Matrix representation of linear transformations

4.5.28 · Maths › Linear Algebra (Full)

WHAT is being represented?

WHY does a basis pin down $T$ ? Koi bhi $v \in V$ lo basis $B = {b_{1}, \dots, b_{n}}$ ke saath. Tab $v = x_{1} b_{1} + \dots + x_{n} b_{n}$ unique coordinates $x_{i}$ ke saath. Linearity se: $T (v) = x_{1} T (b_{1}) + x_{2} T (b_{2}) + \dots + x_{n} T (b_{n}) .$ Toh agar aap $n$ vectors $T (b_{1}), \dots, T (b_{n})$ jaante ho, toh aap $T$ ko saare $V$ par jaante ho. Woh finite data exactly wahi hai jo matrix store karti hai.

HOW to build the matrix (derivation from scratch)

Maano $B = {b_{1}, \dots, b_{n}}$ $V$ ki ek basis hai aur $C = {c_{1}, \dots, c_{m}}$ $W$ ki ek basis hai.

Step 1. Har basis vector $b_{j}$ par $T$ apply karo. Output $T (b_{j})$ $W$ mein rehta hai, toh ise $C$ -basis mein likho: $T (b_{j}) = a_{1 j} c_{1} + a_{2 j} c_{2} + \dots + a_{mj} c_{m} .$ Yeh step kyun? Coordinates sirf ek basis ke relative hote hain; hum output ko $C$ mein express karna hi padega.

Step 2. In coordinates ko matrix ki $j$ -th column ke roop mein collect karo: $[T]_{C}^{B} = [[T (b_{1})]_{C} [T (b_{2})]_{C} \dots [T (b_{n})]_{C}] .$ Columns kyun, rows kyun nahi? Taaki matrix-times-coordinate-vector $T$ ko reproduce kare (agla step).

