Matrix representation of linear transformations: Exercises

4.5.28 · D4 · Maths › Linear Algebra (Full) › Matrix representation of linear transformations

Notation reminder, taaki yahan kuch bhi surprise na ho:

$[v]_{B}$ ka matlab hai "woh numbers ki list jo tum basis vectors of $B$ se multiply karte ho $v$ banane ke liye." Agar $B = {b_{1}, b_{2}}$ aur $v = 3 b_{1} - 2 b_{2}$ ho, toh $[v]_{B} = (- 2 3)$ .
$[T]_{C}^{B}$ ka matlab hai "woh table jo input ke $B$ -coordinates ko output ke $C$ -coordinates mein convert karta hai." Ise right-to-left padho: neeche ka label $B$ woh hai jo andar jaata hai, upar ka label $C$ woh hai jo bahar aata hai.
$e_{1}, e_{2}, \dots$ standard basis arrows hain: $e_{1} = (1, 0, \dots)$ , $e_{2} = (0, 1, 0, \dots)$ , wagera.

Level 1 — Recognition

Recall Solution 1.1

KYA karte hain: har standard basis arrow ko $T$ se hit karo, coordinates padho.

$T (e_{1}) = 2 (1, 0) = (2, 0)$ → pehla column $(0 2)$ .
$T (e_{2}) = 2 (0, 1) = (0, 2)$ → doosra column $(2 0)$ .

KYUN: har column pakadta hai ki ek basis vector kahan jaata hai (mnemonic "Columns Catch Outputs in $C$ "). Yahan $C$ standard basis hai, isliye output already apne coordinates mein hai. $[T] = (2002) .$

Recall Solution 1.2

Size: $3 \times 3$ (teen-dimensional in, teen-dimensional out).

$T (e_{1}) = (1, 0, 0)$ → column $(1, 0, 0)^{⊤}$ .
$T (e_{2}) = (0, 1, 0)$ → column $(0, 1, 0)^{⊤}$ .
$T (e_{3}) = (0, 0, 0)$ → column $(0, 0, 0)^{⊤}$ . Zero kyun? $z$ -axis flat ho jaata hai. $[T] = 100010000 .$ Teesra column poora zero hona ek degenerate direction ki pehchaan hai — dekho Kernel and image: woh column zero hona matlab hai ki $e_{3}$ kernel mein rehta hai.

Recall Solution 1.3

Columns ko wapas padho — woh batate hain ki basis arrows kahan land karte hain.

Column 1 = $(1 0)$ : toh $T (e_{1}) = (0, 1)$ . Right dikhne wala arrow ab upar dikhta hai.
Column 2 = $(0 1)$ : toh $T (e_{2}) = (1, 0)$ . Upar dikhne wala arrow ab right dikhta hai.

Right ↔ up swap ka matlab hai har point $(x, y)$ jaata hai $(y, x)$ par: yeh ==reflection across the line $y = x$ == hai. ✓

Level 2 — Application

Recall Solution 2.1

$180°$ turn har arrow ko uske exact opposite par bhejta hai: $(x, y) \mapsto (- x, - y)$ .

$T (e_{1}) = (- 1, 0)$ → column $(- 1, 0)^{⊤}$ .
$T (e_{2}) = (0, - 1)$ → column $(0, - 1)^{⊤}$ . $[T] = (- 1 0 0 - 1) .$ Verify: $[T] (- 3 4) = (3 - 4)$ , aur actually $(4, - 3)$ ko origin se flip karein toh $(- 4, 3)$ milta hai. ✓

Recall Solution 2.2

Ab output basis poori ${1, x, x^{2}}$ hai, isliye har column mein teen entries hain.

$T (1) = 0$ → $(0, 0, 0)^{⊤}$ .
$T (x) = 1 = 1 \cdot 1$ → $(1, 0, 0)^{⊤}$ .
$T (x^{2}) = 2 x = 2 \cdot x$ → $(0, 2, 0)^{⊤}$ . $[T] = 000100020 .$ Apply: $[p]_{B} = (7, 2, 5)^{⊤}$ , toh $[T] [p]_{B} = (2, 10, 0)^{⊤}$ , yani $2 + 10 x$ . Aur $p^{'} = 2 + 10 x$ . ✓ Note karo ki bottom row poori zeros hai: derivative kabhi $x^{2}$ term produce nahi kar sakta, isliye woh us basis direction ko hit nahi kar sakta — yeh Rank and nullity ka preview hai.

Recall Solution 2.3

Size forecast: $3 \times 2$ (teen out, do in).

