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

4.5.13 · D4 · Maths › Linear Algebra (Full) › Null space (kernel) and column space (image) — basis, dimens

Ladder se pehle, sirf paanch facts ka ek one-screen refresher jo tumhe chahiye. Neeche har solution inhi facts ka careful application hai.

Recall Paanch load-bearing facts (dhako aur recite karo)

Pivot = RREF (reduced row echelon form) mein ek leading $1$ . Pivots ki count $= r = rank (A)$ .
$dim Col (A) = r$ ; Col ka basis = pivot-position columns jo original $A$ se liye gaye hain.
$dim Null (A) = n - r$ (nullity); Null ka basis = special solutions, ek per free column.
$Col (A) \subseteq R^{m}$ (outputs), $Null (A) \subseteq R^{n}$ (inputs). Yahan $A$ hai $m \times n$ : $m$ rows, $n$ columns.
Rank–Nullity: $r + (n - r) = n$ = columns ki total number.

Yeh raha poore exercise flow ka map, taaki tum hamesha jaano tum kahan ho:

L1 — Recognition

Goal: matrix ya uske RREF se structure padho, bina bhaari computation ke.

Recall Solution 1.1

$3$ pivots hain, isliye $r = 3$ . (a) $rank (A) = 3$ . (b) $dim Col (A) = r = 3$ . (c) $dim Null (A) = n - r = 5 - 3 = 2$ (free columns hain $2$ aur $5$ ). (d) $Col (A) \subseteq R^{m} = R^{3}$ — outputs mein utni hi entries hoti hain jitni rows hoti hain.

Recall Solution 1.2

$R_{1}$ : leading $1$ 's columns $1$ aur $3$ mein hain → pivot columns ${1, 3}$ , free column ${2}$ . To $r = 2$ , ek free variable. $R_{2}$ : leading $1$ 's columns $1$ aur $2$ mein hain → pivot columns ${1, 2}$ , free column ${3}$ . To $r = 2$ , ek free variable.

L2 — Application

Goal: ek concrete matrix se dono bases poori tarah compute karo.

Recall Solution 2.1

Step 1 — row reduce (WHAT): pehle pivot ke neeche eliminate karo. $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ : $100200111 .$ Ab $R_{3} \to R_{3} - R_{2}$ , phir doosre pivot ke upar clear karo $R_{1} \to R_{1} - R_{2}$ se: $R = 100200010 .$ RREF kyun: yeh pivots expose karta hai aur pivot-in-terms-of-free formulas seedha deta hai.

Step 2 — pivots columns $1, 3$ mein → $r = 2$ . Col basis = original $A$ ke columns $1, 3$ : ${(1, 2, 3)^{⊤}, (1, 3, 4)^{⊤}}$ . $dim Col = 2$ . Original columns kyun: row operations ne $R$ ke columns ke andar entries move kar di, isliye $R$ ke columns ab sahi output directions mein point nahi karte.

Step 3 — null space. Free variable $x_{2}$ . $R$ se: $x_{1} = - 2 x_{2}$ , $x_{3} = 0$ . $x_{2} = 1$ set karo: special solution $s = (- 2, 1, 0)^{⊤}$ . Null basis $= {(- 2, 1, 0)^{⊤}}$ , $dim Null = 1$ .

Step 4 — check: $r + nullity = 2 + 1 = 3 = n$ . ✓ Sanity: $A s = - 2 (1, 2, 3)^{⊤} + 1 (2, 4, 6)^{⊤} = (0, 0, 0)^{⊤}$ . ✓

Recall Solution 2.2

Row reduce: $R_{2} \to R_{2} - 2 R_{1}$ se ek zero second row milti hai: $R = 000100200 .$ Row $1$ ki pehli nonzero entry column 2 mein hai — to single pivot column $2$ mein hai, column $1$ mein nahi. Pivot column: ${2}$ ; free columns: ${1, 3}$ . $r = 1$ . Col basis = original $A$ ka column $2$ : ${(1, 2, 0)^{⊤}}$ . $dim Col = 1$ . Null space: free $x_{1}, x_{3}$ . Row $1$ : $x_{2} + 2 x_{3} = 0 \Rightarrow x_{2} = - 2 x_{3}$ ; $x_{1}$ unconstrained hai.

$x_{1} = 1, x_{3} = 0$ : $(1, 0, 0)^{⊤}$ .
$x_{1} = 0, x_{3} = 1$ : $(0, - 2, 1)^{⊤}$ . Null basis $= {(1, 0, 0)^{⊤}, (0, - 2, 1)^{⊤}}$ , $dim Null = 2$ . Check: $1 + 2 = 3 = n$ . ✓

L3 — Analysis

Goal: dimension aur structure ke baare mein reason karo, kabhi kabhi bina specific matrix ke.

Recall Solution 3.1

Yahan $m = 4$ , $n = 6$ . (a) Rank–Nullity: $r = n - dim Null = 6 - 2 = 4$ . (b) Injective $⟺ Null (A) = {0} ⟺$ nullity $0$ . Yahan nullity $2 \neq = 0$ hai, to injective nahi hai — alag inputs same output de sakte hain. (c) $R^{4}$ pe surjective $⟺ dim Col (A) = 4 = m$ . Yahan $dim Col = r = 4 = m$ , to haan, yeh surjective hai — har output vector reachable hai. Isko padhna: ek wide matrix ( $n > m$ ) full row rank pe extra input directions ko flatten karta hai (nullity $> 0$ ) lekin phir bhi poore output space ko cover karta hai.

