Exercises — Orthogonal matrices — properties, det = ±1

$\det (Q^{\mathsf{T}})=\det Q$ Kisi matrix ko transpose karne se uska determinant nahi badalta. Quick reason: determinant entries $a_{1\sigma (1)}\cdots a_{n\sigma (n)}$ ke products ke permutations ka sum hota hai; transpose karne se bas har aisi product ko relabel kiya jaata hai (row aur column indices swap ho jaate hain), jo same signs ke saath wohi products ka set deta hai. Isliye $\det (A^{\mathsf{T}})=\det A$ har square $A$ ke liye. Hum iska use Level 5 mein karte hain.

Har ek test karo: kya dono columns unit length hain, aur ek doosre ke perpendicular?

$A$: columns $c_1=\left(\genfrac{}{}{0ex}{}{0}{1}\right)$, $c_2=\left(\genfrac{}{}{0ex}{}{-1}{0}\right)$. Lengths $\sqrt{0+1}=1$ aur $\sqrt{1+0}=1$ ✓. Dot product $c_1\cdot c_2=0\cdot (-1)+1\cdot 0=0$ ✓. Orthogonal.

$B$: columns $\left(\genfrac{}{}{0ex}{}{2}{0}\right)$, $\left(\genfrac{}{}{0ex}{}{0}{1}\right)$. Pehle column ki length $=2\ne 1$ ✗. NOT orthogonal (yeh $x$-axis ko $2$ se stretch karta hai).

$C$: columns $\frac{1}{\sqrt{2}}\left(\genfrac{}{}{0ex}{}{1}{1}\right)$, $\frac{1}{\sqrt{2}}\left(\genfrac{}{}{0ex}{}{1}{-1}\right)$. Length${}^2=\frac{1}{2}(1+1)=1$ har ek ke liye ✓. Dot $=\frac{1}{2}(1\cdot 1+1\cdot (-1))=\frac{1}{2}(0)=0$ ✓. Orthogonal.

Answer: $A$ aur $C$ orthogonal hain; $B$ nahi hai.

$\det A=0\cdot 0-(-1)(1)=1$ → rotation (yeh $90^{\circ }$ ka turn hai, orientation bana rehta hai). $\det C=\frac{1}{2}(1\cdot (-1)-1\cdot 1)=\frac{1}{2}(-2)=-1$ → reflection (handedness flip ho jaati hai).

Pehla column $c_1=\left(\genfrac{}{}{0ex}{}{3/5}{4/5}\right)$ already length $1$ rakhhta hai ($\frac{9}{25}+\frac{16}{25}=1$ ✓; figure mein yellow arrow unit circle par dekho). Hum chahte hain $c_2$ unit-length AUR $c_1$ ke perpendicular.

$90^{\circ }$ rotate kyun karein? ${\mathbb{R}}^2$ mein, $\left(\genfrac{}{}{0ex}{}{a}{b}\right)$ ke perpendicular vector $\left(\genfrac{}{}{0ex}{}{-b}{a}\right)$ (left turn — blue arrow) ya $\left(\genfrac{}{}{0ex}{}{b}{-a}\right)$ (right turn) hote hain — agar original unit hai toh dono unit hain. Yeh $c_1\cdot c_2=0$ guarantee karne ka sabse fast tarika hai (figure mein pink right-angle mark).

Left turn: $c_2=\left(\genfrac{}{}{0ex}{}{-4/5}{3/5}\right)$. $\det$ check: $\frac{3}{5}\cdot \frac{3}{5}-(-\frac{4}{5})\cdot \frac{4}{5}=\frac{9}{25}+\frac{16}{25}=1$ ✓ ($+1$, isliye yeh sahi choice hai).

$Q=\frac{1}{5}\left(\begin{array}{cc}3& -4\\ 4& 3\end{array}\right).$ (Doosra perpendicular $\left(\genfrac{}{}{0ex}{}{4/5}{-3/5}\right)$ — right turn — $\det =-1$ deta hai, ek reflection — reject.)

$Q^{\mathsf{T}}Q=\frac{1}{25}\left(\begin{array}{cc}3& 4\\ -4& 3\end{array}\right)\left(\begin{array}{cc}3& -4\\ 4& 3\end{array}\right)=\frac{1}{25}\left(\begin{array}{cc}9+16& -12+12\\ -12+12& 16+9\end{array}\right)=\frac{1}{25}\left(\begin{array}{cc}25& 0\\ 0& 25\end{array}\right)=I.✓$

Hum kya use karte hain: orthogonal $Q$ ke eigenvalues $|\lambda |=1$ satisfy karte hain (figure mein woh unit circle par baite hain); complex eigenvalues conjugate pairs $e^{\pm i\theta }$ mein aate hain (jinका product $e^{i\theta }{e}^{-i\theta }=1$ hai); aur $\det Q$ = sab eigenvalues ka product.

