Transpose — definition, properties: Question bank

2.6.6 · D5 · Maths › Matrices & Determinants — Introduction › Transpose — definition, properties

Yeh page Transpose — definition, properties ke liye ek rapid-fire trap detector hai. Har line ek claim ya question hai; use padho, apna jawab decide karo, PHIR reveal karo. Point arithmetic nahi hai — balki yeh conceptual slips ko pakadna hai jo yeh topic invite karta hai. Reasoning answer side pe hogi, kabhi bare "yes/no" nahi.

Shuru karne se pehle, ek anchor jo hum har jagah use karte hain: transpose $A^{T}$ ka matlab hai flip across the main diagonal (diagonal jo top-left se neeche ki taraf jaati hai). Neeche ki picture poora idea hai — red diagonal fold line dekho, aur notice karo ki uske upar ke entries apni jagah rehte hain jabki off-diagonal partners uske across swap ho jaate hain.

Figure — Transpose — definition, properties

True or false — justify

m \times n

matrix ka transpose phir se

m \times n

hota hai

False — rows aur columns ko flip karna shape ko swap kar deta hai, isliye

A^{T}

ka shape

n \times m

hota hai. Yeh

m \times n

tabhi hoga jab

m = n

(yaani square matrix).

Agar

A

symmetric hai toh

A

zaroor square hoga

True —

A^{T} = A

force karta hai ki

A

aur

A^{T}

ka shape same ho, lekin

A^{T}

ka shape

n \times m

hai jabki

A

ka shape

m \times n

hai, isliye

m = n

hona padega.

Har diagonal matrix symmetric hota hai

True — off-diagonal entries sab zero hote hain, isliye

a_{ij} = a_{j i} = 0

jab bhi

i \neq = j

, jo exactly symmetry condition hai.

Ek skew-symmetric matrix ki main diagonal par koi nonzero number ho sakta hai

False —

a_{ii} = - a_{ii}

force karta hai ki

2 a_{ii} = 0

ho, isliye har diagonal entry zero honi chahiye.

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

har matrix ke liye hold karta hai, chahe non-square bhi ho

True — diagonal ke across do baar flip karna har entry ko uski starting jagah wapas le aata hai:

[(A^{T})^{T}]_{ij} = [A^{T}]_{j i} = [A]_{ij}

, shape bhi.

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

False — order reverse hota hai:

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

. Galat version aksar legal product bhi nahi hota kyunki inner dimensions match nahi karte.

(A + B)^{T} = A^{T} + B^{T}

ke liye zaruri hai ki

A

aur

B

same shape ke hon

True — tum sirf equal shape ki matrices ko add kar sakte ho, aur har ek ko transpose karna (ab flipped) shared shape ko preserve karta hai, isliye sum dono sides par defined hai.

Agar

A^{T} = A^{- 1}

hai toh

A

orthogonal hai

True — yahi exactly orthogonal matrix ki definition hai: uska transpose uske inverse ke barabar hota hai, isliye

A^{T} A = I

Kisi bhi square matrix ke liye,

A^{T} = A^{- 1}

False — yeh sirf orthogonal matrices ke liye hold karta hai. Transpose ek structural flip hai; inverse woh hai jo multiplication ko undo karta hai. Yeh dono bahut kam hi coincide karte hain.

A + A^{T}

hamesha symmetric hota hai (square

A

ke liye)

True —

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

, isliye yeh apne aap ke transpose ke barabar hai.

A - A^{T}

hamesha skew-symmetric hota hai (square

A

ke liye)

True —

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

, jo exactly skew-symmetric condition hai.

Agar

A^{T} = - A

aur

A

1 \times 1

matrix hai, toh

A

koi bhi real number ho sakta hai

False — ek

1 \times 1

matrix apna khud ka transpose hota hai, isliye

- a = a

force karta hai ki

a = 0

ho; sirf zero scalar qualify karta hai.

Spot the error

(2 A)^{T} = 2^{T} A^{T}

, isliye hume scalar ko bhi transpose karna hoga."

Ek scalar aisi matrix nahi hai jise flip kiya jaye;

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

hota hai. Number

2

ki koi rows ya columns nahi hain jo swap ki ja sakein — "

2^{T}

" meaningless hai.

