Poore note mein, E={e1,e2}R2 ka standard basis hai, matlab e1=(1,0) aur e2=(0,1) — sade "ek step right, ek step up" directions. Jab hum B=[b1∣b2] jaisi matrix likhte hain toh matlab hai: basis vectors b1,b2 ko columns ki tarah stack karo, har ek standard coordinates mein likha hua. Yeh akela convention sab kuch drive karta hai.
Is figure ko kaise padhen (hum baar baar isi ki taraf point karte rahenge): left mein orange box mein B-language mein coordinate list hai, right mein plum box mein wahi information C-language mein hai, aur neeche teal box standard-E language hai jo sabke saath share hoti hai. Center mein chhota circle woh ek physical arrow hai jo kabhi nahi hilta. Teen labelled arrows exactly teen translator matrices hain — ×B (orange, B→E), ×C−1 (teal, E→C), aur unka composite PC←B=C−1B (top edge). Neeche ka har exercise in edges mein se ek ya zyada edges par ek walk hai — hum har baar batayenge kaun si edge hai.
Subscript kya kehta hai:PC←BB-coordinates khata hai (woh right mein baithe hain, arrow ke right mein B se match karte hue) aur C-coordinates return karta hai. Figure mein, yeh top orange-to-plum edge hai — hum orange box se start karte hain aur plum box mein land karte hain.
[v]C=PC←B[v]B=[2103][11]=[24].
Toh output [v]C=(2,4)T hai — naye (C) coordinates. Vector khud kabhi nahi hila (woh figure mein fixed center circle hai); sirf uski description badli.
Recall Solution L1.2
Ek single column b1 ke baare mein kyun batata hai: matrix–vector multiplication columns se padhta hai. Matrix ko (1,0)T feed karo toh column 1 milta hai, (0,1)T feed karo toh column 2 milta hai — yeh sirf product ki definition hai. Toh column 1 ka matlab samajhne ke liye, woh input feed karo jo use isolate kare.
Sahi input [b1]B=(1,0)T hai: yeh pehla purana basis vector apne hi basis B mein described hai (b1 ke along ek step, b2 ke along zero). Kyunki PC←B kisi bhi B-coordinate list ko uski C-coordinate list mein convert karta hai, output b1 ko C mein spoken hona chahiye:
PC←B[10]=[21]=[b1]C.
Toh column 1 hai [b1]C — purana basis vector b1 naye basis C mein likha hua. PC←B ka column j hamesha [bj]C hota hai.
Yahan inverse kyun nahi:B ke columns pehle se standard coordinates mein likhe hain, toh "B→ standard decode" sirf B se multiply karna hai. Yeh figure ka orange edge hai (×B, B→E).
[v]E=B[v]B=[31−12][21]=[54].Arrow rebuild karke verify karo:2(3,1)+1(−1,2)=(6−1,2+2)=(5,4). ✓
Recall Solution L2.2
Inverse kyun: hum standard →B jaana chahte hain, jo figure mein orange edge backwards hai, toh B−1 chahiye.
B=[31−12] ke liye, detB=3⋅2−(−1)⋅1=7, toh
B−1=71[2−113].[w]B=B−1w=71[2−113][54]=71[147]=[21].
L2.1 ke saath consistent — jaise hona chahiye, kyunki humne sirf use reverse kiya. ✓
Recall Solution L2.3
Formula C−1B kyun:B-coords → standard jao (orange edge, B se multiply karo), phir standard →C-coords (teal edge, C−1 se multiply karo). Dono edges chain karne se top edge milti hai PC←B=C−1B.
B=[1011],C=[2001],C−1=[21001].PC←B=C−1B=[21001][1011]=[210211].
Yeh figure ke around ek full trip hai: orange box → teal box (×B) → plum box (×C−1).
