Worked examples — Complex eigenvalues — rotation-scaling interpretation

4.5.32 · D3 · Maths › Linear Algebra (Full) › Complex eigenvalues — rotation-scaling interpretation

Yeh page ek drill hai har tarah ke complex-eigenvalue problem ke through jo tum dekh sakte ho. Hum Complex eigenvalues — rotation-scaling interpretation par build karte hain: yeh idea ki ek real $2\times 2$ matrix jiske complex eigenvalues $\lambda =\alpha \pm i\beta$ hain, woh sahi basis mein rotate by $\theta$, scale by $r$ ki tarah kaam karta hai.

Numbers touch karne se pehle, chalte hain har symbol aur convention ko pin down karte hain jo hum use karte hain, phir list karte hain kaunsi tarah ki cheez galat ho sakti hai ya vary kar sakti hai — taaki koi case tumhe surprise na kare.

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Scenario matrix

Har complex-eigenvalue problem do numbers se decide hota hai jo eigenvalue $\lambda =\alpha -i\beta$ se read kiye jaate hain (plane mein ek point ki tarah draw karo):

• $\alpha =\mathrm{Re}\lambda$ — left/right part.
• $\beta$ — imaginary size, jo spin set karta hai.

In se tum paate ho $r=|\lambda |=\sqrt{{\alpha }^2+{\beta }^2}$ (scale) aur $\theta =\arg \lambda$ (angle). Yeh raha cases ka poora grid, aur kaun sa worked example har cell ko nail karta hai. (Neeche, "$r>1$" wohi statement hai jaise "$|\lambda |>1$".)

Case class	Trigger	Geometrically kya hota hai	Example
$r=1$, pure rotation	$\det A=1$, $(\mathrm{tr}{)}^2<4$	points circle par ride karte hain, koi growth nahi	Ex 1
$r>1$, outward spiral (discrete)	$r=	\lambda	>1$
$r<1$, inward spiral (discrete)	$r=	\lambda	<1$
$\theta$ har quadrant mein	sign of $\alpha$ aur $\beta$	$atan2$ se angle, $\arctan$ se nahi	Ex 4
Degenerate: distinct reals ($\beta =0$, disc $>0$)	discriminant $>0$	real eigenvalues, koi rotation nahi	Ex 5
Degenerate: repeated real root (disc $=0$)	discriminant $=0$	diagonalize fail ho sakta hai (shear)	Ex 5b
Continuous flow $\dot{v}=Av$	sign of $\alpha =\mathrm{Re}\lambda$	growth $\alpha$ se ruled hoti hai, $r$ se nahi	Ex 6
Real-world word problem	ek spinning process diya ho	model banao + spin & scale read karo	Ex 7
Exam twist: $A$ recover karo	$r,\theta ,P$ diya ho	$A=PC{P}^{-1}$ rebuild karo	Ex 8

Figure s01 — eigenvalue ek point ke roop mein: uski length scale $r$ hai, uska angle spin $\theta$ hai.

Khulle rakho ye prerequisites: Trace and Determinant, Characteristic Polynomial, Complex Numbers - Polar Form, Rotation Matrices.

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Ex 1 — $r=1$: pure rotation (circle)

Forecast: aage padhne se pehle scale aur angle guess karo. (Hint: yeh pehle se hi ek rotation matrix hai.)

