Linear independence — formal definition, testing: Question bank

4.5.15 · D5 · Maths › Linear Algebra (Full) › Linear independence — formal definition, testing

Shuru karne se pehle, vocabulary ka ek reminder taaki koi symbol unclear na rahe:

Ek combination $c_{1} v_{1} + \dots + c_{n} v_{n}$ ka matlab hai "har vector ko ek number $c_{i}$ se scale karo aur unhe add karo."
Trivial (boring) combination woh hai jisme sabhi $c_{i} = 0$ hain — yeh hamesha $0$ deta hai, kisi bhi vector ke liye. Yeh kuch prove nahi karta.
Independent = $0$ tak pahunchne ka ek hi tarika hai, woh hai boring wala. Dependent = ek exciting tarika bhi hai (kam se kam ek $c_{i} \neq = 0$ ).

True or false — justify karo

Har item ek statement hai. True/false decide karo, phir reason do, sirf verdict nahi.

Do vectors independent hain jab tak woh equal nahi hain.

False — unhe scalar multiples nahi hona chahiye.

(1, 2)

aur

(2, 4)

unequal hain phir bhi dependent hain kyunki

(2, 4) = 2 (1, 2)

Agar 3 vectors ka ek set independent hai, toh unme se koi bhi 2 ka subset bhi independent hoga.

True — subset mein dependency hogi toh poore set mein bhi dependency hogi (missing coefficients ko

0

se pad karo). Toh independence neeche ki taraf inherit hoti hai.

Agar ek set mein har pair of vectors independent hai, toh poora set independent hai.

False — pairwise independence zyada weak hai.

(1, 0), (0, 1), (1, 1)

har pair mein independent hain, phir bhi

(1, 1) = (1, 0) + (0, 1)

se trio dependent ho jaata hai.

Independent vectors ka ek set phir bhi space ko span karne mein fail ho sakta hai.

True — independence aur spanning alag-alag cheezein hain.

(1, 0, 0), (0, 1, 0)

R^{3}

mein independent hain lekin sirf ek plane span karte hain; dono ke liye tumhe ek basis chahiye.

Ek independent set mein vector add karne se woh aur independent nahi hota, sirf dependent ho sakta hai.

True — ya toh tum genuinely nayi direction add karte ho (phir bhi independent) ya ek redundant direction add karte ho (dependent ho jaata hai). Independence sirf preserve ya break ho sakti hai, "improve" nahi ho sakti.

Agar

det A = 0

hai ek square matrix ke liye jiske columns tumhare vectors hain, toh vectors dependent hain.

True square

A

ke liye — zero determinant ka matlab hai columns ek lower-dimensional space mein collapse ho gaye, toh ek nontrivial combination

0

tak pahunchti hai.

A

ke independent columns guarantee karte hain ki

A

invertible hai.

Generally False — sirf tab jab

A

square ho. Ek

3 \times 2

matrix ke independent columns ho sakte hain phir bhi woh square nahi hai, toh invertibility ka sawaal hi nahi uthta.

Vectors ka empty set linearly dependent hai.

False — empty set convention se independent hai, kyunki koi bhi nontrivial combination likhne ke liye kuch hai hi nahi (vacuously, sirf trivial wala exist karta hai).

Agar

A c = 0

ka sirf trivial solution hai, toh

A

ke columns independent hain.

True — yeh definition ko dobaara bata raha hai: homogeneous system ka sirf trivial null space hona exactly independence hai.

Spot the error

Har item mein ek plausible-sounding argument hai jisme ek hidden flaw hai. Flaw batao.

(1, 2, 3), (4, 5, 6), (7, 8, 9)

teen distinct non-parallel vectors hain, toh yeh independent hain."

Flaw: distinctness aur non-parallelism ek teen-term dependency ko nahi rokti. Yahan teesra

2 \times

doosra

-

pehla hai, toh yeh dependent hain — tumhe poora system solve karna hoga, pairs eyeball nahi karna.

"Maine ek combination

2 v_{1} + 0 v_{2} + 0 v_{3}

compute ki aur yeh zero nahi thi, toh set independent hai."

Flaw: independence ke liye har nontrivial combination ka nonzero hona zaroori hai, ek lucky choice ka nahi. Tumhe dikhana hoga ki koi bhi nontrivial combo

0

tak nahi pahunchti, yaani

A c = 0

fully solve karo.

"Yeh

R^{4}

mein 4 vectors hain, toh yeh independent honge hi."

Flaw: count aur dimension ka match hona necessary hai lekin sufficient nahi.

R^{4}

mein chaar vectors phir bhi dependent ho sakte hain (jaise do equal vectors); tumhe

det \neq = 0

ya full rank chahiye.

"Matrix square nahi hai, lekin main dependence test karne ke liye iska determinant lunga."

Flaw: determinant sirf square matrices ke liye defined hai. Non-square

A

ke liye row reduction use karo aur pivots count karo.

"Row reduction ke baad ek zero row hai, lekin vectors rows hain, aur ek row zero ho gayi — woh zero row sirf leftover hai, toh ignore karo."

Flaw: reduction ke baad zero row exactly yeh signal hai ki rank

<

vectors ki sankhya — yeh prove karta hai ki ek dependency exist karti hai, isse ignore nahi karna chahiye.

"Set

S

mein

0

aur do independent vectors hain, toh

S

independent hai kyunki do real vectors hain."

