Null space (kernel) and column space (image) — basis, dimension: Worked examples

4.5.13 · D3 · Maths › Linear Algebra (Full) › Null space (kernel) and column space (image) — basis, dimens

Yeh page Null & Column space ka "no surprises" drill hai. Neeche ek map hai har tarah ki matrix ka jo is topic mein aa sakti hai. Phir hum har cell ke liye ek example work karte hain, taaki jab exam mein koi strange wala aaye, toh tum uska twin pehle se dekh chuke ho.

The scenario matrix

Hum ek matrix $A \in R^{m \times n}$ ko do sawaalon se classify karte hain jo uske dono spaces ke baare mein sab kuch decide kar dete hain:

Shape: kya yeh tall hai ( $m > n$ , unknowns se zyada equations), square ( $m = n$ ), ya wide ( $m < n$ , equations se zyada unknowns)?
Rank: kya iske paas full column rank hai ( $r = n$ , har input column ek genuinely naya direction hai), full row rank ( $r = m$ , har output direction reachable hai), ya yeh rank-deficient hai (kuch columns dependent hain)?

Yeh raha full grid. Har cell neeche ek example ka naam deta hai jo usse hit karta hai.

Shape \ Rank	Full column rank $r = n$	Rank-deficient $r < min (m, n)$	Full row rank $r = m$
Tall $m > n$	Ex 1 (injective, trivial kernel)	Ex 2 (dependent columns)	— impossible if $m > n$ *
Square $m = n$	Ex 3 (invertible)	Ex 4 (singular)	(same as full col rank)
Wide $m < n$	— impossible if $m < n$ *	Ex 5 (deficient wide)	Ex 6 (surjective, fat kernel)

Neeche wala decision flow-chart wahi taxonomy hai jo visually draw ki gayi hai: shape se shuru karo, rank pe split karo, aur pair $(dim Col, dim Null)$ pe land karo. Compute karne se pehle apni matrix ko isme trace karo — jis leaf pe pahuncho, wahi tumhara forecast hai.

Figure — Null space (kernel) and column space (image) — basis, dimension

Saath mein teen edge/extra cases jo har syllabus ko pasand hain:

Extra case	Example
Zero matrix (degenerate, sab kuch crushed)	Ex 7
Word problem (real-world reachability)	Ex 8
Exam twist (ek parameter $t$ jo rank change kar deta hai)	Ex 9

Ex 1 — Tall, full column rank (injective, trivial kernel)

Worked example Statement

$A = 110011 (m = 3, n = 2) .$ $Col (A)$ aur $Null (A)$ ke bases aur dimensions nikalo.

Forecast: Columns se zyada rows hain. Guess karo: kitne free variables ho sakte hain, aur iska null space ko kya force karta hai?

Step 1 — Row reduce. $R_{2} \to R_{2} - R_{1}$ , phir $R_{3} \to R_{3} - R_{2}$ : $100011 \to 100010 = R .$ Yeh step kyun? RREF expose karta hai ki kaun se columns pivots hold karte hain; yahi ek reliable independence test hai.

Step 2 — Pivots padho. Columns 1 aur 2 pivot columns hain, toh $r = 2$ . Koi free columns nahi hain. Yeh step kyun? Free columns null-space freedom ka source hain; agar koi nahi hai, toh kernel sirf ek single point hone ko forced hai.

Step 3 — Column space. Original $A$ se pivot columns lo: ${(1, 1, 0)^{⊤}, (0, 1, 1)^{⊤}}$ , toh $dim Col (A) = 2$ . Yeh $R^{3}$ ke andar tilted ek 2-D plane hai. Yeh step kyun? Row operations column entries ko move kar dete hain; sirf original columns true output directions mein point karte hain. Dekho Row reduction & RREF.

Step 4 — Null space. Koi free variables nahi ⇒ $A x = 0$ ka ek hi solution hai $x = 0$ . Toh $Null (A) = {0}$ , $dim = 0$ . Yeh map injective hai. Yeh step kyun? Jab har variable ek pivot se pin ho, toh vary karne ke liye kuch nahi bachta; kernel origin ke alawa koi direction contain nahi kar sakta, aur yahi exactly "injective" (koi do inputs ek output share nahi karte) ka matlab hai.

