Orthogonal matrices — properties, det = ±1: Visual walkthrough

4.5.37 · D2 · Maths › Linear Algebra (Full) › Orthogonal matrices — properties, det = ±1

Step 1 — Ek matrix picture ke saath kya karta hai

East arrow ko $e_{1}$ aur North arrow ko $e_{2}$ bolte hain. Yeh standard basis vectors hain: $e_{1} = (10), e_{2} = (01) .$

Upar ka number hai "kitna East", neeche ka number "kitna North". Toh $e_{1}$ hai ek step East, zero North.

Ek $2 \times 2$ matrix $Q$ ke do columns hote hain. Column 1 hai jahan $e_{1}$ land karta hai; column 2 hai jahan $e_{2}$ land karta hai. Isse zyada mysterious kuch nahi.

Figure — Orthogonal matrices — properties, det = ±1

Step 2 — Transpose aur identity, pehle drawn

Isse pehle ki hum orthogonality rule likh sakein, humein notation ke do pieces chahiye. Hum unhe tab tak use nahi karenge jab tak aapne unki pictures nahi dekhi.

Woh aakhri sentence hi poori wajah hai ki transpose hamare liye useful hai: jab hum $A^{T}$ ko $A$ se multiply karte hain, hum literally $A$ ke har column ko (jo ab $A^{T}$ ki row ke roop mein flat pada hai) $A$ ke columns par slide kar rahe hain — aur "row-dotted-into-column" exactly ek dot product hai. Toh $A^{T} A$ saare column-vs-column dot products ki table hai. Step 3 ke liye yeh baat yaad rakhna.

Step 3 — Woh rule jo $Q$ ko orthogonal banata hai

Columns ko $q_{1}$ (jahan $e_{1}$ land karta hai) aur $q_{2}$ (jahan $e_{2}$ land karta hai) naam dete hain. Step 2 ke dot product se symbols mein rule hai: $q_{1}^{T} q_{1} = 1, q_{2}^{T} q_{2} = 1, q_{1}^{T} q_{2} = 0.$

$q_{1}^{T} q_{1}$ hai $q_{1}$ ka khud ke saath dot product (uski flat row-copy times uska column). Matching entries multiply karo, add karo: yeh $q_{1}$ ki squared length ke barabar hai. Isse $1$ set karna matlab length $1$ hai.
$q_{1}^{T} q_{2}$ hai dono alag columns ka dot product. Yeh $∥ q_{1} ∥ ∥ q_{2} ∥ cos θ$ ke barabar hai, toh yeh $0$ hai exactly tab jab unke beech ka angle $θ$ $9 0^{\circ}$ ho.

Step 4 — Orthonormal columns tile ko unit square banane par majboor karte hain

Tile dekho: yeh kabhi stretch nahi hota (sides length $1$ rehte hain) aur kabhi lean nahi karta (corner $9 0^{\circ}$ rehta hai). Yeh sirf turn ya turn-and-flip kar sakta hai.

Step 5 — KYU $det = a d - b c$ signed area hai

Figure dekho. Parallelogram ko $(a + c) \times (b + d)$ rectangle mein band karo. Uska area $(a + c) (b + d)$ hai. Ab woh do chote rectangles aur char triangles subtract karo jo parallelogram ke around bahar nikalte hain. Jab sab settle ho jaata hai, saare cross-terms cancel ho jaate hain aur aap ke paas bacha rehta hai: $area = (a + c) (b + d) - 2 b c - ab - c d = a d - b c .$

$a d$ hai "co-rotating" contribution ( $q_{1}$ ka East-part times $q_{2}$ ka North-part).
$- b c$ "counter" contribution subtract karta hai ( $q_{1}$ ka North-part times $q_{2}$ ka East-part).
Unka difference $a d - b c$ positive hai jab $q_{2}$ , $q_{1}$ se counter-clockwise baitha ho, negative jab clockwise — exactly signed area.

