Worked examples — Orthogonal matrices — properties, det = ±1

4.5.37 · D3 · Maths › Linear Algebra (Full) › Orthogonal matrices — properties, det = ±1

Shuru karne se pehle, teen reminders simple words mein, taaki koi bhi symbol unexplained na lage.

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Scenario matrix

Har orthogonal-matrix problem neeche diye cells mein se ek (ya mix) hoti hai. Jo examples aage aate hain unhe us cell ke saath tag kiya gaya hai, taaki tum dekh sako ki poora territory cover ho raha hai.

Cell	Case class	Isme kya tricky hai	Covered by
A	Pure rotation ($2\times 2$), $\det =+1$	matrix se angle padhna	Ex 1
B	Pure reflection ($2\times 2$), $\det =-1$	flip ki axis, self-inverse	Ex 2
C	Impostor: $\det =\pm 1$ hai par not orthogonal	necessary vs sufficient	Ex 3
D	Degenerate / zero-angle & identity	limiting case $\theta \to 0$	Ex 4
E	Ek single vector se banao (Gram–Schmidt feel)	normalising, $\perp$ partner chunna, sign choice	Ex 5
F	$3\times 3$ rotation about an axis, $\det =+1$	ek fixed eigenvector, block structure	Ex 6
G	$3\times 3$ reflection, $\det =-1$	3D mein orientation flip	Ex 7
H	Word problem (robot / camera)	physical rigid motion ko $Q$ mein translate karna	Ex 8
I	Exam twist: product & "det guess karo"	closure, product ka $\det$	Ex 9
J	Unit circle par eigenvalues	rotation ke complex eigenvalues	Ex 10

Neeche ki figure poori table ko ek picture mein summarize karti hai: left panel har "$+1$" cell ka archetype hai (A, D, F, H, I-rotation — ek spin jo orientation rakhta hai), aur right panel har "$-1$" cell ka archetype hai (B, G, I-reflection — ek flip ek mirror line ke paas). Jab bhi koi example karo, usse mentally is picture ke sahi side par rakh lo.

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Example 1 — Cell A: pure rotation, angle padhna

Steps.

1. Rotation template se compare karo $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. Match karte hain: $\cos \theta =\frac{1}{2}$, $\sin \theta =\frac{\sqrt{3}}{2}$. Ye step kyun? Parent ki rotation matrix ka ek fixed shape hota hai; agar $Q$ fit kare, toh angle directly pad sakte ho.
2. $\theta$ solve karo: $\cos \theta =\frac{1}{2}$ aur $\sin \theta =\frac{\sqrt{3}}{2}$ dono $\theta =60^{\circ }=\frac{\pi }{3}$ par satisfy hote hain. Ye step kyun? Hume ek aisa angle chahiye jo dono equations satisfy kare — ek equation akele ambiguous hoti hai.
3. Orthonormal columns confirm karo. Column 1 hai $\frac{1}{2}(1,\sqrt{3})$, length${}^2=\frac{1}{4}(1+3)=1$ ✓. Column 2 hai $\frac{1}{2}(-\sqrt{3},1)$, length${}^2=\frac{1}{4}(3+1)=1$ ✓. Dot product $=\frac{1}{4}(-\sqrt{3}+\sqrt{3})=0$ ✓. Ye step kyun? Ye $Q^{\mathsf{T}}Q=I$ ka test hi hai, column-by-column kiya gaya.
4. Determinant: $\det Q=\frac{1}{4}(1\cdot 1-(-\sqrt{3})(\sqrt{3}))=\frac{1}{4}(1+3)=1$. Ye step kyun? $\det =+1$ certify karta hai ki ye ek rotation hai (spin), consistent with orientation preserved.

Worked example Verify

${\cos }^{2}60^{\circ }+{\sin }^{2}60^{\circ }=\frac{1}{4}+\frac{3}{4}=1$: unit columns ✓. $Q$ ko $e_1=(1,0)$ par apply karo: milta hai $\frac{1}{2}(1,\sqrt{3})$, ek arrow jo $x$-axis se exactly $60^{\circ }$ upar hai — claimed angle se match karta hai.

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Example 2 — Cell B: pure reflection, mirror line dhundho

Steps.