Verify: $r + nullity = 2 + 0 = 2 = n$ . ✓ Aur $A (0, 0)^{⊤} = (0, 0, 0)^{⊤}$ ek hi crush hai. Kyunki $r = 2 < 3 = m$ , yeh map not surjective hai — yeh $R^{3}$ ka zyada hissa miss kar deti hai.

Neeche wali figure yeh case draw karti hai: do accent-red original columns aur woh plane jo woh $R^{3}$ ke andar sweep karte hain. Origin pe woh akela black dot notice karo — woh lone point hi poora null space hai, "trivial kernel" ka geometric meaning.

Ex 2 — Tall, rank-deficient (dependent columns)

Worked example Statement

$A = 123246 (m = 3, n = 2) .$

Forecast: Kuch bhi touch karne se pehle dono columns dekho. Kya tumhe already koi dependence dikh rahi hai?

Step 1 — Dependence spot karo. Column 2 $= 2 \times$ column 1. Toh columns independent nahi hain. Yeh step kyun? 80/20 trick: ek obvious multiple relation rank instantly bata deta hai aur full reduction bachata hai.

Step 2 — Row reduce to confirm. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ : $100200 = R .$ Ek pivot ⇒ $r = 1$ . Yeh step kyun? Row 1 ke multiples subtract karne se neeche ki rows exactly zero hoti hain kyunki woh uske multiples thin — ek single nonzero row par collapse hona iska visible proof hai ki sirf ek pivot bachta hai.

Step 3 — Column space. Sirf column 1 mein pivot ⇒ $Col (A) = span {(1, 2, 3)^{⊤}}$ , $R^{3}$ mein ek single line. $dim = 1$ . Yeh step kyun? Column 2 kuch naya add nahi karta — yeh usi line par lie karta hai.

Step 4 — Null space. Free variable $x_{2}$ . Row 1 deta hai $x_{1} + 2 x_{2} = 0 \Rightarrow x_{1} = - 2 x_{2}$ . $x_{2} = 1$ set karo: special solution $(- 2, 1)^{⊤}$ . Toh $Null (A) = span {(- 2, 1)^{⊤}}$ , $dim = 1$ . Yeh step kyun? Single free variable ko $1$ set karna woh ek direction generate karta hai jisme kernel move karne ke liye free hai; har null vector uska multiple hai.

Verify: $r + nullity = 1 + 1 = 2 = n$ . ✓ Plug back karo: $A (- 2, 1)^{⊤} = - 2 (1, 2, 3)^{⊤} + (2, 4, 6)^{⊤} = (0, 0, 0)^{⊤}$ . ✓

Ex 3 — Square, full rank (invertible)

Worked example Statement

$A = [2111] (m = n = 2) .$

Forecast: Independent columns wali square matrix — compute karne se pehle dono dimensions kya hain?

Step 1 — Top-left pivot 1 banao. $R_{1} \to R_{1} - R_{2}$ : $[2111] \to [1101] .$ Yeh step kyun? Row 2 ko Row 1 se subtract karna leading entry ko clean $1$ mein turn karta hai aur saath hi $(1, 2)$ entry clear karta hai — ek hi move column 1 mein pivot set up kar deta hai.

Step 2 — Pehle pivot ke neeche clear karo. $R_{2} \to R_{2} - R_{1}$ : $[1101] \to [1001] = R .$ Do pivots ⇒ $r = 2$ . Yeh step kyun? Naye row 1 ko row 2 se subtract karna pehle pivot ke neeche entry zero kar deta hai; bacha hua second row already second pivot carry karta hai, toh hum exactly identity par land karte hain. RREF mein identity tak pahunchna ek invertible square matrix ki pehchaan hai.

Step 3 — Column space. Dono columns pivots hain ⇒ $Col (A) = R^{2}$ , $dim = 2$ . Har output reachable ⇒ surjective. Yeh step kyun? Do independent pivot columns poore 2-D output space ko span karte hain, toh $R^{2}$ mein kuch bhi out of reach nahi hai.

Step 4 — Null space. Koi free variables nahi ⇒ $Null (A) = {0}$ , $dim = 0$ ⇒ injective. Yeh step kyun? Har variable ek pivot variable hai, toh $A x = 0$ sab ko $0$ par pin kar deta hai — kernel ek single point hai.