(A B C)^{T} = A^{T} B^{T} C^{T}

Poori reversal yeh hai:

(A B C)^{T} = C^{T} B^{T} A^{T}

. Transpose poori chain ko ulta kar deta hai jaise use backwards padhna, na ki har term ko apni jagah par.

A

symmetric hai,

B

symmetric hai, isliye

A B

symmetric hai."

Generally nahi —

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

hota hai, jo

A B

ke barabar tabhi hoga jab

A

aur

B

commute karein. Product ki symmetry ek stronger, extra condition hai (dekho Matrix Operations).

[123]

ko transpose karne ke liye main same row ko phir se likhta hoon."

1 \times 3

row vector transpose hokar

3 \times 1

column vector ban jaata hai; numbers vertically stack ho jaate hain. Shape sach mein

1 \times 3

3 \times 1

mein badal jaata hai.

(A^{- 1})^{T} = (A^{- 1})^{- 1} = A

, isliye inverse ka transpose

A

hai."

Sahi identity yeh hai:

(A^{- 1})^{T} = (A^{T})^{- 1}

. Reader ne beech mein "transpose" ko "inverse" se confuse kar liya — do alag operations hain (dekho Matrix Inverse).

"Kyunki transpose diagonal ko flip karta hai, isliye diagonal entries sab naye spots par move ho jaate hain."

Diagonal entries

a_{ii}

exactly fold line par baithe hain, isliye woh apni jagah rehte hain. Sirf off-diagonal entries apne partners se swap karte hain.

"Har matrix ek symmetric matrix aur ek skew-symmetric matrix ka sum hai, isliye har matrix symmetric hai."

Decomposition

A = \frac{1}{2} (A + A^{T}) + \frac{1}{2} (A - A^{T})

sach hai, lekin doosra piece skew-symmetric hai, symmetric nahi — isliye

A

khud zaruri nahi ki symmetric ho.

Why questions

Order

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

mein reverse kyun hota hai?

Kyunki entry

[(A B)^{T}]_{ij} = [A B]_{j i} = \sum_{k} a_{j k} b_{k i}

hoti hai; har factor ko transpose ke zariye rewrite karne par,

a_{j k} = [A^{T}]_{k j}

aur

b_{k i} = [B^{T}]_{ik}

milta hai, jo

\sum_{k} [B^{T}]_{ik} [A^{T}]_{k j} = [B^{T} A^{T}]_{ij}

deta hai — shared summed index tabhi line up hota hai jab

B^{T}

pehle aaye.

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

kyun nahi — ek chhote example par dikhao?

A = [1324], B = [0110]

lo: phir

(A B)^{T} = [2143] = B^{T} A^{T}

aata hai, jabki

A^{T} B^{T} = [3412]

aata hai — clearly alag, isliye sirf reversed order kaam karta hai.

Ek symmetric matrix square kyun honi chahiye, sirf "nicely shaped" nahi?

Equation

A^{T} = A

do matrices ko entry-by-entry compare karti hai, aur yeh tabhi possible hai jab unke dimensions identical hon, jo force karta hai ki

A

ki row count uski column count ke barabar ho.

A A^{- 1} = I

jaise equation ko transpose karna

(A^{- 1})^{T} = (A^{T})^{- 1}

kyun prove karta hai?

Dono sides par transpose apply karne aur reversal rule use karne par

(A^{- 1})^{T} A^{T} = I

milta hai, jo kehta hai ki

(A^{- 1})^{T}

exactly woh matrix hai jo

A^{T}

ko undo karta hai — yahi

(A^{T})^{- 1}

ki definition hai.

Scalars aur sums transpose flip ko kyun "ignore" karte hain, lekin products nahi karte?

Scaling aur addition har entry par alag kaam karte hain:

(k A)_{ij} = k a_{ij}

aur

(A + B)_{ij} = a_{ij} + b_{ij}

mein

i

aur

j

swap karne par kuch reroute nahi hota. Ek product entry

[A B]_{ij} = \sum_{k} a_{ik} b_{k j}

A

ki poori ek row ko

B

ke ek column se mix karti hai, isliye indices flip karna yeh reshuffles karta hai ki kaun si row kaun se column se milti hai — aur isi wajah se order reverse hona padta hai.