1. Orthonormal columns. $c_1=\frac{1}{5}(3,4)$: length${}^2=\frac{9+16}{25}=1$ ✓. $c_2=\frac{1}{5}(4,-3)$: length${}^2=\frac{16+9}{25}=1$ ✓. Dot $=\frac{1}{25}(12-12)=0$ ✓. Ye step kyun? Verify karta hai $R^{\mathsf{T}}R=I$, yani ye orthogonal hai.
2. Determinant: $\det R=\left(\frac{3}{5}\right)(-\frac{3}{5})-\left(\frac{4}{5}\right)\left(\frac{4}{5}\right)=-\frac{9}{25}-\frac{16}{25}=-1$. Ye step kyun? $\det =-1$ matlab ye ek flip hai, spin nahi.
3. Fixed direction dhundho (mirror line). Ek reflection un vectors ko fix karta hai jo mirror par hote hain: $Rv=v$ solve karo, yani $(R-I)v=0$. Row 1: $(\frac{3}{5}-1)x+\frac{4}{5}y=0\Rightarrow -\frac{2}{5}x+\frac{4}{5}y=0\Rightarrow y=\frac{1}{2}x$. Toh mirror hai line $y=\frac{1}{2}x$, direction $(2,1)$. Ye step kyun? Ek reflection ka eigenvalue mirror ke saath $+1$ hota hai aur uske perpendicular $-1$ — parent ka eigenvalue fact.

Worked example Verify

$v=(2,1)$ plug karo: $Rv=\frac{1}{5}(3\cdot 2+4\cdot 1,4\cdot 2-3\cdot 1)=\frac{1}{5}(10,5)=(2,1)=v$ ✓ (fixed). $w=(1,-2)$ plug karo (mirror ke perpendicular): $Rw=\frac{1}{5}(3-8,4+6)=\frac{1}{5}(-5,10)=(-1,2)=-w$ ✓ (flipped).

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Example 3 — Cell C: impostor ($\det =1$ par orthogonal nahi)

Steps.

1. Sirf $\det$ par trust mat karo. $\det =\pm 1$ necessary but not sufficient hai (parent ka mistake box). Asli test chalao. Ye step kyun? Orientation-preserving maps mein shears bhi hote hain, jo stretch karte hain — hume lengths check karni hongi.
2. Columns test karo. $c_1=(2,1)$, length${}^2=4+1=5\ne 1$. Ye step kyun? Orthogonal columns unit length ki honi chahiye; $5\ne 1$ already fail kar deta hai.
3. $A^{\mathsf{T}}A$ compute karo sure hone ke liye. $A^{\mathsf{T}}A=\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}5& 8\\ 8& 13\end{array}\right)\ne I$. Ye step kyun? Definitive certificate: $A^{\mathsf{T}}A\ne I$ ⇒ not orthogonal.

Worked example Verify

$\Vert A(1,0)\Vert =\Vert (2,1)\Vert =\sqrt{5}\ne 1$, isliye $A$ lengths change karta hai — ek isometry kabhi nahi karta. Not orthogonal ✓.

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Example 4 — Cell D: degenerate / zero-angle limit

Steps.

1. $\theta =0$ plug karo: $R(0)=\left(\begin{array}{cc}\cos 0& -\sin 0\\ \sin 0& \cos 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$. Ye step kyun? Identity "kuch na karne wali" rotation hai — limiting/degenerate case.
2. Check karo: $I^{\mathsf{T}}I=I$ ✓, $\det I=1$. Ye orthogonal hai, $SO(2)$ mein ($\det =+1$ rotations). Ye step kyun? Confirm karta hai ki identity orthogonal group ke andar hai (ye group ka neutral element hai).
3. $\theta =180^{\circ }$ plug karo: $R(\pi )=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-I$. $\det (-I)=(-1)(-1)=1$. Ye step kyun? Ye ek subtle case hai: $-I$ dono axes flip karta hai, phir bhi $\det =+1$ — ye ek rotation hai ($180^{\circ }$ ka), reflection nahi, kyunki dono axes flip karna do flips hain = ek spin.

Worked example Verify

$\det (-I_2)=1$ (even dimension), $\det (-I_3)=-1$ (odd dimension). $n$ ki parity decide karti hai $\det (-I_n)$ ka sign.

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Example 5 — Cell E: ek single vector se banao

Steps.

