4.5.19 · Maths › Linear Algebra (Full)
Ek vector v space mein ek cheez hai — ek arrow — jo kisi bhi basis choose karne se pehle exist karta hai. Coordinate vector woh numbers ki list hai jo tum usi arrow ko describe karne ke liye likhte ho jab ek ruler-system (ek basis) choose kar liya ho. Basis badlo → arrow apni jagah rehta hai, lekin numbers badal jaate hain. Change of basis woh dictionary hai jo ek hi arrow ke liye ek list of numbers ko doosri list mein translate karti hai.
Definition Coordinate vector
Maano B = { b 1 , … , b n } ek vector space V ki ordered basis hai. Har v ∈ V ka ek unique expansion hota hai:
v = c 1 b 1 + c 2 b 2 + ⋯ + c n b n .
v ka coordinate vector relative to B yeh hai:
[ v ] B = c 1 ⋮ c n .
Yeh unique KYU hai? Kyunki B linearly independent hai. Agar do expansions hote, unhe subtract karne par ek non-trivial combination 0 ke barabar milti — jo independence ko contradict karta. Uniqueness hi woh asli wajah hai jis se coordinates well-defined hote hain.
Tumhe [ v ] B pata hai. Tum kisi doosri basis C ke liye [ v ] C chahte ho. Humein ek matrix P chahiye jis se:
[ v ] C = C ← B P [ v ] B .
[ v ] B ka matlab kya hai, wahan se shuru karo:
v = c 1 b 1 + ⋯ + c n b n .
Ab dono sides ke coordinates C ke relative lo. Coordinate map [ ⋅ ] C linear hai (neeche prove kiya gaya hai), isliye:
[ v ] C = c 1 [ b 1 ] C + ⋯ + c n [ b n ] C .
Right side bilkul ek matrix times c 1 ⋮ c n hai jahan columns [ b j ] C hain. Isliye:
Coordinate maps linear kyun hote hain: u = ∑ a i b i aur w = ∑ d i b i likhne par, u + α w = ∑ ( a i + α d i ) b i , isliye [ u + α w ] B = [ u ] B + α [ w ] B . Done.
Roles swap karne par B ← C P milta hai, aur:
B ← C P = ( C ← B P ) − 1 .
Kyun? C mein translate karna phir B mein wapas aana kuch nahi karna chahiye: unka product I hai.
R n mein standard basis E ke saath
Define karo B = [ b 1 ⋯ b n ] (basis vectors as columns). Tab:
E ← B P = B , B ← E P = B − 1 .
Toh B "B -numbers" ko "standard numbers" mein turn karta hai, aur B − 1 ulta karta hai. Do non-standard bases ke liye:
C ← B P = C − 1 B .
Worked example Example 1 — basic
R 2
B = { b 1 , b 2 } jahan b 1 = [ 1 1 ] , b 2 = [ 1 − 1 ] . v = [ 4 2 ] ke liye [ v ] B nikalo.
Humein c 1 b 1 + c 2 b 2 = v chahiye, yaani B [ v ] B = v , toh [ v ] B = B − 1 v .
Yeh step kyun? Standard basis mein v ke coordinates bas uske entries hain; B − 1 standard→B convert karta hai.
B = [ 1 1 1 − 1 ] , B − 1 = − 2 1 [ − 1 − 1 − 1 1 ] = 2 1 [ 1 1 1 − 1 ] .
[ v ] B = 2 1 [ 1 1 1 − 1 ] [ 4 2 ] = [ 3 1 ] .
Check: 3 b 1 + 1 b 2 = [ 3 3 ] + [ 1 − 1 ] = [ 4 2 ] . ✓
Worked example Example 2 — do non-standard bases ke beech
B upar jaisa, C = { c 1 , c 2 } jahan c 1 = [ 1 0 ] , c 2 = [ 1 2 ] . C ← B P nikalo.
P = C − 1 B use karo. Kyun? P ke columns [ b j ] C = C − 1 b j hain, stack karne par C − 1 B milta hai.
C = [ 1 0 1 2 ] , C − 1 = [ 1 0 − 2 1 2 1 ] .
P = C − 1 B = [ 1 0 − 2 1 2 1 ] [ 1 1 1 − 1 ] = [ 2 1 2 1 2 3 − 2 1 ] .
