Invertible matrix theorem — 12+ equivalent conditions: Question bank

4.5.25 · D5 · Maths › Linear Algebra (Full) › Invertible matrix theorem — 12+ equivalent conditions

Definition Vocabulary jo is page par assume ki gayi hai (yahan build ki gayi hai taaki bank self-contained rahe)

Square matrix: rows aur columns ki count barabar, likha jaata hai $n \times n$ . Sirf square matrices "invertible" ho sakti hain.
Invertible: ek matrix $A^{- 1}$ exist karti hai jiske liye $A A^{- 1} = A^{- 1} A = I$ , the identity. Ise ek aisi machine samjho jise hamesha ulta bhi chalaya ja sakta hai.
Null space $Nul A$ : sabhi inputs $x$ jinke liye $A x = 0$ — woh vectors jinhe machine origin par crush kar deti hai. Agar sirf $x = 0$ crush hota hai, toh null space "trivial" ${0}$ hai.
Column space $Col A$ : har woh output jo $A$ ke columns ke combination se bana sako — woh set of $b$ jo $A x = b$ actually reach kar sakta hai.
Pivot: row reduction ke baad, ek leading nonzero entry jo ek "real" direction mark karti hai; ek free column mein koi pivot nahi hota aur woh ek free variable create karta hai.
Rank: pivots ki number = column space ki dimension. Nullity = $dim Nul A$ . Yeh rank–nullity obey karte hain: $rank A + dim Nul A = n$ .
Injective (one-to-one): alag-alag inputs kabhi ek output share nahi karte. Surjective (onto): har possible output reach hota hai.

Recall IMT conditions jo is page par cite ki gayi hain (dobara state ki gayi hain taaki page chhodni na pade)

Ek square $n \times n$ matrix ke liye, yeh sab equivalent hain — numbers parent IMT list se match karte hain:

(1) $A$ invertible hai. (2) $A$ , $I_{n}$ ke row-equivalent hai (RREF $= I$ ). (3) $A$ mein $n$ pivots hain. (4) $A x = 0$ ka sirf $x = 0$ solution hai (trivial null space). (5) columns linearly independent hain. (6) map $x \mapsto A x$ injective hai. (7) $A x = b$ har $b$ ke liye solvable hai. (8) columns $R^{n}$ span karte hain. (9) map surjective hai. (10) left inverse exist karta hai ( $C A = I$ ). (11) right inverse exist karta hai ( $A D = I$ ). (12) $A^{T}$ invertible hai. (13) $det A \neq = 0$ . (14) $0$ ek eigenvalue nahi hai. (15) $rank A = n$ . (16) columns ek basis form karte hain. (17) $dim Nul A = 0$ .

Neeche ke do pictures almost har trap ke peeche ka mental model hain — inhe baar baar dekhte rehna.

Figure — Invertible matrix theorem — 12+ equivalent conditions

True or false — justify karo

Kya ek

5 \times 3

matrix ke columns ka linearly independent hona yeh conclude karne ke liye kaafi hai ki woh invertible hai?

False — invertibility (aur poora IMT) sirf square matrices ke liye defined hai; ek

5 \times 3

matrix square nahi hai, toh uske liye "invertible" apply hi nahi hota. Dekho Linear Independence and Span.

Agar

A

4 \times 4

matrix hai aur uske columns linearly independent hain, toh kya

A

invertible hai?

True — ek square matrix ke liye, independent columns har column mein ek pivot force karte hain (condition 3), hence full rank

n

(15), hence RREF

= I

(2), hence invertibility (1); chain 5 ⟹ 3 ⟹ 2 ⟹ 1 kaam karta hai.

Kya yeh sach hai ki

1 0^{- 9}

jaisi bahut chhoti determinant wali matrix "almost not invertible" hoti hai?

False as a theorem statement — invertibility strict condition

det A \neq = 0

(13) hai, aur

1 0^{- 9} \neq = 0

, toh

A

fully invertible hai; chhotaapan ek conditioning issue hai (neeche "condition number" note dekho), invertibility wala nahi.

Kya ek square matrix ke liye, sirf

A D = I

(sirf ek right inverse) prove karne ke liye kaafi hai ki

A

invertible hai?

True — square

A

ke liye, ek one-sided inverse automatically two-sided hota hai (conditions 10 "

C A = I

" aur 11 "

A D = I

" dono IMT mein hain), toh

A D = I

force karta hai

D A = I

Kya yeh sach hai ki

A x = b

ka kisi ek khaas

b

ke liye solution hona batata hai ki

A

invertible hai?

