Transpose — definition, properties: Exercises — Hinglish deep dive

2.6.6 · D4 · Maths › Matrices & Determinants — Introduction › Transpose — definition, properties

Yeh page ek self-test ladder hai. Har problem parent note pe build karti hai aur "kya tum isse pehchaan sakte ho?" se "kya tum isse khud bana sakte ho?" tak climb karti hai. Yahan use kiye gaye har symbol ko wahan define kiya gaya tha — lekin hum har idea ko re-anchor karte hain taaki tum pehli line se follow kar sako.

Hamare ek core rule ki yaad-dahaani, saral shabdon mein: transpose $A^{T}$ woh matrix hai jo tumhe milti hai jab tum $A$ ki har row ko ek column mein badal dete ho. Formally, $A^{T}$ ki row $i$ , column $j$ ki entry, $A$ ki row $j$ , column $i$ ki entry hoti hai: $[A^{T}]_{ij} = [A]_{j i}$

Neeche diya figure pehli hi exercise mein is rule ko action mein dikhata hai — shuru karne se pehle isse dhyan se dekho.

Figure — Transpose — definition, properties

L1 — Recognition

Problem 1.1

Transpose likho is matrix ka: $A = [24 - 1 7 05] .$

Recall Solution 1.1

$A$ mein $2$ rows aur $3$ columns hain, isliye $A^{T}$ mein $3$ rows aur $2$ columns honge.

$A$ ki Row 1 hai $[2, - 1, 0]$ → yeh $A^{T}$ ka column 1 ban jaati hai (figure mein blue block).
$A$ ki Row 2 hai $[4, 7, 5]$ → yeh $A^{T}$ ka column 2 ban jaati hai (green block). $A^{T} = 2 - 1 0 475 .$ Rule check karo: $[A^{T}]_{31} = 0$ ko $[A]_{13} = 0$ ke barabar hona chahiye. ✓

Problem 1.2

Kya yeh matrix symmetric hai, skew-symmetric hai, ya kuch bhi nahi? $S = 0 - 3 2 30 - 4 - 2 40 .$

Recall Solution 1.2

Ek matrix symmetric hoti hai agar use flip karne par wohi matrix wapas milti hai ( $S^{T} = S$ ), aur skew-symmetric hoti hai agar flip karne par negative milta hai ( $S^{T} = - S$ ).

Pehle, diagonal mein sab zeros hain — yeh skew-symmetric ki pehchaan hai (ek symmetric matrix ki koi bhi diagonal ho sakti hai, lekin skew wali ki zaroor zeros honi chahiye kyunki $a_{ii} = - a_{ii}$ se $a_{ii} = 0$ force hota hai).

Flip karo: $S^{T} = 03 - 2 - 3 04 2 - 4 0 = - S .$ Isliye $S$ skew-symmetric hai. Dekho Symmetric Matrices.

L2 — Application

Problem 2.1

$A = [1023]$ aur $B = [42 - 1 5]$ ke liye, dono sides compute karke verify karo ki $(A B)^{T} = B^{T} A^{T}$ .

Recall Solution 2.1

Left side — pehle multiply karo, phir flip karo. Yaad karo matrix product: entry $(i, j)$ = (pehle ki row $i$ ) · (doosre ka column $j$ ). $A B = [1 (4) + 2 (2) 0 (4) + 3 (2) 1 (- 1) + 2 (5) 0 (- 1) + 3 (5)] = [86915] .$ Flip karo: $(A B)^{T} = [89615] .$ Right side — dono ko flip karo, phir reversed order mein multiply karo. $B^{T} = [4 - 1 25], A^{T} = [1203] .$ $B^{T} A^{T} = [4 (1) + 2 (2) - 1 (1) + 5 (2) 4 (0) + 2 (3) - 1 (0) + 5 (3)] = [89615] .$ Dono match karte hain. ✓ Dekho Matrix Operations.

Problem 2.2

$A = [3 - 2 14]$ diya hua hai, $(3 A)^{T}$ compute karo aur confirm karo ki yeh $3 A^{T}$ ke barabar hai.

Recall Solution 2.2

$3 A = [9 - 6 312] ⟹ (3 A)^{T} = [93 - 6 12] .$ Alag se: $A^{T} = [31 - 2 4] ⟹ 3 A^{T} = [93 - 6 12] .$ Barabar. ✓ Scalars flip ke seedha through nikal jaate hain — mirror ko kisi stretch ki parwah nahi hoti.

L3 — Analysis

Problem 3.1

Ek square matrix $M$ , $M^{T} = M$ satisfy karti hai (symmetric). Agar $M = [15 x 3]$ hai, toh $x$ find karo.

Recall Solution 3.1

Symmetric ka matlab hai $[M]_{ij} = [M]_{j i}$ . Position $(1, 2)$ ko position $(2, 1)$ se compare karo: $[M]_{12} = x, [M]_{21} = 5.$ Symmetry ke liye dono equal hone chahiye, isliye $x = 5$ . $M = [1553] .$ Picture: entries $x$ aur $5$ , diagonal ke across mirror images ki tarah baithe hain. Symmetry mirror image ko identical hone ke liye force karti hai.

Problem 3.2

Har square matrix $A$ ko ek symmetric part $S$ aur ek skew-symmetric part $K$ mein split kiya ja sakta hai, is tarah: $S = \frac{1}{2} (A + A^{T}), K = \frac{1}{2} (A - A^{T}), A = S + K .$ $A = [2064]$ ke liye is decomposition ko verify karo, aur check karo ki $S$ symmetric hai aur $K$ skew hai.

Recall Solution 3.2

$A^{T} = [2604] .$ Symmetric part: $S = \frac{1}{2} ([2064] + [2604]) = \frac{1}{2} [4668] = [2334] .$ Skew part: $K = \frac{1}{2} ([2064] - [2604]) = \frac{1}{2} [0 - 6 60] = [0 - 3 30] .$ Symmetry check: $S^{T} = S$ ✓ (mirror image identical). Skew check: $K^{T} = [03 - 3 0] = - K$ ✓ (zero diagonal, off-diagonals negate ho jaate hain). Sum: $S + K = [2064] = A$ ✓

Yeh kyun kaam karta hai: $S^{T} = \frac{1}{2} (A^{T} + (A^{T})^{T}) = \frac{1}{2} (A^{T} + A) = S$ , double-transpose rule $(A^{T})^{T} = A$ use karke. $\frac{1}{2}$ , $A$ ko "mirror-friendly" aur "mirror-flipping" halves ke beech equally share karta hai.

L4 — Synthesis

Problem 4.1

Prove karo ki kisi bhi matrix $A$ ke liye, product $A^{T} A$ hamesha symmetric hota hai.

Recall Solution 4.1

Goal: dikhao ki $(A^{T} A)^{T} = A^{T} A$ . Product-reversal rule $(X Y)^{T} = Y^{T} X^{T}$ apply karo $X = A^{T}$ aur $Y = A$ ke saath: $(A^{T} A)^{T} = A^{T} (A^{T})^{T} .$ Ab double-transpose $(A^{T})^{T} = A$ use karo: $(A^{T} A)^{T} = A^{T} A .$ Kyunki flip same matrix return karta hai, $A^{T} A$ symmetric hai. $■$

Yeh kyun matter karta hai: $A^{T} A$ har jagah appear hota hai — least squares, inner products, aur orthogonal matrices ki definition mein jahan $A^{T} A = I$ hota hai. Iski guaranteed symmetry hi wajah hai ki iske eigenvalues hamesha real hote hain.

Problem 4.2

$A$ invertible hai. Prove karo ki $(A^{- 1})^{T} = (A^{T})^{- 1}$ , aur isse use karo $(A^{T})^{- 1}$ compute karne ke liye, jahan: $A = [1324] .$

Recall Solution 4.2

Proof. Defining identity $A A^{- 1} = I$ se shuru karo (inverse woh hai jo $A$ ko multiply karke identity $I$ deta hai). Dono sides flip karo aur reversal rule use karo: $(A A^{- 1})^{T} = I^{T} ⟹ (A^{- 1})^{T} A^{T} = I .$ (Humne $I^{T} = I$ use kiya: identity symmetric hai.) Yeh kehta hai ki $(A^{- 1})^{T}$ woh matrix hai jo $A^{T}$ ko multiply karke $I$ deti hai — yaani yeh $(A^{T})^{- 1}$ hai. Isliye $(A^{- 1})^{T} = (A^{T})^{- 1}$ . $■$

Computation. Pehle $A^{- 1}$ . $2 \times 2$ matrix $[a c b d]$ ka inverse $\frac{1}{a d - b c} [d - c - b a]$ hota hai. Yahan $a d - b c = 1 (4) - 2 (3) = - 2$ : $A^{- 1} = \frac{1}{- 2} [4 - 3 - 2 1] = [- 2 1.5 1 - 0.5] .$ Theorem se, $(A^{T})^{- 1} = (A^{- 1})^{T}$ — bas flip karo: $(A^{T})^{- 1} = [- 2 1 1.5 - 0.5] .$ Dekho Matrix Inverse aur Determinants ( $- 2$ , $det A$ hai).