1. Trace aur determinant. $\mathrm{tr}A=\sqrt{3}\approx 1.732$, $\det A=\frac{3}{4}+\frac{1}{4}=1$. Yeh step kyun? Characteristic quadratic hai ${\lambda }^2-(\mathrm{tr}A)\lambda +\det A=0$; dono numbers seedha Trace and Determinant se aate hain.
2. Discriminant. $(\mathrm{tr}A{)}^2-4\det A=3-4=-1<0$. Yeh step kyun? Negative discriminant signal hai ki roots ek complex conjugate pair hain — yahan ek rotation chhupa hua hai.
3. Eigenvalues. $\lambda =\frac{\sqrt{3}\pm \sqrt{-1}}{2}=\frac{\sqrt{3}}{2}\pm \frac{1}{2}i$. Yeh step kyun? Quadratic formula $\sqrt{-1}=i$ ke saath.
4. $\lambda =\alpha -i\beta$ pick karo: $\alpha =\frac{\sqrt{3}}{2},\beta =\frac{1}{2}$. Yeh step kyun? Hamara sign convention (upar defined) choice fix karta hai taaki block $C=\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)$ ke lower-left mein $+\beta$ ho.
5. Scale aur angle. $r=\sqrt{{\alpha }^2+{\beta }^2}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1$; $\theta =atan2(\frac{1}{2},\frac{\sqrt{3}}{2})=30°$. Yeh step kyun? $r$ modulus hai (scale), $\theta$ argument hai (spin) — page ke upar se MAS rule.

Verify: $r=1$ matlab koi growth nahi, toh points ek circle par ride karte hain; angle $30°$ us matrix se match karta hai jo hume diya gaya tha. Baara applications = $360°$ = wapas shuru. ✓

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Ex 2 — $r>1$: outward spiral

Forecast: kya points andar spiral karenge ya bahar?

1. Trace/det. $\mathrm{tr}A=2$, $\det A=1-(-4)=5$.
2. Discriminant. $4-20=-16<0$ → complex pair.
3. Eigenvalues. $\lambda =\frac{2\pm \sqrt{-16}}{2}=1\pm 2i$. $\lambda =1-2i$ lo, toh $\alpha =1,\beta =2$. Yeh step kyun? $\sqrt{-16}=4i$, 2 se divide karo.
4. Scale/angle. $r=\sqrt{1+4}=\sqrt{5}\approx 2.236$; $\theta =atan2(2,1)\approx 63.43°$. Yeh step kyun? Modulus = scale, argument = spin.
5. Direction. $r>1$ → har step magnify karta hai → outward spiral (discrete map). Figure s02 pehle steps trace karta hai.

Verify: $|\lambda |=\sqrt{5}$; $\det A=\lambda \bar{\lambda }=(1{)}^2+(2{)}^2=5$ se check karo. ✓ (conjugate eigenvalues ka product determinant ke barabar hota hai.)

Figure s02 — $A$ ko $r>1$ ke saath iterate karna: har teal arrow ek step hai, $\approx 63.4°$ mur ke aur bahar grow karta hua.

Links: Linear Dynamical Systems, Phase Portraits.

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Ex 3 — $r<1$: inward spiral

Forecast: numbers chhote hain — direction par bet lagao.

1. Trace/det. $\mathrm{tr}A=1.2$, $\det A=0.36+0.09=0.45$.
2. Discriminant. $1.44-1.8=-0.36<0$ → complex.
3. Eigenvalues. $\lambda =\frac{1.2\pm \sqrt{-0.36}}{2}=0.6\pm 0.3i$. $\alpha =0.6,\beta =0.3$ lo.
4. Scale/angle. $r=\sqrt{0.36+0.09}=\sqrt{0.45}\approx 0.6708$; $\theta =atan2(0.3,0.6)\approx 26.57°$.
5. Direction. $r<1$ → har step shrink karta hai → origin ki taraf inward spiral.

Verify: $r^2=0.45=\det A$ ✓ (phir se $\det =\lambda \bar{\lambda }=r^2$). Kyunki $r<1$, iterates $0$ par converge karte hain. ✓

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Ex 4 — $\theta$ ka quadrant: $atan2$ kyun, $\arctan$ nahi

Forecast: naive $\arctan (\beta /\alpha )$ dono ke liye same number deta hai. Yeh galat kyun hai?