Verify: $r + nullity = 2 + 0 = 2 = n$ . ✓ Determinant $= 2 \cdot 1 - 1 \cdot 1 = 1 \neq = 0$ invertibility confirm karta hai, jo trivial kernel aur full column space force karta hai.

Ex 4 — Square, rank-deficient (singular)

Worked example Statement

$A = 121241361 (m = n = 3) .$

Forecast: Row 2, Row 1 ka multiple hai. Kya isse tumhara ek pivot jaayega? Nullity predict karo.

Step 1 — Row reduce. $R_{2} \to R_{2} - 2 R_{1}$ (row 2 kill karo), $R_{3} \to R_{3} - R_{1}$ : $100 20 - 1 30 - 2 swap R_{2}, R_{3} 100 2 - 1 0 3 - 2 0 .$ Row 2 normalize karo ( $\times - 1$ ) aur upar clear karo: $R_{1} \to R_{1} - 2 R_{2}$ : $100010 - 1 20 = R .$ Columns 1,2 mein pivots ⇒ $r = 2$ . Yeh step kyun? $2 R_{1}$ subtract karna row 2 zero kar deta hai kyunki woh $2 \times$ row 1 thi; swap phir surviving nonzero row ko upar move karta hai taaki pivots diagonal par line up ho jaayein — yeh standard tidy-up hai jo free-variable formulas readable banata hai.

Step 2 — Column space. Original $A$ se pivot columns: ${(1, 2, 1)^{⊤}, (2, 4, 1)^{⊤}}$ , $dim = 2$ . $R^{3}$ mein ek plane. Yeh step kyun? Column 3 free hai, toh woh pehle do par depend karta hai — koi naya output direction add nahi karta.

Step 3 — Null space. Free variable $x_{3}$ . $R$ se: $x_{1} = x_{3}$ , $x_{2} = - 2 x_{3}$ . $x_{3} = 1$ set karo: $(1, - 2, 1)^{⊤}$ . $dim Null = 1$ . Yeh step kyun? Akela free variable woh single dial hai jo kernel ghuma sakta hai; use $1$ set karna woh ek basis direction read off karta hai.

Verify: $r + nullity = 2 + 1 = 3 = n$ . ✓ $A (1, - 2, 1)^{⊤} = (1 - 4 + 3, 2 - 8 + 6, 1 - 2 + 1)^{⊤} = (0, 0, 0)^{⊤}$ . ✓ Determinant $0$ hai (singular), nontrivial kernel ke saath consistent.

Ex 5 — Wide, rank-deficient

Worked example Statement

$A = [12241135] (m = 2, n = 4) .$

Forecast: Wide matrix, toh $n - r$ kam se kam $n - m = 2$ hai. Kya yeh aur bhi bada ho sakta hai? Nullity guess karo.

Step 1 — Row reduce. $R_{2} \to R_{2} - 2 R_{1}$ : $[1020 1 - 1 3 - 1] .$ Column 3 mein pivot ke upar clear karo: $R_{1} \to R_{1} + R_{2}$ , phir $R_{2} \to - R_{2}$ : $[10200121] = R .$ Columns 1,3 mein pivots ⇒ $r = 2$ . Yeh step kyun? $R_{2}$ se $2 R_{1}$ subtract karna pehla entry cancel karta hai taaki next pivot column 3 mein appear ho (column 2 zero ho gaya, use free bana ke); phir $R_{2}$ ko wapas $R_{1}$ mein add karna us pivot ke upar entry clear karta hai taaki har pivot column ek clean unit column ho — exactly yahi free variables ke formulas readable banata hai $x_{1}, x_{3}$ ko.

Step 2 — Column space. Original columns 1,3: ${(1, 2)^{⊤}, (1, 1)^{⊤}}$ . Yeh independent hain ⇒ $Col (A) = R^{2}$ , $dim = 2$ . Chhote output space par surjective. Yeh step kyun? Do independent columns already 2-D output space fill kar dete hain, toh $R^{2}$ mein har target reachable hai.

Step 3 — Null space. Free variables $x_{2}, x_{4}$ . $R$ se: $x_{1} = - 2 x_{2} - 2 x_{4}$ , $x_{3} = - x_{4}$ .

