Transpose — definition, properties: Worked examples

2.6.6 · D3 · Maths › Matrices & Determinants — Introduction › Transpose — definition, properties

Yeh page Transpose — definition, properties ki "hands dirty" companion hai. Hum parent note ke rules lete hain aur unhe har tarah ki matrix par lagate hain — rectangular, square, zero, degenerate, symmetric, aur ek exam-style trap — taaki koi bhi shape tumhe kabhi surprise na kare.

Kuch bhi compute karne se pehle, ek reminder us ek rule ki jo sab kuch support karta hai:

Scenario matrix

Yeh un sab cases ka pura landscape hai jo ek transpose problem tumhare samne rakh sakti hai. Har cell ko neeche kam se kam ek worked example se cover kiya gaya hai.

Cell	Case class	Kya tricky banata hai	Example
A	Rectangular $m \times n$ , $m \neq = n$	shape badlti hai $m \times n \to n \times m$	Ex 1
B	Square, general	shape same, entries diagonal ke upar reflect hoti hain	Ex 2
C	Row/column vector (degenerate: $1 \times n$ )	ek "flat" matrix ek "tall" matrix ban jaati hai	Ex 3
D	Product with reversal $(A B)^{T}$	order ulta ho jaata hai, non-commutative	Ex 4
E	Chained product $(A B C)^{T}$	reversal cascade hoti hai	Ex 5
F	Symmetric / skew split (sign cases)	$A^{T} = A$ vs $A^{T} = - A$ , zero diagonal	Ex 6
G	Zero matrix & identity (degenerate/limit)	transpose ke fixed points	Ex 7
H	Inverse + transpose saath mein	$(A^{- 1})^{T} = (A^{T})^{- 1}$ , real numbers	Ex 8
I	Word problem (real-world table flip)	rows-vs-columns sahi se padhna	Ex 9
J	Exam twist ( $x^{T} A x$ / trace / trap)	scalar output, dimension trap	Ex 10

Prerequisites jinpar hum depend karte hain: Matrix Operations, Symmetric Matrices, Orthogonal Matrices, Matrix Inverse, Inner Product Spaces.

Cell A — Rectangular, shape badlti hai

Worked example Ex 1 — ek

$2 \times 3$ matrix $3 \times 2$ ban jaati hai $A = [2503 - 1 4]$ $A^{T}$ aur uski size nikalein.

Forecast: Aage padhne se pehle — kya $A^{T}$ , $A$ se zyada wide hogi ya zyada tall? Apna guess likhein.

Step 1. $A$ mein $2$ rows hain aur $3$ columns, isliye yeh $2 \times 3$ hai. Yeh step kyun? Rule $[A^{T}]_{ij} = [A]_{j i}$ donon size numbers ko swap karta hai, isliye $A^{T}$ $3 \times 2$ honi chahiye — wide se zyada tall. (Guess check: taller.)

Step 2. $A$ ki row $1$ , yaani $[2, 0, - 1]$ , lo aur use $A^{T}$ ki column 1 ke roop mein khada kar do. Yeh step kyun? "Row $k$ of $A$ becomes column $k$ of $A^{T}$ " yeh sirf label swap hai jo puri ek line mein padha gaya hai.

Step 3. $A$ ki row $2$ , yaani $[5, 3, 4]$ , lo aur use $A^{T}$ ki column 2 ke roop mein khada kar do. Yeh step kyun? Doosri row ke liye same rule. $A^{T} = 20 - 1 534_{3 \times 2}$

Verify: Ek entry blindly choose karo. $[A^{T}]_{31}$ ko $[A]_{13}$ ke barabar hona chahiye. Hum padhte hain $[A^{T}]_{31} = - 1$ aur $[A]_{13} = - 1$ . ✓ Size $3 \times 2$ hai. ✓

$- 1$ mein negative sign bilkul sahi raha — transpose kisi value ko kabhi nahi badalta, sirf uska address badalta hai.

Cell B — Square, general

Cell C — Degenerate: ek vector

Worked example Ex 3 — ek row vector column vector mein transpose hoti hai

$v = [3 - 1 42]_{1 \times 4}$ $v^{T}$ nikalein.

Forecast: Ek single row "flattest" matrix hai. Uske transpose ki shape kaisi honi chahiye?

Step 1. $v$ $1 \times 4$ hai, isliye $v^{T}$ $4 \times 1$ hai — ek tall column. Yeh step kyun? $1 \times 4$ ke size numbers swap karne par $4 \times 1$ milta hai. Yeh Cell A ka extreme case hai jahan ek dimension $1$ hai.

Step 2. Single row single column ban jaati hai, entry order preserved rehta hai. $v^{T} = 3 - 1 42$

Verify: Yahi exactly woh tarika hai jaise inner products likhe jaate hain: $v v^{T}$ ek $1 \times 1$ number hoga (khud ke saath ek dot product), jabki $v^{T} v$ ek $4 \times 4$ matrix hai. Transpose hi woh cheez hai jo hum ek vector ko "apne aap ke khilaf" do alag tareekon se multiply kar sakte hain. Sanity: $[v^{T}]_{31} = [v]_{13} = 4$ . ✓

Cell D — Product with reversal

Worked example Ex 4 —

$(A B)^{T} = B^{T} A^{T}$ $A = [10 2 - 1], B = [3214]$ $(A B)^{T}$ ko dono tareekon se compute karo aur confirm karo ki order sach mein reverse hona chahiye.

Forecast: Guess karo ki $A^{T} B^{T}$ (galat order) yahan same answer dega ya nahi.

Step 1. $A B$ multiply karo (row-times-column, Matrix Operations se): $A B = [1 \cdot 3 + 2 \cdot 2 0 \cdot 3 + (- 1) \cdot 2 1 \cdot 1 + 2 \cdot 4 0 \cdot 1 + (- 1) \cdot 4] = [7 - 2 9 - 4]$ Yeh step kyun? Pehle product chahiye taaki hum use ek block ke roop mein transpose kar sakein.

Step 2. Us block ko transpose karo (off-diagonals swap karo): $(A B)^{T} = [79 - 2 - 4]$ Yeh step kyun? $A B$ par rule ka direct application.

Step 3. Ab reversed product $B^{T} A^{T}$ : $B^{T} = [3124], A^{T} = [12 0 - 1]$ $B^{T} A^{T} = [3 \cdot 1 + 2 \cdot 2 1 \cdot 1 + 4 \cdot 2 3 \cdot 0 + 2 \cdot (- 1) 1 \cdot 0 + 4 \cdot (- 1)] = [79 - 2 - 4]$ Yeh step kyun? Reversal rule predict karta hai ki yeh $(A B)^{T}$ ke barabar hoga.

Step 4. Contrast ke liye, galat order $A^{T} B^{T}$ : $A^{T} B^{T} = [12 0 - 1] [3124] = [3520]$ Yeh step kyun? Yeh dekhne ke liye ki order rakhne par genuinely alag matrix milti hai.

Verify: $(A B)^{T} = B^{T} A^{T} = [79 - 2 - 4]$ ✓, lekin $A^{T} B^{T} = [3520] \neq =$ woh. Reversal optional nahi hai. ✓

Cell E — Chained reversal

Worked example Ex 5 —

$(A B C)^{T} = C^{T} B^{T} A^{T}$ Diagonal matrices ke saath (taaki arithmetic clean rahe): $A = [2001], B = [1011], C = [1301]$ Dikhao ki $(A B C)^{T} = C^{T} B^{T} A^{T}$ .

Forecast: $A, B, C$ right side par kis order mein aayenge? Compute karne se pehle likhein.

Step 1. $A B = [2001] [1011] = [2021]$ . Yeh step kyun? Product left-to-right build karo.

Step 2. $A B C = [2021] [1301] = [2 + 6 0 + 3 0 + 2 0 + 1] = [8321]$ . Yeh step kyun? Triple product finish karo.

Step 3. Ise transpose karo: $(A B C)^{T} = [8231]$ . Yeh step kyun? Pure block par rule apply karo.

Step 4. Reversal do baar apply karo: $(A B C)^{T} = (B C)^{T} A^{T} = C^{T} B^{T} A^{T}$ . Compute karo $C^{T} = [1031], B^{T} = [1101], A^{T} = [2001]$ . $C^{T} B^{T} = [1031] [1101] = [4131]$ , phir $C^{T} B^{T} A^{T} = [4131] [2001] = [8231]$ . Yeh step kyun? Har transpose sabse baahri factor ko peel karta hai aur use aage land karta hai — order puri tarah reverse ho jaata hai.

Verify: $[8231] = [8231]$ ✓

Cell F — Sign cases: symmetric vs skew

Worked example Ex 6 — ek matrix ko symmetric + skew parts mein split karo

$M = [4172]$ $M = S + K$ likho jahan $S^{T} = S$ (symmetric) aur $K^{T} = - K$ (skew-symmetric) ho.

Forecast: Skew part $K$ ki diagonal kaisi honi chahiye? (Parent ki skew definition dobara padhein.)

Step 1. Standard split use karo: $S = \frac{1}{2} (M + M^{T})$ , $K = \frac{1}{2} (M - M^{T})$ . Yeh step kyun? $(M + M^{T})^{T} = M^{T} + M = M + M^{T}$ , isliye $S$ symmetric hai; $(M - M^{T})^{T} = M^{T} - M = - (M - M^{T})$ , isliye $K$ skew hai. Isme sum rule aur $(A^{T})^{T} = A$ use hota hai.

Step 2. $M^{T} = [4712]$ , isliye $S = \frac{1}{2} [8884] = [4442], K = \frac{1}{2} [0 - 6 60] = [0 - 3 30]$ Yeh step kyun? $M$ aur $M^{T}$ ko formulas mein plug karo.

Step 3. $K$ ki zero diagonal notice karo: kyunki $k_{ii} = - k_{ii} \Rightarrow k_{ii} = 0$ . Yeh degenerate sign case hai jahan ek number apne khud ke negative ke barabar hona chahiye. Yeh step kyun? Yeh confirm karta hai ki $K$ genuinely skew hai, sirf anti-off-diagonal nahi.

Verify: $S + K = [4172] = M$ ✓. $S^{T} = S$ ✓ (off-diagonals dono $4$ ). $K^{T} = [03 - 3 0] = - K$ ✓. Har square matrix is tarah split hoti hai — Symmetric Matrices dekho.

Cell G — Degenerate fixed points: zero aur identity

Worked example Ex 7 — matrices jo apni khud ki transpose hain

$2 \times 3$ zero matrix ke liye $O^{T}$ aur $3 \times 3$ identity ke liye $I^{T}$ nikalein.

Forecast: Inme se ek apni shape maintain karta hai, ek badalta hai. Kaun sa kaun sa hai?

Step 1. Zero matrix $O = [000000]$ $2 \times 3$ hai, isliye $O^{T}$ $3 \times 3$ hai… nahi — yeh $3 \times 2$ hai, aur har entry abhi bhi $0$ hai. Yeh step kyun? Values sab $0$ hain isliye koi value nahi badlti, lekin shape flip hoti hai $2 \times 3 \to 3 \times 2$ . Ek rectangular zero matrix apni transpose ke barabar nahi hai (alag shapes).

Step 2. Identity $I = 100010001$ . Har $1$ diagonal par hai ( $i = j$ ), aur diagonal entries transpose se fixed rahti hain. Off-diagonals sab $0$ hain, jo $0$ hi rahti hain. Yeh step kyun? $I$ symmetric hai: $I^{T} = I$ . Yahi fact Property 5 (transpose of inverse) ko kaam karata hai, kyunki $I^{T} = I$ .

Verify: $O^{T}$ $3 \times 2$ zero matrix hai (shape badli, phir bhi sab zeros) ✓. $I^{T} = I$ exactly ✓. Limit intuition: ek square zero matrix apni khud ki transpose hai; ek rectangular wali nahi hai — shape hi equality decide karta hai.

Cell H — Inverse aur transpose saath mein (real numbers)

Worked example Ex 8 — verify karo

$(A^{- 1})^{T} = (A^{T})^{- 1}$ $A = [2111]$ Dono sides numerically compute karo.

Forecast: Tumhara kya expectation hai — messy fractions ya clean integers? ( $det A$ clue hai.)

Step 1. $det A = 2 \cdot 1 - 1 \cdot 1 = 1$ (Determinants dekho). Kyunki yeh $1$ hai, inverse clean hai: $A^{- 1} = \frac{1}{1} [1 - 1 - 1 2] = [1 - 1 - 1 2]$ Yeh step kyun? $2 \times 2$ ke liye Matrix Inverse: diagonal swap karo, off-diagonal negate karo, $det$ se divide karo.

Step 2. Ise transpose karo: $(A^{- 1})^{T} = [1 - 1 - 1 2]$ (yeh symmetric nikla). Yeh step kyun? Identity ka left-hand side.

Step 3. Ab doosri side. $A^{T} = [2111] = A$ (symmetric!), isliye $(A^{T})^{- 1} = A^{- 1} = [1 - 1 - 1 2]$ . Yeh step kyun? Identity ka right-hand side.

Verify: Dono $[1 - 1 - 1 2]$ ke barabar hain ✓. Extra check: $A^{T} (A^{- 1})^{T} = [2111] [1 - 1 - 1 2] = [1001] = I$ ✓ — confirm karta hai ki $(A^{- 1})^{T}$ sach mein $A^{T}$ ko invert karta hai.