1. Trap. Dono ka $\beta /\alpha =-1$ hai, toh $\arctan (-1)=-45°$ dono ke liye. Yeh step kyun? $\tan$ har $180°$ par repeat hota hai, isliye $\arctan$ kabhi bhi sirf $(-90°,90°)$ mein angle return kar sakta hai — yeh nahi bata sakta ki tum plane ke kis side par ho.
2. $\alpha$ aur $\beta$ ke signs separately use karo. Exactly yahi $atan2(\beta ,\alpha )$ karta hai — yeh dono coordinates dekhta hai aur sahi quadrant pick karta hai.
3. Case (a): $\alpha =-1<0,\beta =1>0$ → point upper-left mein (second quadrant). $\theta =atan2(1,-1)=135°$.
4. Case (b): $\alpha =1>0,\beta =-1<0$ → lower-right (fourth quadrant). $\theta =atan2(-1,1)=-45°$.

Verify: $135°$ aur $-45°$ mein $180°$ ka fark hai — exactly woh ambiguity jo $\arctan$ ne collapse kar di thi. Magnitudes: $r=\sqrt{2}$ dono ke liye. Figure s03 dono points dikhata hai. ✓

Figure s03 — do eigenvalues jinki slope $-1$ same hai lekin opposite quadrants mein; sirf $atan2$ unhe alag karta hai.

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Ex 5 — degenerate: do distinct real roots ($\beta =0$, discriminant $>0$)

Forecast: upper-triangular — aise matrices ke eigenvalues kahan hote hain?

1. Trace/det. $\mathrm{tr}A=5$, $\det A=6$.
2. Discriminant. $25-24=1>0$. Yeh step kyun? Positive discriminant → do distinct real roots → koi rotation nahi, $\beta =0$.
3. Eigenvalues. $\lambda =\frac{5\pm 1}{2}=3$ ya $2$.
4. Interpretation. $\beta =0\Rightarrow \theta =0$; block $C=\left(\begin{array}{cc}\alpha & -\beta \\ \beta & \alpha \end{array}\right)$ plain scaling mein collapse ho jaata hai real eigen-lines ke along. Yeh poore topic ka boundary case hai: koi spin nahi.

Verify: product $3\cdot 2=6=\det A$; sum $3+2=5=\mathrm{tr}A$. ✓ Do distinct real eigenvalues, toh Diagonalization $\mathbb{R}$ par seedha kaam karta hai — koi rotation-scaling block ki zaroorat nahi.

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Ex 5b — degenerate edge: ek repeated real root (discriminant $=0$) jo diagonalize nahi hoga

Forecast: ek eigenvalue "$2$" do baar ke saath, kya hume do eigenvectors milte hain ya sirf ek?

1. Trace/det. $\mathrm{tr}A=4$, $\det A=4$.
2. Discriminant. $16-16=0$. Yeh step kyun? Discriminant $=0$ do real roots aur ek complex pair ke beech ka razor's edge hai — ek repeated real root $\lambda =2$.
3. Eigenvalues. $\lambda =\frac{4\pm 0}{2}=2$ (do baar).
4. Eigenvectors count karo. $(A-2I)v=0$ solve karo: $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)v=0\Rightarrow v_2=0$. Sirf ek independent eigenvector, $\left(\begin{array}{c}1\\ 0\end{array}\right)$. Yeh step kyun? Diagonalize karne ke liye do independent eigenvectors chahiye; yahan sirf ek milta hai, toh $A$ diagonalizable nahi hai — yeh ek shear hai (ek real eigen-line plus ek slide), nahi rotation aur nahi pure stretch.

Verify: $\det (A-2I)=0$ aur rank$(A-2I)=1$, toh eigenspace 1-dimensional hai → defective. Koi complex eigenvalues nahi, koi rotation nahi, aur koi diagonalization nahi: yeh woh case hai jo discriminant-$=0$ boundary chhupata hai. ✓

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Ex 6 — continuous flow: growth $\alpha$ se ruled hoti hai, $r$ se nahi

Forecast: $|\lambda |$ yahan 1 se bada hai. Kya iska matlab yeh hai ki yeh grow karta hai?