$n=3$ ke liye, eigenvalues (multiplicity ke saath) ya toh $\{+1,-1\}$ se teen reals hain, ya ek real ${\lambda }_1\in \{+1,-1\}$ plus ek conjugate pair $e^{\pm i\theta }$ hai.

Case A (ek real + ek pair): product $={\lambda }_1\cdot e^{i\theta }{e}^{-i\theta }={\lambda }_1\cdot 1={\lambda }_1$. Yeh $\det Q=-1$ ke barabar hai, isliye ${\lambda }_1=-1$. ✓ eigenvalue $-1$ present hai.

Case B (teen reals $\{\pm 1\}$ se): unka product $-1$ hai, isliye $-1$'s ki count odd hai — hence kam se kam ek $-1$. ✓

Conclusion: dono cases mein $-1$ ek eigenvalue hai. Geometrically, ek odd-dimensional orientation-flip kisi axis ko reverse zaroor karta hai — woh reversed axis hi $-1$ eigenvector hai. Dekho Eigenvalues and eigenvectors.

Column 1 ${\mathbb{R}}^2$ mein ek unit vector hai, isliye yeh unit circle par hoga: ise $\left(\genfrac{}{}{0ex}{}{\cos \theta }{\sin \theta }\right)$ likhte hain kisi angle $\theta$ ke liye. (Kyun? Plane mein har unit vector $(\cos \theta ,\sin \theta )$ hota hai — yahi unit circle ki definition hai.)

Column 2 ek unit vector hai jo column 1 ke perpendicular hai, isliye yeh $\pm \left(\genfrac{}{}{0ex}{}{-\sin \theta }{\cos \theta }\right)$ hai (column 1 ke do $90^{\circ }$ turns).

• $+\left(\genfrac{}{}{0ex}{}{-\sin \theta }{\cos \theta }\right)$ choose karne par $\det =\cos \theta \cos \theta -(-\sin \theta )\sin \theta ={\cos }^2\theta +{\sin }^2\theta =+1$ ✓ milta hai.
• $-\left(\genfrac{}{}{0ex}{}{-\sin \theta }{\cos \theta }\right)$ choose karne par $\det =-1$ ✗ milta hai (reject).

Toh $\det =+1$ rotation form ko force karta hai. Dekho Rotations and reflections in $\mathbb{R}^2$ and $\mathbb{R}^3$.

Step 1 — pehla column $c_1=\frac{1}{3}(2,2,1{)}^{\mathsf{T}}$. Length check: $\frac{1}{9}(4+4+1)=1$ ✓.

Step 2 — $c_2\perp c_1$ chunno, phir normalize karo. Raw $v=(1,-1,0{)}^{\mathsf{T}}$ chuno; check $c_1\cdot v=\frac{1}{3}(2-2+0)=0$ ✓. Length $\sqrt{2}$, isliye $c_2=\frac{1}{\sqrt{2}}(1,-1,0{)}^{\mathsf{T}}$.

Step 3 — $c_3\perp c_1,c_2$. Clean tarika hai cross product $c_3=c_1\times c_2$ (dono ke perpendicular guaranteed): $c_1\times c_2=\frac{1}{3}\cdot \frac{1}{\sqrt{2}}\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 2& 1\\ 1& -1& 0\end{array}\right|=\frac{1}{3\sqrt{2}}(2\cdot 0-1\cdot (-1),1\cdot 1-2\cdot 0,2\cdot (-1)-2\cdot 1).$ $=\frac{1}{3\sqrt{2}}(1,1,-4{)}^{\mathsf{T}}.$

$c_3$ automatically unit length kyun hai? Cross product ka magnitude formula hai $\Vert a\times b\Vert =\Vert a\Vert \Vert b\Vert \sin \varphi$, jahan $\varphi$, $a$ aur $b$ ke beech ka angle hai. Yahan $a=c_1$ aur $b=c_2$ orthonormal hain: $\Vert a\Vert =\Vert b\Vert =1$ aur $\varphi =90^{\circ }$, isliye $\sin \varphi =1$. Hence $\Vert c_3\Vert =1\cdot 1\cdot 1=1$ — koi extra normalization nahi chahiye. Direct check: $\frac{1}{18}(1+1+16)=\frac{18}{18}=1$ ✓.