$T (e_{1}) = T (1, 0) = (1, 1, 0)$ → column $(1, 1, 0)^{⊤}$ .
$T (e_{2}) = T (0, 1) = (- 1, 1, 2)$ → column $(- 1, 1, 2)^{⊤}$ . $[T] = 110 - 1 12 .$ Test: $[T] (1 3) = (3 - 1, 3 + 1, 0 + 2)^{⊤} = (2, 4, 2)^{⊤}$ , aur $T (3, 1) = (3 - 1, 3 + 1, 2) = (2, 4, 2)$ . ✓

Level 3 — Analysis

Figure — Matrix representation of linear transformations

Recall Solution 3.1

KYA: $T$ ko har naye basis arrow par apply karo, phir answer ko wapas $B^{'}$ mein express karo. Pehle, standard coordinates mein map hai $T (x, y) = (x + 2 y, y)$ .

$T (1, 0) = (1, 0)$ . $(1, 0) = a (1, 0) + b (1, 1)$ likho: doosre slot se $b = 0$ , phir $a = 1$ . → column $(1, 0)^{⊤}$ .
$T (1, 1) = (1 + 2, 1) = (3, 1)$ . $(3, 1) = a (1, 0) + b (1, 1)$ likho: doosra slot deta hai $b = 1$ , phir $a + 1 = 3 \Rightarrow a = 2$ . → column $(2, 1)^{⊤}$ . $[T]_{B^{'}} = (1021) .$ Surprise: yeh same matrix hai! KYUN: $b_{1}^{'} = (1, 0) = e_{1}$ shear ki fixed direction ke saath hai, aur shear $b_{2}^{'}$ ko exactly ek copy of $b_{1}^{'}$ se move karta hai — naya basis jo geometry measure karta hai woh purane se match ho jaata hai. Yeh Change of basis formula $P^{- 1} A P$ hai jo same matrix par land karta hai kyunki $P$ is particular $A$ ke saath commute karta hai.

Recall Solution 3.2

$T (e_{2}) = 3 e_{2}$ . $B^{'} = {e_{2}, e_{1}}$ mein yeh $3 \cdot (pehla B^{'} vector)$ hai → column $(3, 0)^{⊤}$ .
$T (e_{1}) = 2 e_{1}$ . $B^{'}$ mein yeh $2 \cdot (doosra B^{'} vector)$ hai → column $(0, 2)^{⊤}$ . $[T]_{B^{'}} = (3002) .$ Basis reorder karne se bas diagonal entries swap ho jaati hain. Exactly yahi Eigenvalues and diagonalization use karta hai: jab basis eigenvectors se bani ho, toh matrix diagonal hoti hai aur entries stretch factors (eigenvalues) hote hain.

Recall Solution 3.3

Matrices. $[S] = (01 - 1 0)$ , $[T] = (10 0 - 1)$ .

Tarika 1 — multiply (composition rule $[T \circ S] = [T] [S]$ ):

=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}.$$ **Tarika 2 — basis vectors track karo** ($S$ phir $T$ apply karo): - $e_1=(1,0)\xrightarrow{S}(0,1)\xrightarrow{T}(0,-1)$ → column $(0,-1)^\top$. - $e_2=(0,1)\xrightarrow{S}(-1,0)\xrightarrow{T}(-1,0)$ → column $(-1,0)^\top$. $$[T\circ S]=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}. \checkmark$$ Dono agree karte hain — yeh [[Composition of linear maps]] hai: matrix multiplication *defined* hai taaki yeh do routes match hoon.

Level 4 — Synthesis

Recall Solution 4.1

Derivative apply karo, phir har answer ko $C$ mein express karo solve karke.

$T (1) = 0 = 0 \cdot 1 + 0 \cdot (1 + x)$ → column $(0, 0)^{⊤}$ .
$T (1 + x) = 1$ . $1 = a \cdot 1 + b \cdot (1 + x)$ likho: $x$ -term match karo $b = 0$ , phir constant $a = 1$ . → column $(1, 0)^{⊤}$ .
$T (1 + x + x^{2}) = 1 + 2 x$ . $1 + 2 x = a \cdot 1 + b (1 + x)$ likho: $x$ -term match karo $b = 2$ , phir constant $a + b = 1 \Rightarrow a = - 1$ . → column $(- 1, 2)^{⊤}$ . $[T]_{C}^{B} = (0010 - 1 2) .$ Sanity test. $p = 1 + x + x^{2}$ lo; iske $B$ -coords hain $(0, 0, 1)^{⊤}$ (yeh teesra basis vector hai). Toh $[T]_{C}^{B} (0, 0, 1)^{⊤} = (- 1, 2)^{⊤}$ , yani $- 1 \cdot 1 + 2 \cdot (1 + x) = 1 + 2 x$ . Aur directly $p^{'} = (1 + x + x^{2})^{'} = 1 + 2 x$ . ✓

Recall Solution 4.2

Diye gaye vectors ${(1, 2), (0, 1)}$ ek basis banate hain, isliye yeh data $T$ determine karta hai. Lekin hum matrix standard basis mein chahte hain, isliye humein $T (e_{1})$ aur $T (e_{2})$ chahiye.

