4.5.15
Vectors ka ek set linearly independent hai agar set mein koi bhi vector baaki vectors se build nahi kiya ja sakta . Har vector ek genuinely naya direction add karta hai. Agar ek bhi vector redundant hai (baaki ka combination hai), toh set dependent hai.
Intuition Yeh concept exist kyun karta haiJab hum ek space ko vectors ki list se describe karte hain, toh hum chahte hain koi waste na ho . Ek dependent set mein ek "passenger" hota hai — ek vector jo koi nayi information nahi laata. Independence guarantee karta hai:
Vectors ek honest basis banate hain (minimal spanning set).
Coordinates unique hote hain (har vector ko likhne ka sirf ek hi tarika hai).
Unse bane matrices invertible / full rank hote hain.
Toh independence ek gatekeeper hai is sawaal ke liye: "kya yeh ek clean coordinate system hai?"
Definition Linear independenceEk vector space mein vectors v 1 , … , v n linearly independent hain agar inhe solve karne wale akele scalars
c 1 v 1 + c 2 v 2 + ⋯ + c n v n = 0 == c 1 = c 2 = ⋯ = c n = 0 == trivial solution hai).
Agar koi aisa solution exist karta hai jisme kam se kam ek c i = 0 ho, toh set linearly dependent hai.
= 0 nontrivial combination 0 c 1 = 0 v 1 = − c 1 c 2 v 2 − ⋯ − c 1 c n v n . v 1 hai — redundant! "Only trivial solution" condition bilkul algebraic tarika hai yeh kehne ka ki koi bhi vector redundant nahi hai .
Definition ek homogeneous linear system hai. Vectors ko ek matrix A columns ke roop mein stack karo:
A = ∣ v 1 ∣ ∣ v 2 ∣ ⋯ ∣ v n ∣ , A c = 0 .
Yeh isliye sach hai kyunki A c = c 1 v 1 + ⋯ + c n v n hi hai).
Worked example Example 1 — square, determinant use karoKya v 1 = ( 1 , 2 ) , v 2 = ( 3 , 4 ) R 2 A = [ 1 2 3 4 ] , det A = 1 ⋅ 4 − 3 ⋅ 2 = − 2 = 0. Independent.
Determinant kyun? R 2 det = 0 det = 0 ⇒
Worked example Example 2 — inspection se dependence pakadnaa = ( 1 , 1 , 1 ) , b = ( 2 , 2 , 2 ) , c = ( 0 , 1 , 3 ) b = 2 a 2 a − b + 0 c = 0 nontrivial combo hai.
Dependent.
Yeh step kyun? Hamesha row reduction ki zaroorat nahi; ek visible scalar multiple already ek nontrivial dependency hai.
Worked example Example 3 — pakka karne ke liye row reduce karov 1 = ( 1 , 0 , 2 ) , v 2 = ( 0 , 1 , 1 ) , v 3 = ( 1 , 1 , 3 ) A = 1 0 2 0 1 1 1 1 3 R 3 − 2 R 1 1 0 0 0 1 1 1 1 1 R 3 − R 2 1 0 0 0 1 0 1 1 0 . 2 pivots ⇒ ⇒ Dependent .
Yeh step kyun? Ek zero row ka matlab hai rank = 2 < 3 A c = 0 v 3 = v 1 + v 2
Worked example Example 4 — zyada vectorsR 3 automatically dependent hain: A 3 × 4 ≤ 3 < 4
Yeh step kyun? Space ki dimension se zyada independent directions nahi ho sakte.
Recall Compute karne se pehle predict karoTest karne se pehle, forecast karo : vectors ki count vs. dimension; obvious multiples ya sums ke liye scan karo.
Forecast: "( 1 , 2 , 3 ) , ( 2 , 4 , 6 ) 2 × Verify: parallel pair ka det
Common mistake "Unka sum/combination zero nahi hai, toh woh independent hain."Kyun sahi lagta hai: independence ka matlab "feel" hota hai ki vectors alag-alag direction mein hain, aur tumne dekha ki koi kisi ke equal nahi hai.
Fix: Independence is baare mein hai ki kya koi bhi nontrivial combo 0 specific sum deta hai ya nahi. Hamesha A c = 0 ( 1 , 1 ) , ( 2 , 3 ) , ( 3 , 4 ) R 2
Common mistake "Main determinant use karunga" — lekinA Kyun sahi lagta hai: determinant dependence test ka go-to method hai.
Fix: det square matrices ke liye exist karta hai (R n n rank / row reduction use karo.
Zero vector case ke saath confusion.
Kyun sahi lagta hai: "0 Fix: 0 1 ⋅ 0 + 0 ⋅ ( rest ) = 0 c 1 = 1 = 0
Recall Feynman: 12-saal ke bacche ko samjhaoSocho tum directions de rahe ho: "East jao, North jao." Do useful, alag instructions. Ab "Northeast jao" add karo. Yeh teesra useless hai — tum already East + North combine karke Northeast pahunch sakte the. Directions ki ek list independent hai jab har instruction tumhe wahan pahunchati hai jahan baaki nahi pahuncha sakte. Jis moment ek kisi baaki ka leftover combo ho, woh dependent hai — tumhare paas ek passenger hai jo kuch nahi kar raha.
"Zero only the boring way."
Agar c 1 v 1 + ⋯ = 0 sirf tab jab sab c i = 0 independent . Zero hit karne ka koi bhi exciting (nonzero) tarika → dependent .
Aur: R ows reduce karo, P ivots gino: Pivots = #vectors → indePendent .
v 1 , … , v n Sirf c 1 = ⋯ = c n = 0 c 1 v 1 + ⋯ + c n v n = 0
A c = 0 A dependent hain.
n rank ( A ) = n
Determinant test — kab valid hai aur kya kehta hai? Sirf square A ⟺ det A = 0
Kya R 3 Kabhi nahi — dimension se zyada vectors hone par free variable force hota hai.
Kya 0 Nahi, hamesha dependent: 1 ⋅ 0 = 0
Agar v 3 = v 1 + v 2 Dependent — v 1 + v 2 − v 3 = 0
"Only trivial solution" ka matlab koi redundant vector nahi hai, kyun? Ek nontrivial c i = 0 v i
Span and spanning sets — independence + spanning = basis .Basis and dimension — basis ek maximal independent set hai.Rank of a matrix — rank = independent columns/rows ki sankhya.Determinant — zero determinant ⇔ dependent columns (square case).Homogeneous systems and null space — nontrivial null space ⇔ dependence.Invertible matrix theorem — independent columns ⇔ invertible.
Only trivial solution c=0
More vectors than dimension