Recall Solution 3.2

Yahan $n = 3$ columns hain, to rank satisfy karta hai $0 \leq r \leq min (m, n) = min (5, 3) = 3$ . Nullity $= n - r = 3 - r$ . Sabse badi nullity sabse chote rank pe hoti hai, $r = 0$ (zero matrix), jisse nullity $3$ milti hai. To nullity $3$ tak pahunch sakti hai, sirf $2$ tak nahi. Statement GALAT hai. Sabse tight sahi bound hai $0 \leq dim Null (A) \leq 3$ .

L4 — Synthesis

Goal: null space, column space aur $A x = b$ ki solvability ko combine karo.

Recall Solution 4.1

Yaad karo $A = 123246134$ , columns $c_{1} = (1, 2, 3)^{⊤}$ , $c_{2} = (2, 4, 6)^{⊤}$ , $c_{3} = (1, 3, 4)^{⊤}$ , aur $Col (A) = span {c_{1}, c_{3}}$ (Ex 2.1 se). (a) Kya $b$ reachable hai? (solvable iff $b\in\mathrm{Col}(A)$.) Try karo $b = α c_{1} + γ c_{3}$ : $α + γ = 2, 2 α + 3 γ = 5, 3 α + 4 γ = 7.$ Pehle se: $α = 2 - γ$ . Doosre mein substitute karo: $2 (2 - γ) + 3 γ = 5 \Rightarrow 4 + γ = 5 \Rightarrow γ = 1$ , to $α = 1$ . Teesra check karo: $3 (1) + 4 (1) = 7$ . ✓ To haan, $b \in Col (A)$ hai. (b) Particular solution. Humne likha $b = 1 \cdot c_{1} + 0 \cdot c_{2} + 1 \cdot c_{3}$ , yaani $x_{1} = 1, x_{2} = 0, x_{3} = 1$ . To $x_{p} = (1, 0, 1)^{⊤}$ . (Verify karo: $A x_{p} = c_{1} + c_{3} = (1, 2, 3)^{⊤} + (1, 3, 4)^{⊤} = (2, 5, 7)^{⊤} = b$ . ✓) (c) Full solution set = particular + null space: $x = x_{p} + t s = 101 + t - 2 10, t \in R,$ Ex 2.1 se null-space basis $s = (- 2, 1, 0)^{⊤}$ use karke. Yeh $R^{3}$ mein ek line hai: infinitely many solutions kyunki kernel nontrivial hai.

Neeche ki picture solution set ke do flavours dikhati hai — ek single point jab kernel trivial ho, aur poori line jab na ho.

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

Recall Solution 4.2

Null-space basis mein $1$ vector hai, to $dim Null (A) = 1$ . Rank–Nullity: $r = n - 1 = 3 - 1 = 2$ . Invertible $⟺$ full rank $r = n = 3$ $⟺$ nullity $0$ . Yahan nullity $1 \neq = 0$ hai, to $A$ invertible NAHI hai — yeh direction $(1, 1, 1)^{⊤}$ ko zero pe collapse kar deta hai, to isko undo nahi kiya ja sakta.

L5 — Mastery

Goal: ek general statement prove karo aur specification ke hisaab se matrix design karo.

Recall Solution 5.1

Requirements ko column relations ki tarah padho. $A (1, - 1, 0, 0)^{⊤} = 0$ matlab $c_{1} - c_{2} = 0$ , yaani $c_{1} = c_{2}$ . $A (0, 0, 1, - 1)^{⊤} = 0$ matlab $c_{3} - c_{4} = 0$ , yaani $c_{3} = c_{4}$ . To columns do equal pairs mein aate hain. $rank = 2$ ke liye $c_{1}$ aur $c_{3}$ independent chahiye. Choose karo $c_{1} = (1, 0, 0)^{⊤}$ , $c_{3} = (0, 1, 0)^{⊤}$ . Tab: $A = 100100010010 .$ Rank verify karo: RREF $A$ hi hai; columns $1, 3$ mein pivots → $r = 2$ . ✓ Kernel vectors verify karo: $A (1, - 1, 0, 0)^{⊤} = (0, 0, 0)^{⊤}$ ✓; $A (0, 0, 1, - 1)^{⊤} = (0, 0, 0)^{⊤}$ ✓. Rank–Nullity se consistency: nullity $= n - r = 4 - 2 = 2$ , exactly woh do independent kernel vectors jo humne prescribe kiye. ✓

Recall Solution 5.2

Setup. Rank pivots count karta hai, aur har pivot apni khud ki row aur apna khud ka column occupy karta hai, isliye $r \leq m$ aur $r \leq n$ ; khaaskar $r \leq m$ . Core step. Rank–Nullity se, $dim Null (A) = n - r$ . Kyunki $r \leq m$ , $dim Null (A) = n - r \geq n - m .$ Hypothesis use karo. Given $m < n$ , humare paas $n - m > 0$ hai, isliye $dim Null (A) \geq n - m > 0$ . Positive dimension ka ek subspace nonzero vector contain karta hai, to $Null (A) \neq = {0}$ . $■$ Matlab: inputs outputs se zyada hone par machine ko kam se kam $n - m$ independent directions zero pe squash karne padenge — ek wide map kabhi bhi injective nahi ho sakta.

Connections

Rank of a matrix — upar har "find $r$ " step mein.
Rank–Nullity Theorem — L3–L5 ko power dene wali identity.
Row reduction & RREF — Ex 2.1, 2.2, 5.1 ke peeche ka engine.
Linear independence and basis — kyun pivot columns ek basis form karte hain.
Injective and surjective linear maps — Ex 3.1, 4.2, 5.2.
Solving Ax=b — Ex 4.1 ka particular + kernel structure.
Four fundamental subspaces — woh badi picture jisme yeh bases baithte hain.