A^{T} A

kisi bhi (chahe non-square)

A

ke liye hamesha symmetric kyun hota hai?

Compute karo

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

; yeh apne khud ke transpose ke barabar hai. Isliye

A^{T} A

Inner Product Spaces aur least-squares work mein itni baar appear karta hai.

Transpose dot products se related kyun hai?

Column vectors

u, v

ke liye, dot product

u^{T} v

hai — transpose ek column ko row mein badal deta hai taaki matrix multiplication woh single sum-of-products number produce kare jo inner product define karta hai.

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

har square matrix ke liye kyun hold karta hai?

Determinant products ke sums se bana hota hai jo har row aur har column se ek entry chunte hain; transpose bas har "row

i

, column

j

" ko "row

j

, column

i

" ke roop mein relabel karta hai, jo same terms ko unke signs badlaye bina permute karta hai. Minimal check:

A = [a c b d]

ke liye,

det A = a d - b c

aur

det A^{T} = a d - c b

, identical. Dekho Determinants.

Ek symmetric matrix ki eigen-story general matrix se "nicer" kyun ho sakti hai?

Kyunki

A^{T} = A

ek strong constraint hai; yeh real eigenvalues aur orthogonal eigenvectors guarantee karta hai, ek fact jo directly Symmetric Matrices par lean karta hai aur Eigenvalues and Eigenvectors mein feed hota hai.

Edge cases

1 \times 1

matrix

[a]

ka transpose kya hota hai?

Woh

[a]

khud hi hota hai — ek single number diagonal par baitha hota hai jisme swap karne ke liye kuch nahi hota, isliye har

1 \times 1

matrix (trivially) symmetric hota hai.

Ek empty

0 \times n

matrix (koi rows nahi,

n

columns) ka transpose kya hota hai?

Woh

n \times 0

matrix hai — shape-swap rule

m \times n \to n \times m

phir bhi apply hota hai, "koi rows nahi,

n

columns" ko "

n

rows, koi columns nahi" mein badalta hai. Move karne ke liye koi entries exist nahi karti, isliye operation well-defined hai lekin vacuous hai.

Kya zero matrix symmetric hai, skew-symmetric hai, ya dono?

Dono — har entry

0

hai, isliye

a_{ij} = a_{j i}

(symmetric) aur

a_{ij} = - a_{j i}

(skew) simultaneously hold karte hain; square zero matrix aakela aisa matrix hai jo dono hai.

Identity matrix transpose ke under kya hota hai?

I^{T} = I

— identity symmetric hai, kyunki uski sirf nonzero entries diagonal par

1

hain, jo flip ke under fixed rehti hain.

Kya

(A + B)^{T} = A^{T} + B^{T}

phir bhi hold karta hai agar

A

aur

B

2 \times 3

(non-square) hon?

Haan — sum rule ko kabhi squareness ki zarurat nahi hoti, sirf yeh ki

A

aur

B

ek shape share karein; har side valid

3 \times 2

matrix hai aur woh entry by entry agree karte hain.

Kya square matrix ke liye

A

aur

A^{T}

ka determinant hamesha same hota hai?

Haan —

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

hamesha hota hai, kyunki transposing sirf determinant expansion mein same signed products ko permute karta hai; yeh ek clean invariant hai chahe

A

symmetric na ho (dekho Determinants).

Agar

A

m \times n

matrix hai jisme

m \neq = n

ho, toh kya

A^{- 1}

kabhi

A^{T}

ke barabar ho sakta hai?

Nahi — ek non-square matrix ka koi ordinary inverse hota hi nahi, isliye sawaal "

A^{T} = A^{- 1}

" pose hi nahi kiya ja sakta; orthogonality ek square-matrix notion hai.

Kya koi non-square matrix symmetric ho sakti hai?

Nahi — symmetry ke liye

A^{T} = A

chahiye, aur

A^{T}

ke swapped dimensions hain, isliye ek

2 \times 3

matrix kabhi apne

3 \times 2

transpose ke barabar nahi ho sakti.

The one takeaway

Agar tum is poore page se ek single rule yaad rakho, toh yeh rakho: