Null space (kernel) and column space (image) — basis, dimension: Question bank

4.5.13 · D5 · Maths › Linear Algebra (Full) › Null space (kernel) and column space (image) — basis, dimens

True or false — justify

Ek matrix $A \in R^{m \times n}$ ke liye (matlab: $m$ rows, $n$ columns — $n$ inputs, $m$ outputs), decide karo aur kyun batao.

A

ka column space aur uske RREF ka column space hamesha equal hote hain.

False. Row operations har column ke andar entries ko scramble kar dete hain, isliye actual output vectors badal jaate hain; sirf dependence ka pattern (kaun se columns pivot hain) preserve hota hai, vectors khud nahi.

Agar

A

3 \times 5

hai to uske null space mein

5

components wale vectors hote hain.

True. Null space vectors inputs

x

hote hain, aur

A

mein

5

columns hain, isliye

x

5

entries chahiye — har column ko multiply karne ke liye ek ek.

Agar

dim Null (A) = 0

hai to

A

zaroori square matrix hoga.

False. Yeh sirf full column rank force karta hai (

r = n

), jo

m > n

ke saath bhi ho sakta hai (ek tall matrix), jaise parent note ka Example 2

3 \times 2

hai jiska kernel trivial hai.

A

mein ek column add karne se uska rank kabhi decrease nahi ho sakta.

True. Purane columns abhi bhi present hain, isliye unka span abhi bhi reachable hai; ek naya column sirf ek direction add kar sakta hai ya redundant ho sakta hai — koi pehle se existing dimension remove nahi kar sakta.

Nullity RREF mein zero rows ki sankhya ke barabar hoti hai.

False. Zero rows dependent rows count karte hain (yeh

m - r

hai,

n - r

nahi); nullity free columns count karta hai

= n - r

. Ye sirf tab coincide karte hain jab ittifaqan

m = n

ho.

Agar

A

ka har column ek pivot column hai, to map

x \mapsto A x

injective hai.

True. Full column rank ka matlab hai koi free variable nahi, isliye

A x = 0

x = 0

force hota hai; alag inputs se alag outputs milte hain.

Col (A)

aur

Null (A)

ek hi space ke subspaces hain.

Generally False.

Col (A) \subseteq R^{m}

(outputs) aur

Null (A) \subseteq R^{n}

(inputs); ye sirf tab ek hi space mein rehte hain jab

m = n

ho, aur tab bhi inhe overlap karna zaroori nahi.

Agar

rank (A) = m

(full row rank) hai, to

A x = b

har

b

ke liye solvable hai.

True. Full row rank ka matlab hai

Col (A)

poora

R^{m}

fill karta hai, isliye har output

b

reachable hai — map surjective hai.

Do alag matrices ke exactly same null space aur same column space ho sakte hain.

True. Jaise

A

aur

2 A

ke identical column aur null spaces hain; scaling se kabhi nahi badalta ki kaun si directions reachable hain ya crush ho rahi hain.

Spot the error

Har item mein reasoning ka ek piece diya gaya hai. Flaw dhundo.

A

4 \times 6

hai rank

4

ke saath, isliye nullity

= 4 - 6 = - 2

Wrong sign aur wrong subtraction: nullity

= n - r = 6 - 4 = 2

, hamesha

n

(columns) minus rank, kabhi

m - r

nahi. Nullity kabhi negative nahi ho sakti.

"Col(A) ka basis paane ke liye maine row-reduce kiya aur RREF ke pivot columns liye."

Positions sahi hain lekin vectors galat hain. Woh pivot columns original

A

se lo; RREF ke columns ab sahi output directions mein point nahi karte.

A x = 0

ka sirf trivial solution hai, isliye

A

invertible hai."

Yeh sirf tab valid hai jab

A

square ho. Trivial kernel injectivity deta hai (

r = n

); invertibility ke liye additionally

r = m = n

chahiye, matlab full rank ka square matrix.

3

free variables hain, isliye main

3

special solutions likhta hun, phir pivot variable ke liye ek

4

tha."

Pivot variables ko kabhi apna special solution nahi milta — woh free ones se determine hote hain. Basis vectors ki sankhya exactly free variables ki sankhya ke barabar hoti hai, yahan

3

A

ke columns dependent hain, isliye null space poora

R^{n}

hai."

Dependence sirf kam se kam ek free variable guarantee karta hai, isliye

dim Null \geq 1

; yeh

n

tab hi hoga jab

A

zero matrix ho (har column free).

b

A

ka literal column nahi hai, isliye

A x = b

ka koi solution nahi."

b

ko literal column hona zaroori nahi — use columns ke span mein hona chahiye, matlab

Col (A)

mein. Columns ka ek combination

b

tak pahunch sakta hai chahe

b

unme se ek na ho. Dekho Solving Ax=b.