Cell I — Real-world table flip

Worked example Ex 9 — sales spreadsheet

Ek shop units sold record karti hai. Rows stores hain (North, South), columns months hain (Jan, Feb, Mar): $S = [1281511914] \leftarrow North \leftarrow South$ Head office ek aisi table chahta hai jahan rows months hon aur columns stores hon. Use banao aur "February, South" padhke batao.

Forecast: Kaun sa single operation poori table reorganise karta hai? (Tum already jaante ho.)

Step 1. "Rows↔columns" precisely transpose hai. $S^{T}$ compute karo: $S^{T} = 1215981114 \leftarrow Jan \leftarrow Feb \leftarrow Mar$ Yeh step kyun? Har store-row ek store-column ban jaati hai; har month-column ek month-row ban jaata hai.

Step 2. Nayi table mein "February, South" row Feb, column South hai $= [S^{T}]_{22} = 11$ . Yeh step kyun? Original mein wahi number row South ( $= 2$ ), column Feb ( $= 2$ ) par tha: $[S]_{22} = 11$ . Label swap $[S^{T}]_{22} = [S]_{22}$ — identical kyunki dono indices $2$ hain.

Verify: Original mein North-March $[S]_{13} = 9$ hai; flipped table mein yeh $[S^{T}]_{31} = 9$ hai ✓. Data kabhi nahi badla, sirf woh tarika badla jisme hum use address karte hain — yahi data work mein transpose ka pura point hai.

Cell J — Exam twist: scalar output & ek dimension trap

Classic exam object quadratic form $x^{T} A x$ hai, jo ek vector aur ek matrix ko ek single number mein convert karta hai. Yeh Eigenvalues and Eigenvectors aur geometry mein aata hai.

Figure — Transpose — definition, properties

Worked example Ex 10 —

$x^{T} A x$ evaluate karo aur shape trap se bachao $x = [12], A = [3012]$ Scalar $x^{T} A x$ compute karo. Phir explain karo ki $A x^{T}$ illegal kyun hai.

Forecast: Final answer ki size kya hai — ek matrix ya ek single number? Step 1 se pehle guess karo.

Step 1. Sizes: $x^{T}$ $1 \times 2$ hai, $A$ $2 \times 2$ hai, $x$ $2 \times 1$ hai. Chain: $(1 \times 2) (2 \times 2) (2 \times 1) \to 1 \times 1$ . Ek single number. Yeh step kyun? Multiplication ke liye inner sizes match karni chahiye; yahan $2$ – $2$ aur $2$ – $2$ match karte hain, aur surviving outer sizes $1$ aur $1$ hain. Figure dekho: left par row aur right par column ek cell mein pinch ho jaate hain.

Step 2. Pehle inner product compute karo: $A x = [3012] [12] = [3 + 2 0 + 4] = [54]$ . Yeh step kyun? Pehle $A x$ group karne se arithmetic simple rehta hai.

Step 3. Phir $x^{T} (A x) = [12] [54] = 1 \cdot 5 + 2 \cdot 4 = 13$ . Yeh step kyun? Ek row times ek column exactly dot product hai — yeh ek number $13$ mein collapse ho jaata hai.

Step 4. Trap: $A x^{T}$ try karta hai $(2 \times 2) (1 \times 2)$ . Inner sizes $2$ aur $1$ match nahi karte, isliye product exist nahi karta. Yeh step kyun? $x$ ko transpose karne se ek legal column ek illegal-position row ban gayi. Order aur orientation dono matter karte hain.

Verify: $x^{T} A x = 13$ (ek scalar). Cross-check entry formula se $\sum_{i, j} x_{i} A_{ij} x_{j} = 1 \cdot 3 \cdot 1 + 1 \cdot 1 \cdot 2 + 2 \cdot 0 \cdot 1 + 2 \cdot 2 \cdot 2 = 3 + 2 + 0 + 8 = 13$ ✓.

Recall check

Recall Ek hi saansh mein har cell

Rectangular shape flip karti hai ::: $m \times n \to n \times m$ (Ex 1, 3) Transpose ke under diagonal entries ::: fixed rehti hain, kyunki $i = j$ (Ex 2, 7) $(A B)^{T}$ barabar hai ::: $B^{T} A^{T}$ — order reverse hota hai (Ex 4) $(A B C)^{T}$ barabar hai ::: $C^{T} B^{T} A^{T}$ (Ex 5) Skew-symmetric diagonal zaroor honi chahiye ::: sab zeros (Ex 6) $(A^{- 1})^{T}$ barabar hai ::: $(A^{T})^{- 1}$ (Ex 8) $x^{T} A x$ ki shape ::: ek single number $1 \times 1$ (Ex 10)