Flaw:

0

wale kisi bhi set mein dependency hoti hai, kyunki

1 \cdot 0 + 0 \cdot (rest) = 0

nontrivial hai (

c = 1 \neq = 0

). Ek passenger poore set ko poison kar deta hai.

rank (A) = 2

aur

A

ke 2 columns hain, lekin mujhe ek free variable mili, toh yeh dependent hai."

Flaw: contradiction — agar rank, columns ki sankhya ke equal hai (

2 = 2

) toh koi free variable nahi hoti. "Free variable" ek computational slip thi; reduction dobara check karo.

Why questions

Yeh fact nahi, mechanism poochhte hain.

c_{1} v_{1} + \dots + c_{n} v_{n} = 0

ka nontrivial solution kyun matlab hai ki ek vector redundant hai?

Agar koi

c_{i} \neq = 0

hai, toh isse divide karo aur

v_{i} = - \sum_{j \neq = i} \frac{c _{j}}{c _{i}} v_{j}

solve karo,

v_{i}

ko baaki ka combination express karke — ek passenger.

Trivial solution

c_{i} = 0

akele independence prove kyun nahi karta?

Kyunki

0

times kuch bhi

0

hota hai, toh all-zero combination har set ke liye kaam karti hai, dependent ho ya nahi. Yeh zero information carry karta hai.

R^{n}

mein

n

se zyada vectors hamesha dependent kyun hote hain?

Unhe columns ke roop mein stack karne se rows se zyada columns wali matrix milti hai; rank

\leq n <

column count ek free variable force karta hai, isliye ek nontrivial null-space solution milta hai.

Hum test ko homogeneous system

A c = 0

ke roop mein kyun phrase karte hain?

Kyunki

A c

matrix–vector product ki definition se linear combination

c_{1} v_{1} + \dots + c_{n} v_{n}

hi hai, toh "combination equals

0

" ka matlab hai "

A c = 0

solve karo."

Independence kyun guarantee karta hai ki span mein har vector ke liye unique coordinates hote hain?

Agar ek vector ki do representations hoti, toh unhe subtract karne se ek nontrivial combination milti hai jo

0

ke equal hai — independent set ke liye impossible. Toh representation ek-ek-ki-tarah unique hai.

"Pivot in every column" sahi rank condition kyun hai?

Ek pivotless (free) column

A c = 0

mein ek free variable correspond karta hai, jo nontrivial solutions produce karta hai; har column mein pivot hona isse rule out karta hai, independence force karta hai.

det A \neq = 0

sirf square case mein kyun kaam karta hai, aur geometrically yeh kya dekhta hai?

Determinant columns se bane parallelepiped ka signed volume measure karta hai; yeh sirf

R^{n}

mein

n

vectors ke liye exist karta hai. Zero volume ka matlab hai columns lower dimension mein hain — dependent.

Edge cases

Degenerate aur boundary scenarios jinhe log bhool jaate hain.

Kya ek single nonzero vector

{v}

independent hai?

Haan —

c v = 0

jab

v \neq = 0

hoga toh

c = 0

force hoga, ek hi (trivial) solution, toh akela nonzero vector hamesha independent hota hai.

Kya single zero vector

{0}

independent hai?

Nahi —

c \cdot 0 = 0

har

c

ke liye hold karta hai, including

c = 1 \neq = 0

, ek nontrivial solution. Toh

{0}

dependent hai.

Kya sirf 2 vectors ka set dependent ho sakta hai bina kisi ke zero hue?

Haan — agar ek doosre ka nonzero scalar multiple ho, jaise

(1, 3)

aur

(2, 6)

2 (1, 3) - (2, 6) = 0

nontrivially, dependent.

Agar

n

vectors

R^{m}

mein hain jahan

n < m

, kya woh phir bhi dependent ho sakte hain?

Haan — dimension se kam vectors hona independence force nahi karta; jaise

R^{3}

mein

(1, 0, 0)

aur

(2, 0, 0)

do vectors hain, phir bhi dependent hain (parallel).

Ek matrix ka rank kya hoga jiske columns size

n

ka maximal independent set hain?

Exactly

n

— har column ek pivot column hai, toh rank vectors ki sankhya ke equal hai, woh borderline jahan koi bhi added vector dependency create karega.

Kya

R^{n}

ke standard basis vectors kisi bhi

n

ke liye dependent hain?

Kabhi nahi —

c_{1} e_{1} + \dots + c_{n} e_{n} = (c_{1}, \dots, c_{n}) = 0

directly har

c_{i} = 0

force karta hai, toh yeh har

n \geq 1

ke liye independent hain.

Recall Har trap ki one-line summary

Independence decide hoti hai ek honest sawaal se — kya koi nontrivial combination $0$ tak pahunchti hai? Distinctness se nahi, ek lucky combination se nahi, counts match karne se nahi, pairwise checks se nahi. $A c = 0$ solve karo aur pivots count karo.

Connections

Linear independence — formal definition, testing — woh parent jise yeh bank drill karta hai.
Span and spanning sets — independence + spanning milke basis dete hain.
Basis and dimension — independence basis definition ka ek aadha hissa hai.
Rank of a matrix — pivot count independence decide karta hai.
Determinant — square-case shortcut.
Homogeneous systems and null space — test ka algebraic ghar.
Invertible matrix theorem — independent square columns ⇔ invertible.