"Row rank aur column rank alag ho sakte hain, isliye main safe rehne ke liye dono compute karunga."

Ye hamesha equal hote hain — woh single number hi the rank hai. Yeh ek theorem hai (Rank of a matrix), koi coincidence nahi, isliye ek computation kaafi hai.

Why questions

Hum Col(A) original matrix se kyun padhte hain lekin Null(A) reduced se?

Row operations

A x = 0

ka solution set preserve karte hain (isliye null space unchanged rehta hai aur RREF se safe padhna theek hai), lekin column entries distort kar dete hain (isliye Col untouched original se lena zaroori hai). Alag spaces, alag rules.

dim Col (A) + dim Null (A)

n

ke barabar kyun hona chahiye,

m

ke nahi?

Yeh columns ki count hai:

n

columns mein se har ek ya to pivot hai (rank mein count) ya free hai (nullity mein count), isliye dono dimensions

n

inputs ko partition karte hain. Dekho Rank–Nullity Theorem.

Ek free variable null space ke liye exactly ek basis vector kyun produce karta hai?

Us free variable ko

1

aur baaki sabhi free variables ko

0

set karne se har pivot variable uniquely determine ho jaata hai, ek specific solution milta hai; yeh har free variable ke liye karne se independent vectors milte hain jo saare solutions span karte hain.

Column space kabhi "curved" ya origin se shifted kyun nahi ho sakta?

Kyunki

A \cdot 0 = 0

(to origin contain karta hai) aur

A (α x + β y) = α A x + β A y

(combinations ke under closed); linearity ek flat sheet force karti hai

0

se guzarti hui — ek subspace, kabhi affine ya bent surface nahi.

Full column rank (

r = n

) map ko injective kyun banata hai lekin necessarily surjective nahi?

r = n

null space ko khatam karta hai (injective), lekin agar

m > n

hai to column space ek bade

R^{m}

ke andar

n

-dimensional slab hai, isliye zyaadatar outputs unreachable hain (surjective nahi).

Pivot positions jaanna independent columns identify karne ke liye kyun kaafi hai, chahe RREF ne values badal diye hon?

Columns ke beech dependence relations exactly null space ke vectors hain, aur null space row reduction se unchanged rehta hai — isliye RREF wahi independence pattern report karta hai jo original columns mein tha.

Edge cases

Zero matrix

0_{m \times n}

: uska column space, null space, rank, nullity kya hain?

Col = {0} \subseteq R^{m}

with

dim 0

, isliye rank

0

;

Null = R^{n}

(sab kuch crush ho jaata hai) with nullity

n

. Check:

0 + n = n

. ✓

Ek single column matrix

A \in R^{m \times 1}

jo nonzero hai: dono spaces describe karo.

Col (A)

us ek column se spanned line hai (dim

1

, rank

1

);

Null (A) = {0} \subseteq R^{1}

kyunki

x

ek single scalar hai aur

A x = 0

x = 0

force hota hai (nullity

0

Ek wide matrix jisme columns rows se zyada hain (

n > m

): kya uska null space kabhi trivial ho sakta hai?

Nahi. Rank zyada se zyada

min (m, n) = m < n

ho sakti hai, isliye nullity

= n - r \geq n - m > 0

; hamesha kam se kam ek free variable hoga, isliye kernel mein ek nonzero vector hoga.

Ek matrix jiske sirf diagonal entries nonzero hain, unme se ek zero hai (jaise

diag (3, 0, 5)

): nullity kya hai?

Nullity

1

— doosra column ek zero column hai (free variable

x_{2}

), jisse special solution

(0, 1, 0)^{⊤}

milta hai; baaki do diagonal entries pivot hain, isliye rank

2

A

square hai aur

det A \neq = 0

: ek saanch mein Col, Null, rank, nullity batao.

Col (A) = R^{n}

(surjective),

Null (A) = {0}

(injective), rank

= n

, nullity

= 0

— map ek bijection hai, matlab invertible.

Agar

b = 0

hai, to

A x = b

ke kitne solutions hain, aur iska null space se kya relation hai?

Poora null space hi complete solution set hai — particular solution

0

liya ja sakta hai, isliye saare solutions

0 + Null (A)

hain. Agar nullity

> 0

hai to infinitely many hain; agar nullity

= 0

hai, to sirf

x = 0

Connections

Rank of a matrix — woh single number jis par dono bases depend karte hain.
Rank–Nullity Theorem — counting identity jo yahan har trap test karta hai.
Row reduction & RREF — kyun positions transfer hoti hain lekin values nahi.
Linear independence and basis — pivots $\Leftrightarrow$ independent columns.
Injective and surjective linear maps — trivial kernel vs full row rank.
Four fundamental subspaces — har space kahan rehta hai.
Solving Ax=b — solvability column space mein rehti hai.

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