$e_{2} = (0, 1)$ already diya hua hai: $T (e_{2}) = (1, 1)$ .
$e_{1} = (1, 0)$ : $e_{1} = 1 \cdot (1, 2) - 2 \cdot (0, 1)$ likho. Linearity se $T (e_{1}) = 1 \cdot T (1, 2) - 2 \cdot T (0, 1) = (3, 0) - 2 (1, 1) = (1, - 2)$ . $[T] = (1 - 2 11) .$ Dono conditions check karo: $[T] (2 1) = (- 2 + 2 1 + 2) = (0 3)$ ✓ aur $[T] (1 0) = (1 1)$ ✓.

Level 5 — Mastery

Recall Solution 5.1

Plane ka unit normal hai $n = \frac{1}{3} (1, 1, 1)$ . Plane par projection $n$ ke saath wala component hata deta hai: $T (v) = v - (v \cdot n) n = v - \frac{( v \cdot ( 1 , 1 , 1 ))}{3} (1, 1, 1) .$ Normal part kyun subtract karte hain? plane se $n$ ke saath jo kuch baahar niklata hai wahi exactly woh shadow hai jo remove ho raha hai; baaki plane mein flat rehta hai.

$T (e_{1}) = (1, 0, 0) - \frac{1}{3} (1, 1, 1) = (\frac{2}{3}, - \frac{1}{3}, - \frac{1}{3})$ .
$T (e_{2}) = (0, 1, 0) - \frac{1}{3} (1, 1, 1) = (- \frac{1}{3}, \frac{2}{3}, - \frac{1}{3})$ .
$T (e_{3}) = (0, 0, 1) - \frac{1}{3} (1, 1, 1) = (- \frac{1}{3}, - \frac{1}{3}, \frac{2}{3})$ . $[T] = \frac{1}{3} 2 - 1 - 1 - 1 2 - 1 - 1 - 1 2 .$ (b) Image poora plane hai (2-dimensional), isliye rank $= 2$ . Normal $(1, 1, 1)$ ke saath sab kuch $0$ par collapse ho jaata hai, isliye kernel line hai jo $(1, 1, 1)$ se span hoti hai, nullity $= 1$ . Check: $[T] (1, 1, 1)^{⊤} = \frac{1}{3} (0, 0, 0)^{⊤} = 0$ ✓, aur rank $+$ nullity $= 2 + 1 = 3 = dim R^{3}$ , yeh Rank and nullity theorem hai. Dekho Kernel and image.

Recall Solution 5.2

(a) $T$ ko har basis vector par apply karo.

$T (1) = 1 + 0 = 1$ → column $(1, 0, 0)^{⊤}$ .
$T (x) = x + 1 = 1 + x$ → column $(1, 1, 0)^{⊤}$ .
$T (x^{2}) = x^{2} + 2 x$ → column $(0, 2, 1)^{⊤}$ . $[T] = 100110021 .$ (b) Yeh upper-triangular hai aur sari diagonal entries $1$ hain, isliye $det = 1 \neq = 0$ : invertible. Invert karo (triangular system back-substitute karo): $[T]^{- 1} = 100 - 1 10 2 - 2 1 .$ Check: $[T] [T]^{- 1} = I$ . Ek map ki tarah, $T^{- 1} (p) = p - p^{'} + p^{''}$ (ek finite version $\sum (- 1)^{k} p^{(k)}$ ka, jo terminate ho jaata hai kyunki $P_{2}$ par teesre aur usse upar ke derivatives zero hain). Quick test: $T^{- 1} (x^{2})$ ke coords $[T]^{- 1} (0, 0, 1)^{⊤} = (2, - 2, 1)^{⊤}$ hain, yani $2 - 2 x + x^{2}$ ; $T$ apply karo: $(2 - 2 x + x^{2}) + (- 2 + 2 x) = x^{2}$ ✓.

Recall Self-check: har problem ke peeche ek-line recipe

$[T]_{C}^{B}$ banane ke liye ::: $B$ ke har vector par $T$ apply karo, har output ko basis $C$ mein likho, aur un coordinate lists ko columns banao. $T : V \to W$ ke liye matrix size ::: $dim W$ rows aur $dim V$ columns. $[T]$ kab invertible nahi hota ::: exactly jab $T$ ka nonzero kernel ho (koi direction $0$ par map ho), yani equivalently $det = 0$ .

Connections

Coordinate vectors and bases — har "ise basis $C$ mein likho" step yahan hai.
Change of basis — Exercises 3.1 aur 3.2 disguise mein $P^{- 1} A P$ hain.
Composition of linear maps — Exercise 3.3 ka do-route check.
Rank and nullity — zero columns/rows se padho (2.2, 5.1).
Kernel and image — 5.1 mein flatten hua direction.
Eigenvalues and diagonalization — 3.2 mein eigenbasis matrix.