Rank of a matrix — definition, row rank = column rank theorem: Question bank

4.5.12 · D5 · Maths › Linear Algebra (Full) › Rank of a matrix — definition, row rank = column rank theore

Shuru karne se pehle, rank ke teen roop yaad karo jo yahan chahiye: row rank = row space ki dimension, column rank = column space ki dimension, aur pivots = elimination ke baad leading nonzero entries. Central theorem kehta hai teeno ek hi number hain.

Figure — Rank of a matrix — definition, row rank = column rank theorem

Jahan "Spot the error" aur "Why" sections mein ek symbol baar baar aata hai, hum rank factorization $A = C R$ use karte hain. Yahan ek definition hai jo in sabke liye chahiye:

True or false — justify karo

Har

m \times n

matrix mein

rank (A) = rank (A^{T})

hota hai.

TRUE.

A

ka row rank,

A

ke column rank ke barabar hota hai, aur transpose karna rows aur columns ko swap kar deta hai, toh

rank (A^{T})

dusri taraf se wahi number count karta hai.

Agar ek matrix mein saare zeros ki ek row ho, toh uska rank rows ki sankhya se kam hona chahiye.

TRUE. Ek zero row row space mein kuch contribute nahi karti, toh zyada se zyada

m - 1

rows independent ho sakti hain, jo

rank < m

force karta hai.

5 \times 3

matrix ka rank

4

ho sakta hai.

FALSE. Rank

\leq min (m, n) = min (5, 3) = 3

hota hai, kyunki sirf

3

columns picture mein hain aur wo zyada se zyada ek

3

-dimensional space span kar sakte hain.

Agar

A

ka har column nonzero ho, toh

rank (A)

columns ki sankhya ke barabar hota hai.

FALSE. Nonzero columns phir bhi dependent ho sakte hain; jaise

(1, 1)

aur

(2, 2)

dono nonzero hain lekin sirf ek line span karte hain, jo rank

1

deta hai.

Row operations ek matrix ke actual column space ko badal sakti hain.

TRUE — aur yahi subtle part hai. Row ops change karti hain ki columns kaunse vectors hain, toh column space (reachable outputs ka set) genuinely move ho sakta hai; sirf iska dimension preserve hota hai.

Nonzero matrix

A

ke liye

rank (2 A) = 2 rank (A)

hota hai.

FALSE.

2

se scale karna har column ko

2

se scale karta hai lekin independence kabhi nahi badalta, toh

rank (2 A) = rank (A)

; rank ek magnitude nahi hai.

Agar

A

n \times n

hai aur

rank (A) = n

hai, toh

A

invertible hai.

TRUE. Full column rank matlab columns independent hain aur

R^{n}

span karte hain, jo exactly invertible condition hai.

rank (A + B) = rank (A) + rank (B)

tab hota hai jab

A, B

mein koi zero rows na hon.

FALSE. Generally sirf

rank (A + B) \leq rank (A) + rank (B)

hi hota hai; overlapping column spaces additivity ko destroy kar dete hain chahe zero rows hon ya na hon.

Ek matrix aur uska reduced echelon form hamesha same rank rakhte hain.

TRUE. Elimination rows ko rows ke linear combinations se replace karta hai, jo row space ya uski dimension kabhi nahi badalta, toh pivot count preserved rehta hai.

Ek matrix ka rank zero ho sakta hai.

TRUE — lekin sirf zero matrix ke liye. Agar koi bhi entry nonzero hai, toh woh row (aur column) independent hai, jo rank

\geq 1

deta hai.

Spot the error

"Maine

A

ko echelon form mein reduce kiya; pivots columns 1 aur 3 mein hain, toh reduced matrix ke columns 1 aur 3 column space ka basis banate hain."

Positions sahi hain lekin vectors galat hain: aapko original

A

ke columns 1 aur 3 lene chahiye, kyunki row ops ne column vectors khud ko badal diya.

A

mein do identical rows hain, toh main ek delete karta hun; yeh rank change kar deta hai kyunki maine ek row remove ki."

Duplicate row delete karna row space se kuch remove nahi karta (woh already span mein thi), toh rank unchanged rehta hai — woh row kabhi contribute hi nahi kar rahi thi.

A

mein do identical columns hain, toh main ek delete karta hun; yeh rank shrink kar deta hai kyunki matrix mein ab fewer columns hain."

Rows wali wajah se hi yahan bhi galat hai: ek duplicate column already doosron ke span mein hai, toh column space aur uski dimension unchanged rehti hai — ise delete karna matrix ka size badalta hai, rank nahi.

"Kyunki

rank (A B) \leq rank (A)

hai, jab

B

nonzero ho toh hamesha

rank (A B) = rank (A)

hota hai."

Bound sirf

\leq

hai. Agar

B

un directions ko collapse kar de jo

A

ko chahiye thi, toh product ka rank strictly smaller ho sakta hai;

B

ke nonzero hone se equality guarantee nahi hoti.

"Matrix mein 4 nonzero rows hain, isliye uska rank 4 hai."

