Matrix representation of linear transformations: Worked examples

4.5.28 · D3 · Maths › Linear Algebra (Full) › Matrix representation of linear transformations

Sab se pehle: ek coordinate vector $[v]_{B}$ ka matlab sirf yeh hai — "woh mixing amounts ki list jo tumhe $v$ ko basis $B$ se banane ke liye chahiye." Agar $B = {b_{1}, b_{2}}$ aur $v = 3 b_{1} - 2 b_{2}$ , toh $[v]_{B} = (- 2 3)$ . Bas itna hi matlab hai $[\cdot]_{B}$ notation ka — dekho Coordinate vectors and bases. Ek basis ek chhooti se chhooti "ingredient arrows" ki set hoti hai jo har vector ko mixing se bana sakti hai.

Scenario matrix

Har representation problem in mein se kisi ek cell mein aata hai. Neeche ke examples us cell ke tag ke saath aate hain jise woh cover karte hain.

Cell	Kya cheez alag banati hai	Example
A. Square, standard basis	$n = m$ , dono bases standard — sabse aasaan	Ex 1
B. Non-square (zyada inputs)	$dim V > dim W$ , matrix "wide" hai	Ex 2
C. Non-square (zyada outputs)	$dim V < dim W$ , matrix "tall" hai	Ex 3
D. Non-standard codomain basis	coordinates ke liye $C$ mein SOLVE karna padta hai	Ex 4
E. Degenerate / collapsing map	$T$ ek dimension ko crush kar deta hai (zero column)	Ex 5
F. Zero / limiting input	origin aur scaling behaviour	Ex 5 ke andar
G. Composition	do machines chain mein → matrix product	Ex 6
H. Real-world word problem	ek story ko matrix mein translate karo	Ex 7
I. Exam twist (matrix se $T$ recover karo)	matrix diya hua hai, rule/action dhundho	Ex 8

Sign aur quadrant coverage geometric examples mein built-in hai (Ex 1 rotation ke chaaon quadrants sweep karta hai; Ex 5 har quadrant mein projected component ka sign dikhata hai).

A · Square, standard basis (chaaon quadrants)

Figure — Matrix representation of linear transformations

$T (e_{1})$ dhundho. Yeh step kyun? Columns pakadtay hain ki basis vectors kahaan jaate hain. Point $(1, 0)$ horizontal axis par hai; $y = x$ ke paas reflect karne se yeh $(0, 1)$ par jaata hai. Toh pehla column hai $(1 0)$ .
$T (e_{2})$ dhundho. Kyun? Doosre ingredient ke liye same recipe. $(0, 1)$ reflect hokar $(1, 0)$ ban jaata hai, doosra column $(0 1)$ deta hai.
Assemble karo. Columns kyun, rows kyun nahi? Kyunki $[T] e_{j}$ ko column $j$ pick out karna chahiye (parent ka boxed formula): $[T] = (0110) .$
Chaaon quadrants sweep karo (figure mein coloured dots dekho). Reflection coordinates swap karta hai, toh $(x, y) \mapsto (y, x)$ :
- Q1: $(3, 1) \mapsto (1, 3)$
- Q2: $(- 2, 4) \mapsto (4, - 2)$
- Q3: $(- 3, - 2) \mapsto (- 2, - 3)$
- Q4: $(2, - 5) \mapsto (- 5, 2)$

Verify: $[T] (4 - 2) = (- 2 4)$ ✓. Do baar reflect karne par original wapas aana chahiye, toh $[T]^{2}$ identity honi chahiye: $(0110)^{2} = (1001)$ ✓.

B · Wide matrix (outputs se zyada inputs)

Shape. Kyun? rows $= dim W = 2$ , columns $= dim V = 3$ . Toh $[T]$ hai $2 \times 3$ — "wide."
$e_{1} = (1, 0, 0)$ ke liye column. Kyun? Pehla ingredient feed karo. $T (1, 0, 0) = (2 \cdot 1 - 0, 1 + 0 + 0) = (2, 1)$ → $(1 2)$ .
$e_{2} = (0, 1, 0)$ ke liye column. $T (0, 1, 0) = (0 - 0, 0 + 1 + 0) = (0, 1)$ → $(1 0)$ .
$e_{3} = (0, 0, 1)$ ke liye column. $T (0, 0, 1) = (0 - 1, 0 + 0 + 1) = (- 1, 1)$ → $(1 - 1)$ .
Assemble: $[T] = (2101 - 1 1) .$

Verify: $(3, - 1, 2)$ lo. Directly $T (3, - 1, 2) = (2 \cdot 3 - 2, 3 - 1 + 2) = (4, 4)$ . Matrix se: $(2101 - 1 1) 3 - 1 2 = (3 - 1 + 2 6 - 2) = (4 4)$ ✓. Ek wide matrix ka matlab hai inputs mil kar collapse ho sakte hain — Rank and nullity aur Kernel and image se related.

