Linear transformations — definition, kernel, image: Exercises

4.5.27 · D4 · Maths › Linear Algebra (Full) › Linear transformations — definition, kernel, image

Level 1 — Recognition

Exercise 1.1

In mein se kaun se maps $T : R^{2} \to R^{2}$ linear hain?

(a) $T (x, y) = (3 x, x - y)$
(b) $T (x, y) = (x + 2, y)$
(c) $T (x, y) = (x y, 0)$

Recall Solution

"Linear" ki demand: map ko $0 \to 0$ bhejna chahiye, aur adding aur scaling ko respect karna chahiye (dekho the definition). Sabse fast pehla filter hai yeh sanity check: $T (0) = 0$ .

(a) $T (0, 0) = (0, 0)$ ✓. Ab homogeneity aur additivity check karo — lekin notice karo ki har output $x$ aur $y$ ke constant-multiples ka sum hai, yaani degree-1 bina kisi constant ke. $T (c (x, y)) = (3 c x, c x - cy) = c (3 x, x - y) = c T (x, y)$ ✓, aur sum ko expand karne par cleanly split ho jaata hai. Linear. ✓

(b) $T (0, 0) = (2, 0) \neq = (0, 0)$ . Kyun fail hota hai: $+ 2$ ek constant shift hai (affine), aur shifting origin ko move kar deta hai. Linear nahi hai.

(c) $T (0, 0) = (0, 0)$ ✓ — lekin yeh sasta test pass kar leta hai aur phir bhi fail karta hai. Homogeneity try karo: $T (c (x, y)) = (c x \cdot cy, 0) = (c^{2} x y, 0)$ , jabki $c T (x, y) = (c x y, 0)$ . Ye alag hain ( $c^{2} \neq = c$ for e.g. $c = 2$ ). Linear nahi hai — product $x y$ degree 2 ka hai.

Exercise 1.2

$T (x, y, z) = (x - z, y)$ ke liye, matrix $A$ likhoo jisme $T (v) = A v$ ho.

Recall Solution

Hum kya karte hain: standard basis vectors ko andar daalo aur outputs ko columns ki tarah stack karo — yahi tarika hai jisse ek linear map matrix ban jaati hai (parent note §1). $T (e_{1}) = T (1, 0, 0) = (1, 0)$ , $T (e_{2}) = T (0, 1, 0) = (0, 1)$ , $T (e_{3}) = T (0, 0, 1) = (- 1, 0)$ . Columns kyun: koi bhi $x = x_{1} e_{1} + x_{2} e_{2} + x_{3} e_{3}$ deta hai $T (x) = \sum x_{i} T (e_{i})$ , exactly matrix-times-vector. $A = (1001 - 1 0) .$ Yeh Matrix multiplication se connect hota hai.

Level 2 — Application

Exercise 2.1

$ker T$ aur uski dimension (nullity) nikalo: $T : R^{3} \to R^{2}, T (x, y, z) = (x + 2 y + z, 2 x + 4 y + 2 z) .$

Recall Solution

Kernel kya hai: woh saare inputs jo $(0, 0)$ par squash ho jaate hain (parent §2). Toh solve karo: $x + 2 y + z = 0, 2 x + 4 y + 2 z = 0.$ Solve karne se pehle dekho: doosri equation pehli ki exactly twice hai, toh usme koi nayi information nahi — ek genuine equation, teen unknowns. $x + 2 y + z = 0 \Rightarrow x = - 2 y - z$ set karo. Do free choices $y, z$ : $(x, y, z) = (- 2 y - z, y, z) = y (- 2, 1, 0) + z (- 1, 0, 1) .$ $ker T = span {(- 2, 1, 0), (- 1, 0, 1)}, nullity = 2.$ Ye do vectors independent hain (koi ek doosre ka multiple nahi hai), toh dimension sach mein $2$ hai. Dekho Column space and null space.

Exercise 2.2

Usi $T$ ke liye, $im T$ aur rank nikalo.

Recall Solution

Image kya hai: woh sab jo $T$ output kar sakta hai = uski matrix ka column space. $A = (122412)$ ke columns hain $(1, 2), (2, 4), (1, 2)$ . Yeh shortcut kyun kaam karta hai: $A x = \sum x_{i} (column_{i})$ , toh outputs exactly columns ke combinations hain. Teeno columns $(1, 2)$ ke multiples hain. Toh woh ek hi line span karte hain: $im T = span {(1, 2)} \subseteq R^{2}, rank = 1.$ Rank–Nullity Theorem se check karo: $dim V = 3$ , nullity $2$ + rank $1 = 3$ ✓.