Nonzero rows dependent ho sakti hain; aapko echelon form mein reduce karke pivots count karne chahiye, visible nonzero rows nahi.

"Column rank columns count karta hai, row rank rows count karta hai, toh ek

3 \times 5

matrix mein dono equal nahi ho sakte — totals alag hain."

Wo spans ki dimensions count karte hain, raw totals nahi; central theorem dono spans ko same dimension share karne ke liye force karta hai chahe

5

columns aur

3

rows hon.

A = C R

mein inner dimension

r

rank ko

r

prove karta hai sirf tab jab

r

m, n

mein smaller ho."

Nahi — yaad rakho

C

hai

m \times r

independent columns ke saath aur

R

hai

r \times n

independent rows ke saath, toh

r

forced hai ki shared rank ho; yeh

min (m, n)

tak koi bhi value ho sakta hai, necessarily smaller dimension khud nahi.

Why questions

A = C R

factorization row rank

\leq

column rank kyun prove karta hai?

Yaad rakho

C

hai

m \times r

aur

R

hai

r \times n

A

ki row

i

equals (C ki row

i

) times

R

, toh

A

ki har row,

R

r

rows ka combination hai, jo row rank ko

r

= column rank par cap kar deta hai.

Hum column-space basis ke liye pivot columns original matrix se kyun lete hain, lekin count reduced wale se kyun padhte hain?

Count (dimension) row ops ke under invariant hai, lekin actual column vectors badal jaate hain; reduced pivots sirf reveal karte hain kaunsi positions independent hain, aur wo positions untouched original columns ki taraf point karti hain.

rank (A) \leq min (m, n)

kyun hota hai aur isse zyada nahi?

Columns

R^{m}

mein rehne wale

n

vectors hain, toh unka span zyada se zyada

n

vectors use karta hai aur zyada se zyada

m

dimensions mein fit hota hai — jo bhi smaller ho wahi ceiling hai.

Rank map

x \mapsto A x

ki "output dimension" kyun measure karta hai?

Har output

A x

columns ka combination hai, toh reachable set exactly column space hai, jiski dimension rank hai.

Hum row rank = column rank sirf "ek baar pivots count karke" kyun prove nahi kar sakte?

Pivots count karna ek number deta hai, lekin yeh explain nahi karta ki dono spaces kyun agree karte hain;

A = C R

argument wahi hai jo prove karta hai ki dono spans genuinely us dimension mein hain.

Linearly dependent row add karna rank ko unchanged kyun rakhta hai?

Ek dependent row already doosron ke span mein hai, toh woh row space mein koi nayi direction add nahi karti aur dimension wahi rehti hai.

Edge cases

3 \times 3

zero matrix ka rank kya hai?

Zero. Koi nonzero rows ya columns nahi hain, toh dono spaces sirf origin hain, ek

0

-dimensional space.

Kisi bhi matrix

A

ke liye

rank (0 \cdot A)

kya hai, aur yeh

rank (2 A)

se kaise contrast karta hai?

Zero — scalar

0

se multiply karna

A

ko zero matrix mein badal deta hai, har direction collapse kar deta hai, jabki

2

(ya koi bhi nonzero scalar) se multiply karna independence preserve karta hai aur rank ko

rank (A)

ke barabar rakhta hai.

n \times n

identity matrix ka rank kya hai?

n

. Iske columns standard basis vectors hain — poori tarah independent — toh iska full rank hai aur yeh invertible hai.

Ek single nonzero column vector (ek

m \times 1

matrix) ka rank kya hai?

Ek. Ek single nonzero vector ek line span karta hai, toh column rank aur row rank dono

1

ke barabar hote hain.

Us matrix ka rank kya hai jiske saare entries same nonzero constant

c

ke barabar hain?

Ek. Har row

c \cdot (1, 1, \dots, 1)

hai, toh saari rows ek vector ke scalar multiples hain, jo ek

1

-dimensional row space deta hai.

rank (A) = n

(full column rank) wale

m \times n

matrix ke liye

dim ker (A)

kya hai?

Zero, Rank–Nullity Theorem ke zariye:

rank (A) + dim ker (A) = n

nullity ko vanish hone par force karta hai, toh

A x = 0

ka sirf trivial solution hai.

Agar

A

hai

m \times n

jahan

m < n

, toh kya

A

ka trivial kernel ho sakta hai?

Nahi. Rank

\leq m < n

hai, toh rank–nullity se

dim ker (A) = n - rank (A) \geq n - m > 0

; hamesha

A x = 0

ka ek nonzero solution hota hai.

Kya ek nonsquare matrix "full rank" ho sakti hai?

Haan — full rank ka matlab sirf

rank = min (m, n)

hai; ek

2 \times 5

matrix full rank

2

rakhti hai agar uski do rows independent hain, chahe woh square se door ho.

Connections

Row Space and Column Space
Gaussian Elimination & Echelon Form
Rank–Nullity Theorem
Linear Independence and Basis
Matrix Multiplication as Linear Combination
Invertible Matrix Theorem