$x_{2} = 1, x_{4} = 0$ : $s_{1} = (- 2, 1, 0, 0)^{⊤}$ .
$x_{2} = 0, x_{4} = 1$ : $s_{2} = (- 2, 0, - 1, 1)^{⊤}$ . $dim Null = 2$ . Yeh step kyun? Do free variables ka matlab do independent dials hain; ek ek ko akele turn karna (baaki $0$ ) do basis null vectors produce karta hai.

Verify: $r + nullity = 2 + 2 = 4 = n$ . ✓ $s_{2}$ check karo: $A s_{2} = - 2 (1, 2)^{⊤} + 0 + (- 1) (1, 1)^{⊤} + (3, 5)^{⊤} = (- 2 - 1 + 3, - 4 - 1 + 5)^{⊤} = (0, 0)^{⊤}$ . ✓

Ex 6 — Wide, full row rank (surjective, fat kernel)

Worked example Statement

$A = [100123] (m = 2, n = 3) .$

Forecast: Already almost RREF mein hai. Yeh sabse clean wide case hai: ek nazar mein dono spaces predict karo.

Step 1 — Pivots padho. Already RREF: columns 1,2 mein pivots ⇒ $r = 2 = m$ . Full row rank. Yeh step kyun? Koi row operation zaroori nahi — leading 1's already apne columns mein akele baith ti hain, toh hum directly pivots padh sakte hain.

Step 2 — Column space. Pivot columns $(1, 0)^{⊤}, (0, 1)^{⊤}$ sab $R^{2}$ span karte hain ⇒ $Col (A) = R^{2}$ , $dim = 2$ . Har output reachable ⇒ surjective. Yeh step kyun? $r = m$ exactly surjectivity ki condition hai — dekho Injective and surjective linear maps.

Step 3 — Null space. Free variable $x_{3}$ . $x_{1} = - 2 x_{3}$ , $x_{2} = - 3 x_{3}$ . $x_{3} = 1$ set karo: $(- 2, - 3, 1)^{⊤}$ . $dim Null = 1$ . Yeh step kyun? Single free variable kernel ki woh ek direction furnish karta hai; har crushed input uska multiple hai.

Verify: $r + nullity = 2 + 1 = 3 = n$ . ✓ $A (- 2, - 3, 1)^{⊤} = (- 2 + 2, - 3 + 3)^{⊤} = (0, 0)^{⊤}$ . ✓

Neeche wali figure yeh null space dikhati hai: direction $(- 2, - 3, 1)$ mein $R^{3}$ mein origin se guzarti ek single accent-red line. Us line par har point $A$ mein feed hota hai aur zero vector nikalti hai — poori line "wasted input" hai.

Ex 7 — Degenerate: zero matrix

Ex 8 — Word problem (real-world reachability)

Worked example Statement

Ek juice bar do drinks banata hai. Har Green mein $2$ apples aur $1$ kale lagte hain; har Gold mein $4$ apples aur $2$ kale. Agar tum $g$ Greens aur $d$ Golds banate ho, toh kul use hua (apples, kale) hai $A [g d], A = [2142] .$ (a) Kaun se (apples, kale) totals achievable hain? (b) Kaun si recipes kuch waste nahi karti doosre ke comparison mein — yaani same total deti hain?

Forecast: Gold mein Green se exactly do guna sab kuch lagta hai. Kya tum expect karte ho ki achievable totals poora plane fill karein, ya ek line?

Step 1 — Rank. Column 2 $= 2 \times$ column 1 ⇒ $r = 1$ . Yeh step kyun? Dependent columns ka matlab hai ki dono drinks resource space mein ek hi direction mein kheenchti hain.

Step 2 — (a) Column space. $Col (A) = span {(2, 1)^{⊤}}$ : ek line. Sirf woh totals achievable hain jinka apple-to-kale ratio $2 : 1$ hai. Tum kabhi $(3, 3)$ ka total produce nahi kar sakte — yeh $Col (A)$ mein nahi hai. (Solvable iff $b \in Col (A)$ ; dekho Solving Ax=b.) Yeh step kyun? Achievable set by definition columns ka span hai; rank $1$ ke saath woh span ek single line hai, toh line se bahar koi bhi target impossible hai chahe kitni bhi drinks banao.