B=[1011],C=[111−1],detC=−2,C−1=[212121−21].PC←B=C−1B=[212121−21][1011]=[212110].[v]C=PC←B[v]B=[212110][42]=[42].Verify (teal box ke through): actual arrow hai v=4(1,0)+2(1,1)=(6,2). C se rebuild: 4(1,1)+2(1,−1)=(6,2). ✓ Wahi arrow, do descriptions.
Recall Solution L3.2
Inverse kyun:C→B translate karne ke liye figure ki top edge backwards chalate hain — translator card flip karo: PB←C=(PC←B)−1.
detPC←B=21⋅0−1⋅21=−21, toh
PB←C=(PC←B)−1=−211[0−21−121]=[012−1].[u]B=PB←C[u]C=[012−1][12]=[4−1].Verify:C se arrow: u=1(1,1)+2(1,−1)=(3,−1). B se rebuild: 4(1,0)−1(1,1)=(3,−1). ✓
Yeh formula kyun:T ko B-coordinates mein kaam karte hue apply karne ke liye, B→ standard decode karo (×B, orange edge), A apply karo, phir standard →B re-encode karo (×B−1, orange edge backwards). Toh A′=B−1AB — Similar Matrices relation.
B=[1011],B−1=[10−11].AB=[2013][1011]=[2033].A′=B−1(AB)=[10−11][2033]=[2003].
Map is basis mein diagonal dikhta hai — kyunki (1,0) aur (1,1)A ke eigenvectors nikle (eigenvalues 2 aur 3).
Recall Solution L4.2
Change of basis kyun: diagonalizing = apna naya basis eigenbasis choose karna; phir T sirf har axis ke along stretch karta hai.
Eigenvalues: det(A−λI)=(2−λ)(3−λ)=0⇒λ=2,3.
Strategy: sab kuch standard basis ke through route karo — orange edge phir teal edge. PC←B=C−1B.
B=[2111],C=[1012],detC=2,C−1=[10−2121].PC←B=C−1B=[10−2121][2111]=[23212121].[v]C=PC←B[v]B=[23212121][13]=[32].Verify: arrow v=1(2,1)+3(1,1)=(5,4). C se rebuild: 3(1,0)+2(1,2)=(5,4). ✓
Recall Solution L5.2
[v]C=PC←B[00]=[00].Har basis mein zero kyun hona zaroori hai: zero vector woh ek arrow hai jiska length 0 hai; uski koi direction describe karne ke liye nahi hai, toh har basis use coordinate list (0,…,0) assign karti hai. Koi bhi change of basis matrix 0↦0 map karti hai (ek linear map hamesha origin fix karta hai — figure ka center circle exactly origin par baitha hai). Yeh boundary case confirm karta hai ki tumhari matrix sahi behave kar rahi hai.
Recall Solution L5.3
C=[100−1]. Kyunki C−1=C yahan ((−1)−1=−1),
[w]C=C−1w=[100−1][35]=[3−5].Matlab kya hai: doosra C-axis neeche point karta hai, toh ek aisi point tak pahunchne ke liye jo 5 units upar hai, tumhe downward axis ke along −5 steps lene padte hain. Geometrically C−1vertical component ko reflect karta hai — horizontal axis ke across ek mirror flip. Sign handling matter karta hai: basis direction ulta lo aur har doosra coordinate apna sign flip kar leta hai.
Linear Transformations — passive (coordinates change hote hain) vs active (arrows move karte hain) views.
Recall Feynman check — poori ladder ek saanp mein bolo
Ek basis mein coordinates do, C−1B se multiply karo doosre basis mein coordinates paane ke liye; card flip karo (B−1C) jab bhi wapas jaana ho; aur ek linear map ko ek friendlier basis mein dekhne ke liye uski matrix ko P−1AP ke roop mein sandwich karo jahan P mein naye basis vectors hain — aur agar woh friendlier basis eigenbasis nikle, toh sandwich diagonal aaega. Poore time, figure ke center mein arrow kabhi nahi hila; sirf uski language badli.
