.

Step 8 — Mixed case, poori tarah worked (parent ka Example 1)

KYA HAI. $A = 1312622634126$ RREF pe reduce hota hai jisme pivots columns 1 aur 3 mein hain. Toh $r = 2$ , aur columns 2 aur 4 free hain, nullity $2$ dete hain.

PICTURE KE ROOP MEIN KYUN REVISIT KAREIN. Teeno characters ek saath dekhne ke liye: 2 surviving directions (green pivots → $R^{3}$ ke andar ek 2-D plane $Col (A)$ ), 2 crushed directions (red free → $R^{4}$ ke andar ek 2-D plane $Null (A)$ ), jinka sum $n = 4$ hai.

PICTURE. Left half: 4 tokens ki input strip, columns 1,3 green aur 2,4 red. Right half: do special solutions $s_{1} = (- 2, 1, 0, 0)$ , $s_{2} = (0, 0, - 2, 1)$ null-space basis ke roop mein, har ek verify kiya gaya ki $0$ pe land karta hai.

Ek-picture summary

Sab kuch ek saath: input space $R^{n}$ ek surviving part (dimension $r$ , green drawn, faithfully $Col (A)$ pe mapped) aur ek crushed part (dimension $n - r$ , red drawn, origin pe squash) mein split hoti hai. Output space $R^{m}$ sirf green shadow receive karta hai. Input ke do coloured pieces count karo — unhe milke poora $n$ banana chahiye.

Recall Feynman retelling: poora walkthrough seedhe shabdon mein

Ek matrix ek aisi machine hai jo ek arrow andar leta hai aur ek arrow bahar deta hai. Jo bhi bahar deta hai woh hamesha uske column-arrows ka mix hota hai (Step 2), toh reachable outputs exactly column blends hain — column space. Kuch columns genuine nayi directions hain (pivots) aur kuch fakes hain jo doosron ko copy karte hain (Step 3). Har fake column secretly aapko ek aisa input batata hai jise machine kuch nahi karta — kyunki "col 3 equals 2·col 1 minus col 2" rearrange hota hai "koi input zero pe map hota hai" mein (Step 4). Ab sab $n$ columns line up karo aur har ek ko colour karo: green agar genuinely nayi direction hai, red agar fake hai. Beech mein kuch nahi hota (Step 5). Greens rank count karte hain (kitni output directions survive karti hain); reds nullity count karte hain (kitni input directions crush hoti hain). Kyunki har column ko exactly ek colour mila, greens + reds = sab columns = $n$ . Woh ek akela sentence hi Rank–Nullity hai. Aur yeh kabhi nahi tootha: agar kuch crush nahi hota toh map injective hai (Step 6), agar sab kuch crush hota hai toh zero matrix hai (Step 7), aur Example 1 jaisa normal mixed matrix 2 aur 2 mein split hoke 4 banaata hai (Step 8).

Connections

Rank–Nullity Theorem — yeh page uska visual derivation hai.
Rank of a matrix — green count $r = dim Col (A)$ .
Row reduction & RREF — hum columns ko green/red mein kaise sort karte hain.
Linear independence and basis — pivots = independent = green.
Injective and surjective linear maps — Step 6 ka trivial kernel.
Four fundamental subspaces — Col aur Null chaar mein se do hain.
Solving Ax=b — solutions = ek particular arrow + poori red crushed plane.
↑ Parent topic