Step 3 — (b) Null space. Solve karo $2 g + 4 d = 0 \Rightarrow g = - 2 d$ . Special solution $(- 2, 1)^{⊤}$ . Interpreted: " $- 2$ Greens ke badle $+ 1$ Gold" swap karna resource total mein kuch nahi badlata — ek Gold do Greens ka resource use karta hai. Woh direction null space hai. Yeh step kyun? Null space exactly woh recipe changes ka set hai jo resource total fix rakhti hain; $A x = 0$ solve karna unhe dhundh ta hai, aur free variable $d$ woh ek trade direction deta hai.

Verify: $r + nullity = 1 + 1 = 2 = n$ . ✓ $A (- 2, 1)^{⊤} = (- 4 + 4, - 2 + 2)^{⊤} = (0, 0)^{⊤}$ : resource-neutral trade. ✓

Neeche wali figure resource plane draw karti hai: accent-red achievable totals ki line (sab $2 : 1$ ratio mein) aur black dot $(3, 3)$ par jo us line se bahar hai — geometric reason ki $(3, 3)$ kabhi produce nahi ho sakta.

Ex 9 — Exam twist: ek parameter rank change karta hai

Worked example Statement

$A (t) = [12 2 t] (m = n = 2) .$ Kis value of $t$ par null space nontrivial ho jaata hai? Dono regimes mein bases do.

Forecast: Zyaatar $t$ ke liye columns independent hain. Exactly ek "bura" $t$ hai jahan second row collapse ho jaati hai. Use guess karo.

Step 1 — Parameter ke saath row reduce. $R_{2} \to R_{2} - 2 R_{1}$ : $[10 2 t - 4] .$ Yeh step kyun? Row 2 mein pivot tabhi exist karta hai jab entry $t - 4$ nonzero ho. Parameters aisi entries par jeete ya marte hain, toh $t - 4$ isolate karna poori kahaani bata deta hai.

Step 2 — Case $t \neq = 4$ . Toh $t - 4 \neq = 0$ ⇒ do pivots ⇒ $r = 2$ . $Col = R^{2}$ , $Null = {0}$ , nullity $0$ . Invertible. Yeh step kyun? Nonzero $(2, 2)$ entry second pivot ban jaati hai, toh koi variable free nahi hai aur map ek saath injective aur surjective hai.

Step 3 — Case $t = 4$ . Toh row 2 all zeros ho jaati hai ⇒ $r = 1$ . Column space $= span {(1, 2)^{⊤}}$ . Free variable $x_{2}$ : $x_{1} + 2 x_{2} = 0 \Rightarrow x_{1} = - 2 x_{2}$ . Special solution $(- 2, 1)^{⊤}$ . Nullity $= 1$ . Yeh step kyun? Critical $t$ par second row gaayab ho jaati hai, rank drop karta hai; ab-free variable $x_{2}$ ek one-dimensional kernel khol deta hai.

Verify: $t = 4$ par, $det A = 1 \cdot 4 - 2 \cdot 2 = 0$ (singular) ⇒ nontrivial kernel, consistent. $A (4) (- 2, 1)^{⊤} = (- 2 + 2, - 4 + 4)^{⊤} = (0, 0)^{⊤}$ . ✓ Maano $t = 5$ : $det = 5 - 4 = 1 \neq = 0$ , kernel trivial. ✓

Recall

Recall Kaun sa scenario kaun sa space force karta hai?

Full column rank ( $r = n$ ) ::: trivial null space, map injective. Full row rank ( $r = m$ ) ::: column space poora $R^{m}$ hai, map surjective. Wide matrix ( $m < n$ ) ::: nullity $\geq n - m > 0$ , toh HAMESHA ek nonzero null vector hoga. Zero matrix ::: rank $0$ , null space poora input space hai. Parameter jo pivot kill karta hai ::: rank girta hai, nullity badhta hai, determinant $0$ ho jaata hai (agar square ho).

Connections

Rank of a matrix — woh $r$ jo har example mein count kiya gaya.
Rank–Nullity Theorem — har worked case ki check line.
Row reduction & RREF — woh engine jo pivots dhundha.
Linear independence and basis — kyun pivot columns ek basis hain.
Injective and surjective linear maps — Ex 1/3 (injective), Ex 3/6 (surjective).
Four fundamental subspaces — Col aur Null chaar mein se do hain.
Solving Ax=b — Ex 8 ka reachability sawal.