False — condition 7 chahti hai ki

R^{n}

ke har

b

ke liye solution ho; ek lucky

b

jo column space mein happen to lie karta hai, baaki ke baare mein kuch prove nahi karta.

Agar

A

0

ek eigenvalue hai, toh kya

A

non-invertible hai?

True — ek zero eigenvalue matlab hai ki koi

x \neq = 0

satisfy karta hai

A x = 0 x = 0

, ek nonzero null vector, jo trivial-null-space condition (4) aur condition 14 tod deta hai. Dekho Eigenvalues and Eigenvectors.

Kya

A^{T}

ka invertible hona guarantee karta hai ki

A

invertible hai?

True — condition 12; kyunki

det A^{T} = det A

, ek determinant nonzero hai exactly tab jab doosra bhi ho.

Kya ek aisi matrix possible hai jiske columns

R^{n}

span karte hain lekin linearly dependent hain, ek square matrix ke liye?

False — ek square matrix ke liye,

R^{n}

span karne ke liye

n

pivots chahiye (condition 8 forces 3), aur

n

pivots independence bhi force karte hain (5); square matrix ke liye in dono properties ko alag nahi kiya ja sakta. Dekho Basis and Dimension.

Error dhundho

A

3 \times 5

matrix hai rank

3

ke saath, toh yeh

R^{3}

span karta hai, toh yeh invertible hai." — galti kahan hai?

R^{3}

span karna theek hai, lekin "invertible" ek

3 \times 5

non-square matrix ke liye undefined hai; map onto hai lekin one-to-one nahi hai (iske

5 - 3 = 2

free variables hain), toh koi two-sided inverse exist nahi karta.

"Kyunki

A x = 0

ka trivial solution

x = 0

hai, toh columns independent hain." — kya galat hai?

Har homogeneous system mein trivial solution hota hai; independence (condition 5, equivalently 4) chahti hai ki trivial solution only solution ho. Speaker hamesha-true case ko galat label kar raha hai.

A

ki row 2, row 1 ke barabar hai, toh columns dependent hain." — kya yeh reasoning sound hai?

Conclusion (not invertible) sahi hai, lekin ek step skip ho raha hai. Do equal rows rows ko dependent banate hain, toh row rank

n

se kam ho jaata hai. Kyunki row rank hamesha column rank ke barabar hoti hai (

rank A = rank A^{T}

), column rank bhi

n

se kam ho jaati hai, toh ek column free ho jaata hai aur columns dependent hain — yeh link, na ki "rows columns jaisi dikhti hain," reason hai. Dekho Rank and Nullity.

det A = 0

, toh

A

ka koi eigenvalue nahi hai." — theek karo.

det A = 0

matlab

0

ek eigenvalue hai (condition 14 fail ho raha hai), yeh nahi ki eigenvalues absent hain;

A

ke phir bhi

n

eigenvalues hain (multiplicity count karte hue) jinka product

det A = 0

hai.

A

invertible hai, toh uske RREF mein kuch free columns hain." — galti kahan hai?

Invertible ⟹ RREF exactly

I_{n}

hai (condition 2), jisme har column mein ek pivot hai aur isliye koi free column nahi hai. Dekho Elementary Matrices and Row Reduction.

C A = I

dikhata hai ki

C

ek left inverse hai, toh

A

phir bhi invertible nahi ho sakta." — square

A

ke liye sahi hai?

Nahi — square

A

ke liye, ek left inverse (condition 10) already IMT list mein hai, toh

C A = I

prove karta hai ki

A

fully invertible hai aur

C = A^{- 1}

hai.

"Columns dependent hain, isliye

det A

ek chhota nonzero number hai." — theek karo.

Dependent columns

det A

ko exactly zero banate hain, sirf chhota nahi; dependence aur

det A \neq = 0

(condition 13) mutually exclusive hain. Dekho Determinant.

Why questions

"One-to-one" (injective) square matrices ke liye "onto" (surjective) ke saath kyun coincide karta hai, lekin generally nahi?

Kyunki ek square

A

ke liye,

n

pivots simultaneously independence deliver karte hain — injective (condition 6) — aur spanning — onto (9); ek non-square matrix mein zyaada se zyaada inme se ek ho sakti hai, toh dono properties alag ho jaati hain. Dekho Linear Transformations — Injective and Surjective.