Column 1 verify karo = [ b 1 ] C : kya 2 1 c 1 + 2 1 c 2 = 2 1 [ 1 0 ] + 2 1 [ 1 2 ] = [ 1 1 ] = b 1 ? ✓
Worked example Example 3 — Forecast-then-Verify (polynomials)
V = P 2 , B = { 1 , t , t 2 } , C = { 1 , 1 + t , 1 + t + t 2 } . Lo p ( t ) = 2 + 3 t + t 2 .
Forecast: C terms ko "stack" karta hai, toh main expect karta hoon chote coordinates. Verify karte hain.
Solve karo a ( 1 ) + b ( 1 + t ) + c ( 1 + t + t 2 ) = 2 + 3 t + t 2 .
t 2 : c = 1 . Kyun? Sirf c 3 mein t 2 hai.
t : b + c = 3 ⇒ b = 2 .
const: a + b + c = 2 ⇒ a = − 1 .
Toh [ p ] C = − 1 2 1 . Forecast (chote numbers) sahi nikla; note karo ek negative hai — coordinates positive hone zaroori nahi.
P ke columns naye basis vectors c j hain."
Kyun sahi lagta hai: tum C mein ja rahe ho, toh lagta hai C ke vectors jaayenge. Fix: columns purane vectors b j hain, naye coordinates mein expressed: [ b j ] C . Matrix B -numbers consume karta hai, isliye B se banana parega.
C − 1 B ki jagah B − 1 C use karna.
Kyun sahi lagta hai: dono letters aate hain; order arbitrary lagta hai. Fix: units check karo. C ← B P = C − 1 B . Right-to-left padho: B B →standard bhejta hai, C − 1 standard→C bhejta hai. Beech ka "standard" cancel ho jaata hai.
Common mistake Yeh sochna ki vector khud badal jaata hai.
Kyun sahi lagta hai: numbers clearly badal jaate hain. Fix: sirf description badlti hai. v , geometric arrow, invariant hai — wahi invariance exactly [ v ] C = P [ v ] B force karti hai.
Recall Quick self-test (answers cover karo)
Q: C ← B P ke columns kya hain? → [ b j ] C .
Q: R n mein bases-as-columns B , C ke saath, C ← B P kya hai? → C − 1 B .
Q: C ← B P aur B ← C P ka relation kya hai? → inverses.
Q: [ v ] B unique kyun hai? → basis linearly independent hoti hai.
Coordinate vector [ v ] B kya hota hai? Scalars c i ka unique column jis se v = ∑ c i b i .
Coordinates unique kyun hote hain? Kyunki basis linearly independent hoti hai, isliye expansion unique hoti hai.
Change-of-basis matrix P C ← B ke columns kya hain? Purane basis vectors naye basis mein expressed: [ b j ] C .
R n mein B , C ke saath jo basis vectors columns mein rakhe hain, P C ← B = ? C − 1 B .
P C ← B aur P B ← C ka relation kya hai?Yeh ek doosre ke inverses hain.
Kya basis change karne se vector badalta hai? Nahi — sirf uski coordinate description badlti hai; vector invariant rehta hai.
Coordinate map linear kyun hai? Vectors ko add/scale karne par unke B -coefficients termwise add/scale ho jaate hain.
Recall Feynman: ek 12-saal ke bachche ko samjhao
Socho ek treasure map par ek toy rakha hai. Ek dost door se maap karta hai, doosra khidki se. Toy nahi hila, lekin dono doston ne alag-alag "steps left, steps up" numbers likhe. Change of basis woh chota rulebook hai jo ek dost ke numbers doosre ke numbers mein turn karta hai — sirf directions translate karta hai, toy kabhi nahi hilata.
"New columns of Old, then C-inverse-B."
Columns = naye kapdon mein old vectors b j . Aur C − 1 B : C − 1 tumhe C mein "land" karta hai, B B se "launch" karta hai.
Basis and dimension — coordinates tabhi exist karte hain jab basis fix ho.
Linear independence — coordinate uniqueness guarantee karta hai.
Invertible matrices — B , C invertible hain kyunki unke columns bases hain.
Similar matrices and diagonalization — A ′ = P − 1 A P operators ke liye change of basis hai.
Linear transformations and their matrices — coordinate maps abstract maps ko concrete banate hain.
old vectors in new coords
Columns are b_j in C coords
Standard basis shortcut P equals B