1. Normalize karo. $\Vert (1,1)\Vert =\sqrt{2}$, isliye $u_1=\frac{1}{\sqrt{2}}(1,1)$. Ye step kyun? $Q$ ke columns unit vectors hone chahiye (Gram–Schmidt step 1).
2. Perpendicular unit partner chunna. $u_1$ ko $+90^{\circ }$ rotate karne par $u_2=\frac{1}{\sqrt{2}}(-1,1)$ milta hai; $-90^{\circ }$ par $u_2^{\prime }=\frac{1}{\sqrt{2}}(1,-1)$ milta hai. Ye step kyun? Plane mein exactly do unit vectors $u_1$ ke perpendicular hote hain (opposite directions) — yehi sign freedom hai.
3. Assemble karo aur rotation pick karo. $Q_{+}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$, $\det Q_{+}=\frac{1}{2}(1+1)=1$ → rotation. $Q_{-}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$, $\det Q_{-}=\frac{1}{2}(-1-1)=-1$ → reflection. Ye step kyun? Partner ka sign control karta hai ki orientation rakhi jaaye ($+1$) ya flip ho ($-1$).

Worked example Verify

$Q_{+}^{\mathsf{T}}{Q}_{+}=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)=I$ ✓. $\det Q_{+}=1$, $\det Q_{-}=-1$ ✓.

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Example 6 — Cell F: ek $3\times 3$ rotation about an axis

Steps.

1. $\theta =90^{\circ }$ dalo: $\cos 90^{\circ }=0,\sin 90^{\circ }=1$, isliye $Q=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right)$. Ye step kyun? Concrete numbers checks ko unambiguous banate hain.
2. Orthonormal columns: $c_1=(1,0,0)=e_1$, $c_2=(0,0,1)=e_3$, $c_3=(0,-1,0)=-e_2$: har ek unit length, pairwise dot products $0$ ✓. Ye step kyun? 3D mein $Q^{\mathsf{T}}Q=I$ ka direct verification.
3. Block structure se determinant: top-left $1$ ek $2\times 2$ rotation block ko multiply karta hai jiska det hai ${\cos }^2\theta +{\sin }^2\theta =1$. Toh $\det Q=1\cdot 1=1$. Ye step kyun? $\det =+1$ ⇒ ek genuine 3D rotation, isliye $Q\in SO(3)$ (space mein $\det =+1$ rotations).
4. Axis = fixed direction: $Q{e}_1=(1,0,0)=e_1$, isliye $x$-axis fixed hai; rotation $y$–$z$ plane mein hoti hai. Ye step kyun? Ek 3D rotation hamesha ek line fix karti hai (uska eigenvector eigenvalue $+1$ ke saath) — wo line axis hai.

Worked example Verify

$Q(0,1,0)=(0,0,1)$ aur $Q(0,0,1)=(0,-1,0)$: $y$–$z$ plane ke vectors $90^{\circ }$ rotate hote hain, jabki $e_1$ fixed rehta hai ✓. $\det =1$ ✓.

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Example 7 — Cell G: ek $3\times 3$ reflection

Steps.

1. Orthogonal check: columns hain $e_1,e_2,-e_3$ — orthonormal ✓, aur $S^{\mathsf{T}}S=S^2=I$ (do baar flip karne par ghar wapas). Ye step kyun? Confirm karta hai $S\in O(3)$.
2. Determinant: diagonal matrix ⇒ $\det =1\cdot 1\cdot (-1)=-1$. Ye step kyun? $\det =-1$ ⇒ reflection (orientation flip), isliye $S$ $SO(3)$ mein nahi hai — sirf ek axis flip = odd count.

Worked example Verify

Right-handed triple $(e_1,e_2,e_3)$ lo. $S$ ke baad: $(e_1,e_2,-e_3)$ — ek left-handed triple. Handedness flip, $\det =-1$ se match ✓.

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Example 8 — Cell H: word problem (camera gimbal)

Steps.