A^{- 1}

directly compute karne ke comparison mein

det A \neq = 0

check karna often kyun prefer kiya jaata hai?

det A

(condition 13) ek single scalar test hai usi yes/no sawaal ke liye, jabki

A^{- 1}

compute karna kaafi zyaada expensive hai; IMT aapko sabse sasta equivalent condition choose karne deta hai aur yeh jaante hue ki baaki sab follow karengi.

Ek free (non-pivot) column invertibility ko turant kyun khatam kar deta hai?

Ek free column ek free variable correspond karta hai, jo

A x = 0

ka ek nonzero solution deta hai; woh nonzero null vector condition 4 aur isliye har IMT condition tod deta hai. Dekho Rank and Nullity.

A \sim I

(identity ke row equivalent) kyun imply karta hai ki

A

invertible hai?

Har row operation wahi hai jaise left mein ek elementary matrix

E_{i}

se multiply karna — identity ke saath woh ek operation applied, aur har

E_{i}

invertible hai kyunki har row operation undo ki ja sakti hai (swaps swap back karte hain, scalings divide karte hain, additions subtract karte hain). Agar

E_{k} \dots E_{1} A = I

toh

B = E_{k} \dots E_{1}

invertibles ka product hai (hence invertible) aur

B A = I

, toh

B = A^{- 1}

(condition 2 ⟹ 1). Dekho Elementary Matrices and Row Reduction.

Square matrix ke liye

dim Nul A = 0

full rank ke equivalent kyun hai?

Rank–nullity se,

rank A + dim Nul A = n

; toh nullity

0

(condition 17) rank

n

(15) force karta hai, aur vice versa. Dekho Rank and Nullity.

"Condition number" ka matlab kya hai, aur ek tiny determinant not invertible hona ke saath same kyun nahi hai?

Condition number measure karta hai ki

b

(ya

A

) mein ek chhoti si wiggle

A x = b

ke solution

x

mein kitni badi wiggle mein amplify ho sakti hai — ek "sensitivity to error" gauge, koi yes/no property nahi. Ek matrix perfectly invertible ho sakti hai (

det \neq = 0

) phir bhi ill-conditioned ho (huge condition number), toh numerically solve karna shaky hai even though theorem use khushi se invertible kehta hai.

Edge cases

Kya

1 \times 1

matrix

[0]

invertible hai?

Nahi — uski determinant

0

hai, uska single column zero vector hai (dependent), aur yeh sab kuch

0

par map karta hai; har IMT condition fail hoti hai.

Kya zero matrix

0_{n \times n}

kabhi invertible hoti hai?

Kabhi nahi — uske columns sab

0

hain, toh

det = 0

, rank

= 0

, aur uska null space poora

R^{n}

hai; yeh har

n \geq 1

ke liye IMT fail karta hai.

Kya IMT ek

0 \times 0

(empty) matrix par apply hota hai?

Convention ke hisaab se empty matrix invertible hai jisme

det = 1

hai, kyunki yeh vacuously trivial space

{0}

ko bijectively khud par map karta hai — yeh ek real trap se zyaada ek harmless boundary case hai.

Agar ek square

A

ke do identical columns hain, toh IMT turant kya kehta hai?

Do equal columns dependent hain, toh condition 5 fail hoti hai; isliye

det A = 0

(13), rank

< n

(15),

Nul A \neq = {0}

(4/17),

0

ek eigenvalue hai (14), aur

A

not invertible hai — sab ek saath.

Kya ek square matrix onto ho sakti hai (

Col A = R^{n}

) lekin one-to-one na ho?

Nahi — ek square matrix ke liye onto

n

pivots force karta hai, jo trivial null space bhi force karta hai; non-square world ke upar ke unlike, dono ko alag nahi kiya ja sakta.

Recall Har item ke peeche ek hi sawaal

Upar ke almost har trap actually do errors mein se ek hai: (1) yeh bhool jaana ki IMT invoke karne se pehle matrix square honi chahiye, ya (2) "there exists" aur "for all" ko confuse karna — ek solution versus har solution, ek $b$ versus har $b$ . Pucho "kya yeh square hai?" aur "kya main ek case ya sab cases par quantify kar raha hoon?" aur zyaadatar traps apne aap disarm ho jaate hain.