Step 6 — Do determinant facts, dekhe gaye sirf bataye nahi

Fact 1 — areas multiply karte hain: $det (A B) = det A \cdot det B$ . Machine $B$ apply karo: yeh har area ko $det B$ se scale karta hai. Phir machine $A$ apply karo: yeh us pehle-se-scale hue area ko phir $det A$ se scale karta hai. Unit square ko dono mein se feed karna dono factors ko multiply karta hai — tile ka final area $det A \cdot det B$ hai. (Left panel dekho: ek $\times 2$ stretch ke baad ek $\times 3$ stretch ek $\times 6$ tile deta hai.)

Fact 2 — transpose area rakhta hai: $det (A^{T}) = det A$ . Transpose karna (Step 2 ka diagonal flip) off-diagonal entries $b \leftrightarrow c$ swap karta hai. Lekin determinant $a d - b c$ sirf product $b c$ par depend karta hai, aur $b c = c b$ — $b$ aur $c$ swap karna product ko untouched chhod deta hai. Toh signed area identical hai. (Right panel dekho: flipped tile ka area same hai.)

Step 7 — Algebra: $Q^{T} Q = I$ se sign nikalo

Orthogonality rule ke dono sides ka determinant lete hain: $det (Q^{T} Q) = det I .$

Left side, Fact 1 se (areas multiply karte hain): $det (Q^{T}) det (Q)$ .
Fact 2 se (transpose area rakhta hai): $det (Q^{T}) = det Q$ , toh left side $(det Q)^{2}$ hai.
Right side: identity tile ko untouched chhod deta hai, area $1$ , toh $det I = 1$ (Step 2).

Sab milake: $(det Q)^{2} = 1 ⟹ det Q = \pm 1.$

Ek real number jiska square $1$ hai woh sirf $+ 1$ ya $- 1$ ho sakta hai. Ho gaya.

Step 8 — Case $+ 1$ : rotation (orientation kept)

Clean example hai rotation matrix $Q_{θ} = (cos θ sin θ - sin θ cos θ), det Q_{θ} = cos^{2} θ + sin^{2} θ = 1.$

Column 1 hai $e_{1}$ jo angle $θ$ se ghuma; column 2 hai $e_{2}$ jo same $θ$ se ghuma — poora square ek piece ki tarah spin karta hai.

Step 9 — Case $- 1$ : har reflection, sirf ek nahi

Sabse simple example hai $x$ -axis ke paas flip: $R_{0} = (10 0 - 1), det R_{0} = (1) (- 1) - 0 = - 1.$

Column 1 $e_{1}$ hi rehta hai (mirror line par points move nahi karte); column 2 hai $e_{2}$ jo apni mirror image $- e_{2}$ par bheja gaya.

General reflection. Mirror line ko tilt karo taaki woh East axis ke saath angle $ϕ$ banaye. Us us line ke paas reflect karna hai orthogonal matrix $R_{ϕ} = (cos 2 ϕ sin 2 ϕ sin 2 ϕ - cos 2 ϕ), det R_{ϕ} = - cos^{2} 2 ϕ - sin^{2} 2 ϕ = - 1.$

Mirror angle $ϕ$ chahe jo bhi ho, determinant hamesha $- 1$ hai: saari reflections orientation flip karti hain, koi stretch nahi karta. Figure ka left panel teen alag mirror lines dikhata hai, har ek apne tile ke saath flipped.

Reflection = rotation × flip. Koi bhi $det = - 1$ orthogonal matrix ek ek fixed flip ke baad ek rotation ki tarah likha ja sakta hai: $R_{ϕ} = Q_{2 ϕ} R_{0}$ . Toh " $- 1$ world" bas " $+ 1$ world" (rotations) hai jisme ek mirror stapled hai — right panel yeh composition dikhata hai. Yahi $O (2)$ ka poora doosra half hai.