C · Tall matrix (inputs se zyada outputs)

Shape. $dim W = 3$ rows, $dim V = 2$ columns → $3 \times 2$ , "tall."
$e_{1} = (1, 0)$ ke liye column. $T (1, 0) = (1, 1 + 0, 0) = (1, 1, 0)$ → pehla column.
$e_{2} = (0, 1)$ ke liye column. $T (0, 1) = (0, 0 + 1, 1) = (0, 1, 1)$ → doosra column.
Assemble: $[T] = 110011 .$

Verify: $(5, 2) \mapsto (5, 7, 2)$ directly; matrix deta hai $110011 (2 5) = (5, 7, 2)^{⊤}$ ✓. Ek tall matrix kabhi bhi $R^{3}$ par onto nahi ho sakta — iska image zyada se zyada ek 2-D plane hai (Kernel and image).

D · Non-standard codomain basis (tumhe SOLVE karna hoga)

Raw outputs compute karo. $T (e_{1}) = (1 \cdot 1 + 0, 0) = (1, 0)$ ; $T (e_{2}) = (0 + 1, 2) = (1, 2)$ .
$T (e_{1})$ ko $C$ mein express karo. Yeh step kyun? Parent ka doosra "mistake" callout: tumhe SOLVE karna CHAHIYE $(1, 0) = a c_{1} + b c_{2} = a (1, 1) + b (1, - 1)$ . Isse $a + b = 1$ aur $a - b = 0$ milta hai, toh $a = b = \frac{1}{2}$ . Pehla column $= (1/2 1/2)$ .
$T (e_{2})$ ko $C$ mein express karo. Solve karo $(1, 2) = a (1, 1) + b (1, - 1)$ : $a + b = 1, a - b = 2 \Rightarrow a = \frac{3}{2}, b = - \frac{1}{2}$ . Doosra column $= (- 1/2 3/2)$ .
Assemble: $[T]_{C}^{E} = (\frac{1}{2} \frac{1}{2} \frac{3}{2} - \frac{1}{2}) .$

Verify: $v = e_{1} = (1, 0)$ test karo, toh $[v]_{E} = (0 1)$ . Phir $[T]_{C}^{E} (0 1) = (1/2 1/2)$ , claim karta hai $T (v) = \frac{1}{2} c_{1} + \frac{1}{2} c_{2} = \frac{1}{2} (1, 1) + \frac{1}{2} (1, - 1) = (1, 0)$ . Aur directly $T (1, 0) = (1, 0)$ ✓. Step 2 skip karna (raw $(1, 0)$ seedha column mein daalna) ek alag, galat map dega — dekho Change of basis.

E · Degenerate map + F · Zero/limiting input

$e_{1}$ ke liye column. $T (1, 0) = (1, 0)$ → $(0 1)$ . Horizontal ingredient bina badlaav ke bachta hai.
$e_{2}$ ke liye column. $T (0, 1) = (0, 0)$ → $(0 0)$ . Zero column kyun? Vertical ingredient poori tarah flatten ho jaata hai; uska kuch bhi nahi bachta. Woh zero column ek degenerate (non-invertible) map ki pehchaan hai.
Assemble: $[T] = (1000) .$
Har quadrant mein sign (figure mein coloured projection arrows dekho): output $x$ ka sign rakhta hai aur $y$ ko khatam kar deta hai.
- Q1 $(4, 3) \mapsto (4, 0)$ — positive $x$
- Q2 $(- 4, 3) \mapsto (- 4, 0)$ — negative $x$
- Q3 $(- 4, - 3) \mapsto (- 4, 0)$ — negative $x$
- Q4 $(4, - 3) \mapsto (4, 0)$ — positive $x$
Cell F, zero/limiting input. Origin examine kyun karte hain? Har linear map $0 \mapsto 0$ bhejta hai ( $T (c v) = c T (v)$ mein $c = 0$ set karo): $T (0, 0) = (0, 0)$ ✓. Aur input ko scale karne se output scale hota hai: $T (t \cdot (4, 3)) = (4 t, 0) = t \cdot T (4, 3)$ , toh jaise $t \to 0$ output smoothly origin par shrink hota hai — koi jump nahi.

Verify: kernel (jo zero par map hota hai) poori $y$ -axis hai: $T (0, c) = (0, 0)$ for all $c$ ✓. $T$ do baar lagaane se kuch nahi badalta (pehle se flat hai): $[T]^{2} = (1000)$ ✓ — ek projection. Ek zero column ⟹ non-trivial kernel ⟹ invertible nahi.