Result: $Q=\left(\begin{array}{ccc}\frac{2}{3}& \frac{1}{\sqrt{2}}& \frac{1}{3\sqrt{2}}\\ \frac{2}{3}& -\frac{1}{\sqrt{2}}& \frac{1}{3\sqrt{2}}\\ \frac{1}{3}& 0& -\frac{4}{3\sqrt{2}}\end{array}\right).$ Kyunki $c_3=c_1\times c_2$ right-hand rule follow karta hai, yeh ek right-handed frame hai: $\det Q=+1$. ($c_2\leftrightarrow c_3$ swap karne par $-1$ milta.)

Claim: $+1$ ek eigenvalue hai, aur uska eigenvector rotation axis hai. $n=3$ ke liye, eigenvalues ek real ${\lambda }_1\in \{\pm 1\}$ plus ya toh do aur reals $\{\pm 1\}$ se ya ek pair $e^{\pm i\theta }$ hain. Product $=\det Q=+1$.

• Agar conjugate pair present hai: ${\lambda }_1\cdot 1=+1\Rightarrow {\lambda }_1=+1$. ✓
• Agar teen real hain: unka product $+1$ hai, isliye $-1$'s ki count even hai (0 ya 2), kam se kam ek $+1$ bacha rehta hai. ✓

Toh ek nonzero $v$ exist karta hai jisme $Qv=v$: woh axis hai. $v$ ke perpendicular vectors $e^{\pm i\theta }$ pair ke angle $\theta$ se rotate kiye jaate hain. Yeh exactly $v$ ke around rotation hai. Links to Spectral theorem aur Eigenvalues and eigenvectors.

Determinant part. $\det (AB)=\det A\det B$ aur $\det (Q^{\mathsf{T}})=\det Q$ (page ke top par reminder) use karte hue: $\det (Q^{\mathsf{T}}MQ)=\det (Q^{\mathsf{T}})\det (M)\det (Q)=(\det Q{)}^2\det M.$ Kyunki $Q$ orthogonal $\Rightarrow (\det Q{)}^2=1$: $\det (Q^{\mathsf{T}}MQ)=\det M.✓$

Eigenvalue part. Maano $A=Q^{\mathsf{T}}MQ$. Toh kisi bhi scalar $\lambda$ ke liye: $A-\lambda I=Q^{\mathsf{T}}MQ-\lambda Q^{\mathsf{T}}Q=Q^{\mathsf{T}}(M-\lambda I)Q,$ $I=Q^{\mathsf{T}}Q$ use karte hue. Determinants lete hue aur upar wala fact $M\to M-\lambda I$ ke saath apply karte hue: $\det (A-\lambda I)=\det (M-\lambda I).$ Dono characteristic polynomials identical hain, isliye $A$ aur $M$ ke same eigenvalues hain. Yeh Spectral theorem aur QR decomposition ki backbone hai (orthogonal similarity spectrum preserve karti hai). Dekho bhi Inner product spaces.

$Q^2=I$ use karo eigenvalues pin karne ke liye. Agar $Qx=\lambda x$ toh $Q^{2}x={\lambda }^{2}x=x$, isliye ${\lambda }^2=1$, yaani $\lambda \in \{+1,-1\}$ — koi genuine complex pair survive nahi karta (ek pair $e^{\pm i\theta }$ ko $e^{2i\theta }=1$ chahiye hoga, jo $\theta \in \{0,\pi \}$ force karta hai, jo again real $\pm 1$ hai).

Ab $\det Q=-1$ laago. $n=3$ eigenvalues ke saath har ek $\pm 1$ aur unka product $=\det Q=-1$, $-1$'s ki count odd honi chahiye: ya toh ek $-1$ ya teen $-1$. Do possibilities hain:

• Eigenvalues $\{+1,+1,-1\}$ → ek plane across reflection: do $+1$ directions mirror plane span karte hain (fixed), single $-1$ direction flip hoti hai.
• Eigenvalues $\{-1,-1,-1\}$ → point inversion $Q=-I$: har direction origin ke through flip hoti hai.

Explicit example ($xy$-plane across reflection): $Q=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right),Q^2=I,\det Q=-1.✓$ Yahan mirror plane $z=0$ hai ($\{+1,+1\}$ eigenspace) aur flipped axis $z$ hai ($-1$ eigenvector). Doosra case, $Q=-I$, bhi $Q^2=I$ aur $\det (-I)=(-1{)}^3=-1$ satisfy karta hai.