Boxed formula ki derivation. Maano $[v]_{B} = (x_{1}, \dots, x_{n})^{⊤}$ . Tab

= \sum_i \Big(\underbrace{\textstyle\sum_j a_{ij} x_j}_{i\text{-th coord of output}}\Big) c_i.$$ Bracket exactly row $i$ of $A$ dotted with $x$ hai — yaani $(Ax)_i$. Isliye $[T(v)]_C = A[v]_B$. $\blacksquare$ --- ## Worked Examples > [!example] 1. $\mathbb{R}^2$ mein $90°$ ka Rotation (standard basis) > $T$ vectors ko $90°$ counterclockwise rotate karta hai. Basis $e_1=(1,0)$, $e_2=(0,1)$. > - $T(e_1) = (0,1)$ → pehla column $\binom{0}{1}$. *Kyun?* $(1,0)$ ko 90° rotate karo toh upar point karta hai. > - $T(e_2) = (-1,0)$ → doosra column $\binom{-1}{0}$. > $$[T] = \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}.$$ > Check: $T(3,2) = [T]\binom{3}{2} = \binom{-2}{3}$, jo $(3,2)$ ko 90° rotate kiya hua hai. ✓ > [!example] 2. Polynomials $P_2 \to P_1$ par Derivative > $T = \dfrac{d}{dx}$, domain basis $B=\{1,x,x^2\}$, codomain basis $C=\{1,x\}$. > - $T(1)=0 = 0\cdot1 + 0\cdot x$ → column $\binom{0}{0}$. *Kyun?* constant ka derivative 0 hota hai. > - $T(x)=1 = 1\cdot 1 + 0\cdot x$ → column $\binom{1}{0}$. > - $T(x^2)=2x = 0\cdot 1 + 2\cdot x$ → column $\binom{0}{2}$. > $$[T]_C^B = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2\end{pmatrix}.$$ > $p=4+5x+6x^2$ par test karo: $[p]_B=(4,5,6)^\top$, $[T]_C^B[p]_B = (5,12)^\top$, yaani > $5 + 12x$. Aur wakai $p'=5+12x$. ✓ *$2\times3$ kyun?* $\dim P_1=2$ rows, $\dim P_2=3$ cols. > [!example] 3. Same map, DIFFERENT basis se matrix badal jaati hai > Rotation map lekin basis $B'=\{(1,1),(1,-1)\}$ in/out dono ke liye. > $T(1,1)=(-1,1)$. Solve karo $(-1,1)=a(1,1)+b(1,-1)$: $a=0,\ b=-1$. → column $\binom{0}{-1}$. > $T(1,-1)=(1,1)$. Solve karo: $a=1,b=0$. → column $\binom{1}{0}$. > $$[T]_{B'} = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix} \ne \begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$ > *Yeh kyun matter karta hai:* transformation same hai, lekin iska matrix **basis choice par depend karta hai**. --- ## Forecast-then-Verify > [!recall] Compute karne se pehle predict karo > Maano $T:\mathbb{R}^3\to\mathbb{R}^2$, $T(x,y,z)=(x+z,\ y-x)$, standard bases. > **Aage padhne se pehle matrix ka size aur entries forecast karo.** > > Size: $2\times 3$. Columns hain $T(e_1),T(e_2),T(e_3)$: > $T(e_1)=(1,-1),\ T(e_2)=(0,1),\ T(e_3)=(1,0)$. > $$[T]=\begin{pmatrix} 1 & 0 & 1 \\ -1 & 1 & 0\end{pmatrix}.$$ > Verify karo: $[T](2,3,4)^\top=(6,1)^\top$, aur $T(2,3,4)=(2+4,\,3-2)=(6,1)$. ✓ --- ## Composition = Matrix Multiplication **WHY does matrix multiplication look so weird?** Yeh *define* kiya gaya hai taaki composition ki matrix, matrices ke product ke barabar ho. Agar $S:U\to V$ aur $T:V\to W$, tab $$[T\circ S]_{D}^{B} = [T]_D^C \,[S]_C^B.$$ *Derivation:* $[T(S(u))]_D = [T]_D^C [S(u)]_C = [T]_D^C ([S]_C^B [u]_B)$, aur associativity inhe ek matrix mein group karta hai. "Row times column" rule consistency ki keemat hai. --- ## Common Mistakes > [!mistake] $T(b_j)$ ko columns ki jagah rows mein rakhna > **Kyun theek lagta hai:** hum left-to-right padhte hain, toh outputs ko row mein list karna natural lagta hai. > **Fix:** formula $[T(v)]_C = A[v]_B$ force karta hai ki $T(b_j)$ $j$-th **column** ho, > kyunki $A e_j$ column $j$ pick karta hai. Rows use karne se transpose $A^\top$ milta hai — ek alag map. > [!mistake] Output ko codomain basis $C$ mein express karna bhool jaana > **Kyun theek lagta hai:** standard basis mein $T(b_j)$ already apne coordinates jaisa *dikhta* hai. > **Fix:** jab $C$ non-standard ho toh aapko $a_{ij}$ ke liye $T(b_j)=\sum a_{ij} c_i$ zaroor solve karna hoga. > Example 3 dikhata hai ki agar aap yeh skip karo toh answer completely badal jaata hai. > [!mistake] Yeh maanna ki matrix unique hai > **Kyun theek lagta hai:** ek map "ek cheez" lagti hai. **Fix:** matrix DONO chosen bases par depend karti hai. Same $T$ + alag bases = similar/equivalent lekin alag matrices. --- > [!mnemonic] Recipe yaad rakho > **"Columns Catch Outputs in $C$"** — har **C**olumn **C**atch karta hai ki ek basis vector kahaan jaata hai, > **C**odomain basis $C$ mein likha hua. > [!recall]- Feynman: ek 12-saal ke bacche ko explain karo > Socho ek machine jo arrows ko stretch aur turn karti hai. Tum har possible arrow nahi dekh sakte. > Lekin har arrow kuch khaas "ingredient arrows" ko mix karke banta hai. Toh tum bas > machine ko un thodi si ingredients par test karte ho aur likh lete ho ki har ek kahaan jaati hai ek chhoti si table mein. Kisi BHI arrow ko dhundhne ke liye, bas apni table mein ingredients dekho aur results ko usi tarah mix karo. Woh table hi matrix hai! --- ## #flashcards/maths Kya cheez ek linear transformation ko completely determine karti hai? ::: Domain ki basis par iska action (jahaan har basis vector map hota hai). Basis vectors $T$ define karne ke liye kyun kaafi hain? ::: Har $v$ ek unique combination $\sum x_i b_i$ hai, aur linearity deta hai $T(v)=\sum x_i T(b_i)$. $[T]_C^B$ ki $j$-th column mein kya jaata hai? ::: $T(b_j)$ ke coordinates codomain basis $C$ mein expressed. Matrix representation ka fundamental relationship batao. ::: $[T(v)]_C = [T]_C^B \,[v]_B$. $T:V\to W$ ke liye $\dim V=n,\dim W=m$, matrix ka size kya hai? ::: $m \times n$ (rows = $\dim W$, columns = $\dim V$). Basis change karne par $T$ ki matrix kyun badal jaati hai? ::: Coordinates basis-dependent hote hain, toh bookkeeping table badal jaati hai chahe $T$ same ho. Composition $T\circ S$ ki matrix? ::: $[T\circ S]_D^B = [T]_D^C\,[S]_C^B$ (matrix multiplication). $\{1,x,x^2\}\to\{1,x\}$ par $d/dx$ ki matrix? ::: $\begin{pmatrix}0&1&0\\0&0&2\end{pmatrix}$. Common error: $T(b_j)$ ko rows mein list karne se kya milta hai? ::: Transpose $A^\top$, jo ek alag map represent karta hai. --- ## Connections - [[Change of basis]] — $[T]$ naye bases ke under kaise transform hota hai ($P^{-1}AP$). - [[Composition of linear maps]] — isliye matrix multiplication define ki gayi hai. - [[Coordinate vectors and bases]] — $[v]_B$ ka notion jis par yeh sab tika hai. - [[Rank and nullity]] — matrix representation se padhe jaate hain. - [[Eigenvalues and diagonalization]] — ek aisi basis choose karna jo $[T]$ ko simplify kare. - [[Kernel and image]] — $[T]_C^B$ ka null space aur column space. ## 🖼️ Concept Map ```mermaid flowchart TD T[Linear transformation T V to W] LIN[Linearity axioms] BV[Action on basis vectors] BASISV[Basis B of V] BASISW[Basis C of W] COORD[Unique coordinates of v] COL[Columns store T of b_j in C] MAT[Matrix representation] FORMULA[Output coords = matrix times input coords] EX[Examples rotation and derivative] T -->|satisfies| LIN LIN -->|implies| BV BASISV -->|gives| COORD COORD -->|expand v then apply T| BV BV -->|express in| BASISW BASISW -->|coords become| COL COL -->|assembled into| MAT MAT -->|yields| FORMULA FORMULA -->|verified by| EX MAT -->|size m by n from| BASISV ```