L5 — Mastery

Problem 5.1

Dikhao ki agar $A$ skew-symmetric hai ( $A^{T} = - A$ ) aur $n \times n$ hai, toh diagonal mein sab zeros hain, aur degenerate $1 \times 1$ case check karo.

Recall Solution 5.1

Skew ka matlab hai $[A]_{ij} = - [A]_{j i}$ sabhi $i, j$ ke liye. $i = j$ set karo (ek diagonal entry): $[A]_{ii} = - [A]_{ii} ⟹ 2 [A]_{ii} = 0 ⟹ [A]_{ii} = 0.$ Isliye har diagonal entry $0$ hai. ✓

Degenerate $1 \times 1$ case: $A = [a]$ . Iski transpose khud hi hai, $A^{T} = [a]$ . Skew ke liye zaroorat hai $[a] = - [a]$ , jo force karta hai $a = 0$ . Toh sirf $1 \times 1$ skew-symmetric matrix $[0]$ hai — "diagonal zero honi chahiye" ke consistent, kyunki $1 \times 1$ matrix poori tarah diagonal hoti hai.

Problem 5.2

Maano $A$ , $2 \times 3$ hai. $A^{T} A$ aur $A A^{T}$ ke sizes determine karo, decide karo kaun sa (agar koi) invertible ho sakta hai, aur dono compute karo: $A = [100121] .$

Recall Solution 5.2

Sizes. $A$ , $2 \times 3$ hai, isliye $A^{T}$ , $3 \times 2$ hai.

$A^{T} A$ : $(3 \times 2) (2 \times 3) = 3 \times 3$ .
$A A^{T}$ : $(2 \times 3) (3 \times 2) = 2 \times 2$ .

Dono square hain aur (Problem 4.1 ke argument se) symmetric hain.

$A A^{T}$ compute karo (chhota wala):

A A^T = \begin{bmatrix} 1{\cdot}1+0{\cdot}0+2{\cdot}2 & 1{\cdot}0+0{\cdot}1+2{\cdot}1 \\ 0{\cdot}1+1{\cdot}0+1{\cdot}2 & 0{\cdot}0+1{\cdot}1+1{\cdot}1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}.$$ Iski determinant hai $5(2)-2(2) = 6 \ne 0$, isliye $A A^T$ **invertible hai**. **$A^T A$ compute karo:** $$A^T A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 5 \end{bmatrix}.$$ **Kya yeh invertible ho sakta hai?** Nahi. $A$ mein sirf $2$ rows hain, isliye iski rows zyada se zyada $2$-dimensional space span kar sakti hain — $3\times 3$ result ki full rank $3$ nahi ho sakti. Actually row 3 $=2\cdot(\text{row 1}) + 1\cdot(\text{row 2})$ hai, isliye $\det(A^T A) = 0$: **not invertible**. **Sabak:** ek "wide" matrix ($m<n$) ke liye, sirf $A A^T$ (small square) invertible ho sakta hai; bada square $A^T A$ hamesha singular hota hai. Yeh govern karta hai ki tum kaun sa normal-equation form solve kar sakte ho. Dekho [[Determinants]].

Recall Self-audit checklist

Kya tum ab, bina notes ke... Ek non-square matrix ko flip kar sakte ho aur uska naya size bata sakte ho? ::: Haan — rows columns ban jaati hain, $m \times n \to n \times m$ . Bata sakte ho ki $(A B)^{T} = B^{T} A^{T}$ order kyun reverse karta hai? ::: Summed inner index sides swap kar leta hai, jo factors ko swap hone ke liye force karta hai. Prove kar sakte ho ki $A^{T} A$ hamesha symmetric hota hai? ::: $(A^{T} A)^{T} = A^{T} (A^{T})^{T} = A^{T} A$ . Bata sakte ho ki $A^{T} = A^{- 1}$ kab hota hai? ::: Sirf orthogonal matrices ke liye, jahan $A^{T} A = I$ hota hai.