Figure — Linear transformations — definition, kernel, image

Level 3 — Analysis

Exercise 3.1

Ek linear map $T : R^{4} \to R^{3}$ mein $dim (ker T) = 2$ hai. Kya $T$ injective hai? Kya surjective hai? Justify karo.

Recall Solution

Injective? $T$ injective hai $⟺ ker T = {0}$ (parent §2). Yahan $dim (ker T) = 2 > 0$ , toh kernel mein nonzero vectors hain. Injective nahi hai — dekho Injective surjective bijective.

Surjective? Rank–Nullity Theorem use karo: $dim V = 4 = nullity + rank = 2 + rank$ , toh $rank = 2$ . Image $W = R^{3}$ mein baithti hai lekin uski dimension sirf $2 < 3$ hai. Yeh kyun settle karta hai: 2-dimensional image ek 3-dimensional space ko fill nahi kar sakti. Surjective nahi hai.

Exercise 3.2

Projection $P : R^{3} \to R^{3}$ , $P (x, y, z) = (x, y, 0)$ , space ko $x y$ -plane par flatten kar deta hai. $ker P$ , $im P$ nikalo, aur rank–nullity verify karo.

Recall Solution

Kernel — kya $0$ par crush hota hai: $(x, y, 0) = (0, 0, 0)$ solve karo, jaisse $x = 0, y = 0$ , $z$ free milta hai. $ker P = span {(0, 0, 1)} = the z -axis, nullity = 1.$ Kaisa dikhta hai: poori vertical $z$ -axis woh shadow-line hai jo origin par land hoti hai — origin ke directly upar/neeche ke har point us par collapse ho jaata hai.

Image — kya bahar aa sakta hai: har output ka third coordinate $0$ hota hai, aur koi bhi $(a, b, 0)$ $P (a, b, kuch bhi)$ se reachable hai. $im P = {(a, b, 0)} = the x y -plane, rank = 2.$ Rank–nullity: $dim V = 3 = 1 + 2$ ✓.

Level 4 — Synthesis

Exercise 4.1

Maano $D : P_{2} \to P_{2}$ degree $\leq 2$ polynomials par differentiation hai, yaani $D (a + b x + c x^{2}) = b + 2 c x$ . Basis ${1, x, x^{2}}$ use karke $D$ ki matrix nikalo, phir $ker D$ , $im D$ , aur rank–nullity check karo.

Recall Solution

Differentiation linear kyun hai: $\frac{d}{d x} (f + g) = f^{'} + g^{'}$ aur $\frac{d}{d x} (c f) = c f^{'}$ — exactly additivity aur homogeneity. Toh $D$ polynomials ke vector space par ek genuine linear transformation hai.

Matrix (basis images ke zariye): har basis polynomial ko andar daalo. $D (1) = 0$ , $D (x) = 1$ , $D (x^{2}) = 2 x$ . Outputs ko coordinates $(a, b, c)$ mein likhein $a + b x + c x^{2}$ ke liye: $D (1) = (0, 0, 0), D (x) = (1, 0, 0), D (x^{2}) = (0, 2, 0) \Rightarrow M = 000100020 .$ Kernel: $D (p) = 0$ ka matlab hai $b + 2 c x = 0$ for all $x$ , jo force karta hai $b = 0, c = 0$ ; $a$ free. $ker D = span {1} = {constants}, nullity = 1.$ Kyun: constant ka derivative zero hota hai — constants exactly woh information hain jo derivative kho deta hai. Image: outputs $b + 2 c x$ range karte hain degree $\leq 1$ ke saare polynomials par. $im D = span {1, x}, rank = 2.$ Rank–nullity: $dim P_{2} = 3 = 1 + 2$ ✓.

Exercise 4.2

Ek linear map $T : R^{3} \to R^{3}$ banao jiska kernel line $span {(1, 1, 1)}$ ho aur image plane $z = 0$ ho. Ek explicit matrix do, ya explain karo kyun impossible hai.

Recall Solution

Pehle feasibility (rank–nullity): hum chahte hain nullity $= 1$ aur rank $= 2$ ; $1 + 2 = 3 = dim R^{3}$ ✓, toh possible hai. Strategy: humein ek matrix $A$ chahiye jisme (i) $A (1, 1, 1)^{⊤} = 0$ ho aur (ii) column space $= {z = 0}$ ho, yaani saare outputs ka third coordinate $0$ ho → $A$ ki poori bottom row zero hai. Try karo: $A = 100 - 1 10 0 - 1 0 .$ Kernel check: $A (1, 1, 1)^{⊤} = (1 - 1, 1 - 1, 0) = (0, 0, 0)$ ✓. $A x = 0$ solve karo: $x - y = 0, y - z = 0 \Rightarrow x = y = z$ , toh $ker T = span {(1, 1, 1)}$ ✓, nullity 1. Image check: top do rows independent hain, third row zero, toh outputs saare $(a, b, 0)$ hain = plane $z = 0$ ✓, rank 2. Ho gaya. (Bahut saare valid answers exist karte hain; Change of basis se connect hota hai agar tum $T$ banao yeh choose karke ki basis vectors kahan jaate hain.)