G · Composition = matrix product

Pieces yaad karo. Parent note se $[S] = (01 - 1 0)$ (rotation), aur Ex 1 se $[T] = (0110)$ (reflection).
Product ka order. $[T] [S]$ kyun, $[S] [T]$ kyun nahi? Parent ka composition rule: $[T \circ S] = [T] [S]$ — sabse baad apply hone waali machine left par hoti hai, kyunki $[T (S (v))] = [T] ([S] [v])$ .
Multiply karo: $=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$$
Interpret karo. Yeh kaisa dikhta hai? $(10 0 - 1)$ $x$ rakhta hai, $y$ flip karta hai: yeh $x$ -axis ke paas reflection hai. Rotate-then-reflect $=$ ek single reflection.

Verify: $(2, 0)$ track karo. $S (2, 0) = (0, 2)$ (rotate up); phir $T (0, 2) = (2, 0)$ (swap back). Net $(2, 0) \mapsto (2, 0)$ . Matrix: $(10 0 - 1) (0 2) = (0 2)$ ✓. $(0, 3)$ check karo: $S \to (- 3, 0)$ , $T \to (0, - 3)$ ; matrix $(10 0 - 1) (3 0) = (- 3 0)$ ✓. Dekho Composition of linear maps.

H · Real-world word problem

$b_{1} =$ ek Applejoy ke liye column. Kyun? Pehle "ingredient" ki ek unit feed karo (ek bottle Applejoy) aur output record karo. Ise chahiye $(2 apples, 1 orange)$ → $(1 2)$ .
$b_{2} =$ ek Sunburst ke liye column. Chahiye $(1 apple, 3 oranges)$ → $(3 1)$ .
Assemble ( $M$ bottles → fruit leta hai): $M = (2113) .$
Order $(4 5)$ par apply karo: $M (4 5) = (1 \cdot 5 + 3 \cdot 4 2 \cdot 5 + 1 \cdot 4) = (17 14) .$

Verify (units ke saath): apples $= 2 (5) + 1 (4) = 14$ apples; oranges $= 1 (5) + 3 (4) = 17$ oranges ✓. Units consistent hain: (fruit per bottle) $\times$ (bottles) $=$ fruit. Matrix bilkul mixing rule hai.

I · Exam twist — map ko matrix se recover karo

Matrix se formula padho. Kyun? Kyunki $T (x, y) = [T] (y x)$ : $(30 - 1 2) (y x) = (2 y 3 x - y) \Rightarrow T (x, y) = (3 x - y, 2 y) .$
Fixed-point condition set up karo $T (v) = v$ . Kyun? Ek "fixed vector" satisfy karta hai $T (x, y) = (x, y)$ : $3 x - y = x and 2 y = y .$
Solve karo. $2 y = y$ se milta hai $y = 0$ . $3 x - y = x$ mein substitute karo: $3 x = x \Rightarrow x = 0$ . Ruko — kuch aur bhi hai? Diagonal entries check karo: $2 \neq = 1$ aur $3 \neq = 1$ , toh koi eigenvalue $1$ ke barabar nahi siwaaye origin ke. Ek hi fixed vector hai $(0 0)$ .

Verify: $T (0, 0) = (3 \cdot 0 - 0, 2 \cdot 0) = (0, 0) = (0, 0)$ ✓. Formula ko $(1, 1)$ par spot-check karo: $T (1, 1) = (3 - 1, 2) = (2, 2)$ , aur $[T] (1 1) = (2 2)$ ✓. "Fixed vectors" eigenvalue $1$ ke saath eigenvectors hain — yahaan siwaaye origin ke koi nahi, dekho Eigenvalues and diagonalization.

Recall Quick self-test

$R^{4} \to R^{2}$ map ki matrix ka shape kya hoga? ::: $2 \times 4$ (rows $= dim W = 2$ , columns $= dim V = 4$ ). Kaunsa cell har column ke liye ek linear system solve karne par majboor karta hai? ::: Cell D — ek non-standard codomain basis. Matrix mein kya cheez ek collapsing (degenerate) map signal karti hai? ::: Ek zero column (iska basis vector $0$ par map hota hai), jo ek non-trivial kernel deta hai. $T$ ke baad $S$ compose karo: product ka order kya hai? ::: $[T] [S]$ — last-applied map left par hota hai.

Connections

Coordinate vectors and bases — har column ek coordinate vector $[T (b_{j})]_{C}$ hai.
Change of basis — kyun Ex 4 ki non-standard basis matrix ko reshuffle karti hai.
Composition of linear maps — Ex 6 ke peeche ka engine.
Rank and nullity / Kernel and image — wide/tall/degenerate matrices se padha jaata hai (Ex 2, 3, 5).
Eigenvalues and diagonalization — Ex 8 mein fixed-vector question.

Worked examples — Matrix representation of linear transformations