Step 10 — Sign kabhi $0$ (ya $\pm 1$ ke alawa kuch bhi) kyun nahi ho sakta

Area $0$ matlab hoga ki do columns ek hi line par hain (ek doosre ka multiple hai). Lekin orthonormal columns perpendicular hain, toh woh kabhi parallel nahi ho sakte — koi collapse nahi. Isliye $det \neq = 0$ , aur actually $Q$ hamesha invertible hai.
Koi bhi area $\neq = 1$ matlab hoga ki ek column $1$ se longer ya shorter hai, jo "unit length" ke khilaf hai. Toh $∣ det ∣ = 1$ exactly.
Step 7 ke $(det Q)^{2} = 1$ ke saath combine karke, sirf survivors hain $+ 1$ (Step 8) aur $- 1$ (Step 9). Koi teesra option nahi hai.

Recall Kyun

$(det Q)^{2} = 1$ sirf do values chhod ta hai Real numbers mein, $x^{2} = 1$ ke solutions kya hain? ::: exactly $x = + 1$ aur $x = - 1$ — koi doosra real number $1$ tak square nahi karta. Kya $det Q$ ek imaginary number jaise $i$ ho sakta hai? ::: Nahi — $Q$ ek real matrix hai, toh $det Q$ ek real number hai; $i$ exclude hai.

Ek-picture summary

Ek canvas par poora argument: orthonormal columns ⟶ unit square ki image ek unit square hai (area magnitude $1$ ) ⟶ $Q^{T} Q = I$ ka $det$ lene par $(det Q)^{2} = 1$ milta hai ⟶ spin sign rakhta hai $(+ 1)$ , flip isse reverse karta hai $(- 1)$ .

Recall Feynman retelling — walkthrough plain words mein

Ek graph paper ki sheet ka picture karo jisme do arrows bane hain: ek East point karta hai, ek North, har ek ek square lamba. Ek matrix in do arrows ko uthata hai aur kahi naya drop karta hai — aur poora grid follow karta hai.

Chota " $T$ " (transpose) bas sheet hai jo apne diagonal par flip hui hai; columns ko rows ki tarah flat laana aur unhe columns par slide karna ek saath har dot product spell karta hai. "Do-nothing" matrix $I$ ke diagonal par ones hain aur square ko exactly wahi chhod deta hai jahan tha — isliye uska area factor $1$ hai.

Ek orthogonal matrix ka ek promise hai: do dropped arrows abhi bhi har ek exactly ek square lamba hai, aur woh abhi bhi ek perfect right angle par milte hain. Toh woh jo chota square banate hain woh abhi bhi ek perfect unit square hai — machine sirf square ko spin kar sakti hai ya flip kar sakti hai, kabhi squash ya stretch nahi.

Determinant bas "us square ka signed area" hai, aur parallelogram ko uske bounding rectangle se kaat ke dikhata hai ki woh area $a d - b c$ hai. Area hamesha $1$ hai kyunki yeh ek unit square hai, toh number sirf $+ 1$ ya $- 1$ ho sakta hai. Agar square sirf ghuma, humne same handedness rakhi: $+ 1$ . Agar humein isse pancake ki tarah flip karna pada match karne ke liye — kisi bhi mirror line ke paas — handedness reverse hui: $- 1$ .

Algebra wahi idea hai jo suit pehne hua hai: $Q^{T} Q = I$ ke dono sides ka $det$ lo, "areas multiply karte hain" aur "diagonal ke paas flip karna area rakhta hai" use karo, aur aap $(det Q)^{2} = 1$ par pohunch jaate ho, toh $det Q = \pm 1$ . Same truth, do languages — picture aur equation agree karte hain.

Yeh bhi dekho: Rotations and reflections in $\mathbb{R}^2$ and $\mathbb{R}^3$, Determinants — properties, Orthonormal bases & Gram–Schmidt, Eigenvalues and eigenvectors.