Level 5 — Mastery

Exercise 5.1

Maano $T : V \to V$ finite-dimensional space par linear hai aur $T \circ T = T$ (ek projection, ya idempotent map). Prove karo ki $V = ker T \oplus im T$ — yaani har $v$ uniquely split hota hai (kernel part) + (image part) mein.

Recall Solution

" $\oplus$ " ka matlab: do subspaces $A, B$ satisfy karte hain $V = A \oplus B$ jab (1) $A \cap B = {0}$ aur (2) $A + B = V$ (har vector dono mein se ek ka sum hai). Hume dono show karne hain $A = ker T$ , $B = im T$ ke liye.

Step 1 — spanning ( $A + B = V$ ). Kisi bhi $v$ ke liye, yeh clever identity likho: $v = isse k kaho (v - T (v)) + isse i kaho T (v) .$ Yeh split kyun: $T (v)$ manifestly image mein hai. Check karo $k \in ker T$ : $T$ apply karo aur $T \circ T = T$ use karo: $T (k) = T (v) - T (T (v)) = T (v) - T (v) = 0. ✓$ Toh $v = k + i$ jisme $k \in ker T$ , $i \in im T$ . Har vector split ho jaata hai.

Step 2 — trivial intersection ( $A \cap B = {0}$ ). Maano $w \in ker T \cap im T$ . Image mein hone ki wajah se, $w = T (u)$ kisi $u$ ke liye. Kernel mein hone ki wajah se, $T (w) = 0$ . Tab: $0 = T (w) = T (T (u)) = T (u) = w .$ Beech wali equality kyun: $T \circ T = T$ $T (T (u))$ ko $T (u)$ par collapse kar deta hai. Isliye $w = 0$ , toh shared vector sirf $0$ hai.

Dono conditions hold karti hain, isliye $V = ker T \oplus im T$ . $■$ (Isliye ek projection cleanly "jo rakhta hai" use "jo khatam karta hai" se alag kar deta hai — Exercise 3.2 ki echo. Linked idea: Eigenvalues and eigenvectors, kyunki idempotents ke eigenvalues sirf $0$ aur $1$ hote hain.)

Exercise 5.2

Maano $T : R^{n} \to R^{n}$ linear aur injective hai. Prove karo ki woh automatically surjective (hence bijective) hoga. Phir ek counterexample do jo dikhaye ki yeh fail hota hai jab map $R^{n} \to R^{m}$ ke liye ho aur $m \neq = n$ ho.

Recall Solution

Tool kyun: rank–nullity. Injective hone ka matlab $ker T = {0}$ (parent §2), toh nullity $= 0$ . Tab: $n = dim R^{n} = nullity 0 + rank \Rightarrow rank = n .$ Image $R^{n}$ ka ek $n$ -dimensional subspace hai, aur ek $n$ -dimensional space ka sirf ek $n$ -dimensional subspace hota hai — poora space. Toh $im T = R^{n}$ → surjective, hence bijective (dekho Injective surjective bijective).

Equal dimensions kyun zaroori hain: argument ne $dim V = dim W = n$ use kiya. Woh todo aur yeh collapse ho jaata hai. Counterexample $T : R^{2} \to R^{3}$ : $T (x, y) = (x, y, 0), A = 100010 .$ $ker T = {0}$ (injective), phir bhi $im T = {z = 0} \neq = R^{3}$ (rank 2 < 3), toh surjective nahi hai. Injectivity surjectivity force nahi karti jab spaces ka size alag ho.

Recap

Recall Har level ke ek-line takeaways

L1 — $T (0) = 0$ necessary hai, sufficient nahi; hamesha scaling test karo. L2 — image = column space; repeated columns rank nahi badhate. L3 — inject/surject decide karo kernel ko ${0}$ se aur rank ko $dim W$ se compare karke. L4 — kernel aur image rank–nullity se chained hain; dono ko dhyan mein rakh kar construct karo. L5 — "injective ⇒ surjective" ko equal dimensions chahiye; projections $V = ker \oplus im$ split karte hain.