1. $z$ ke around rotation $z$ ko fixed rakhta hai, $x,y$ par act karta hai: $Q=\left(\begin{array}{ccc}\cos 30^{\circ }& -\sin 30^{\circ }& 0\\ \sin 30^{\circ }& \cos 30^{\circ }& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}\frac{\sqrt{3}}{2}& -\frac{1}{2}& 0\\ \frac{1}{2}& \frac{\sqrt{3}}{2}& 0\\ 0& 0& 1\end{array}\right)$. Ye step kyun? Vertical axis ke around spin karna $z$-axis rotation block hai, Example 6 ka mirror image.
2. Naya forward vector: $Q(1,0,0)=(\frac{\sqrt{3}}{2},\frac{1}{2},0)\approx (0.866,0.5,0)$. Ye step kyun? Ye physical answer hai — lens ab kidhar point kar raha hai.
3. No scaling check: $\Vert Q(1,0,0)\Vert =\sqrt{\frac{3}{4}+\frac{1}{4}}=1$, same as $\Vert (1,0,0)\Vert =1$. Ye step kyun? Ek orthogonal map ek isometry hai — drone ka "zoom" (length) pure rotation se change nahi ho sakta. Units: ek unit direction unit direction hi rehti hai.

Worked example Verify

$\det Q=1\cdot ({\cos }^{2}30^{\circ }+{\sin }^{2}30^{\circ })=1$ ✓. $0.866^2+0.5^2=0.75+0.25=1$ ✓.

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Example 9 — Cell I: exam twist (product & det guess karo)

Steps.

1. Closure se orthogonality: $(Q_1{Q}_2{)}^{\mathsf{T}}(Q_1{Q}_2)=Q_2^{\mathsf{T}}{Q}_1^{\mathsf{T}}{Q}_1{Q}_2=Q_2^{\mathsf{T}}I{Q}_2=I$. Toh $Q_1{Q}_2$ orthogonal hai. Ye step kyun? Orthogonal group products ke under closed hai — products orthogonal rehte hain (parent ka closure box).
2. Determinant multiplicative hai: $\det (Q_1{Q}_2)=\det Q_1\cdot \det Q_2=(+1)(-1)=-1$. Ye step kyun? $\det (AB)=\det A\det B$ — matrices ko actually multiply karne ki zarurat nahi.
3. Interpret karo: $\det =-1$ ⇒ combined map ek reflection hai (ye $SO(2)$ se bahar chala jaata hai). Ek rotation ke baad ek reflection = reflection (flips ki total odd count). Ye step kyun? $\det$ ka sign "flip parity" hai; even = rotation, odd = reflection.

Worked example Verify

Concretely: $Q_1=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ ($\det 1$), $Q_2=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ ($\det -1$). $Q_1{Q}_2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, $\det =-1$ ✓ — parent ka swap/reflection matrix.

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Example 10 — Cell J: unit circle par eigenvalues

Steps.

1. Characteristic polynomial: $\det (Q-\lambda I)=\det \left(\begin{array}{cc}-\lambda & -1\\ 1& -\lambda \end{array}\right)={\lambda }^2+1=0$. Ye step kyun? Eigenvalues $\det (Q-\lambda I)=0$ ke roots hote hain (dekho Eigenvalues and eigenvectors).
2. Solve karo: ${\lambda }^2=-1\Rightarrow \lambda =\pm i$. Ye step kyun? $90^{\circ }$ rotation ka koi real eigenvector nahi hota (koi bhi vector same direction mein point nahi karta), isliye eigenvalues complex hain — bilkul jaisa parent ne predict kiya tha, $e^{\pm i\theta }$ with $\theta =90^{\circ }$.
3. Magnitude check karo: $|\pm i|=\sqrt{0^2+1^2}=1$. Ye step kyun? Parent ne prove kiya ki orthogonal matrix ke har eigenvalue ka $|\lambda |=1$ hota hai; yahan $\lambda =e^{\pm i\pi /2}$, unit circle par.

Worked example Verify

Eigenvalues ka product $=(i)(-i)=1=\det Q$ ✓; sum $=i+(-i)=0=\text{trace}=\cos \theta +\cos \theta$ with $\theta =90^{\circ }$ ✓.

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Recall Har cell ka one-line summary

Spin = $\det +1$; flip = $\det -1$; impostor = $\det$ sahi par $Q^{\mathsf{T}}Q\ne I$; identity/$-I_2$ rotations hain; $-I_3$ reflection hai; product ka det multiply hota hai; eigenvalues unit circle par rehte hain.

Active recall