1. Eigenvalues. $\mathrm{tr}A=-0.4,\det A=0.04+9=9.04$; discriminant $0.16-36.16=-36<0$. $\lambda =-0.2\pm 3i$, toh $\alpha =-0.2,\beta =3$.
2. Modulus. $r=\sqrt{0.04+9}=\sqrt{9.04}\approx 3.007$ — 1 se kaafi bada.
3. Kaun sa rule? Discrete map $v\mapsto Av$ ke liye, $r>1$ matlab grow. Continuous ODE $\dot{v}=Av$ ke liye, solution mein $e^{\alpha t}$ hota hai, toh growth $\mathrm{Re}\lambda =\alpha$ se decide hoti hai. Yeh step kyun? $\dot{v}=Av$ ke solutions $e^{\lambda t}$ ki tarah behave karte hain; real part exponent mein hota hai, imaginary part sirf spin karta hai.
4. $\alpha$ padho. $\alpha =-0.2<0\Rightarrow e^{-0.2t}\to 0$: trajectory andar spiral karti hai aur decay hoti hai, $r>3$ hone ke bawajood.

Verify: $\alpha =\mathrm{Re}\lambda =\frac{1}{2}\mathrm{tr}A=-0.2$ ✓ (real part = conjugate pair ke liye trace ka aadha). Decay $\alpha <0$ se confirm. ✓

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Ex 7 — word problem: ek spinning, cooling turntable

Forecast: spin (deg/sec) aur shrink rate guess karo.

1. Form pehchano. $A=0.9\cdot (\text{rotation by }40°)$, toh $r=0.9,\theta =40°$ seedha per second. Yeh step kyun? $A=rR(\theta )$ pehle se hi rotation-scaling block $C$ hai; eigenvalue hai $\lambda =r{e}^{-i\theta }$ (hamara $\lambda =\alpha -i\beta$ convention), $|\lambda |=0.9$ ke saath.
2. Spin rate. $\theta =40°$ per second.
3. $n$ seconds baad radius. Radius $\times r^n=0.9^n$. $0.9^n=0.5$ set karo.
4. Solve karo. $n=\frac{\ln 0.5}{\ln 0.9}=\frac{-0.6931}{-0.10536}\approx 6.579$ seconds. Yeh step kyun? Discrete iteration → scale $r^n$ hai; solve karne ke liye logs lo.

Verify: $0.9^{6.579}\approx 0.5$ ✓. Units: $n$ dimensionless step count hai = yahan seconds. Spin $40°/\text{s}$ toh half-life approximately $6.58\times 40°\approx 263°$ turning ke baad hoti hai.

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Ex 8 — exam twist: $r$, $\theta$, $P$ se $A$ rebuild karo

Forecast: kya $A$ phir bhi ek clean rotation matrix jaisa lagega, ya $P$ se distort ho jaayega?

1. $C$ banao. $C=r\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)=2\left(\begin{array}{cc}\cos 60°& -\sin 60°\\ \sin 60°& \cos 60°\end{array}\right)=\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)$. Yeh step kyun? $C$ upar defined rotation-scaling block hai; $2\cos 60°=1,2\sin 60°=\sqrt{3}$.
2. $P$ invert karo. $\det P=2$, toh $P^{-1}=\frac{1}{2}\left(\begin{array}{cc}2& -1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& -0.5\\ 0& 0.5\end{array}\right)$.
3. $A=PC{P}^{-1}$ assemble karo. $PC=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)\left(\begin{array}{cc}1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)=\left(\begin{array}{cc}1+\sqrt{3}& 1-\sqrt{3}\\ 2\sqrt{3}& 2\end{array}\right).$ $A=(PC)P^{-1}=\left(\begin{array}{cc}1+\sqrt{3}& 1-\sqrt{3}\\ 2\sqrt{3}& 2\end{array}\right)\left(\begin{array}{cc}1& -0.5\\ 0& 0.5\end{array}\right)=\left(\begin{array}{cc}1+\sqrt{3}& -\sqrt{3}\\ 2\sqrt{3}& 1-\sqrt{3}\end{array}\right).$ Yeh step kyun? Similarity $A=PC{P}^{-1}$ clean block $C$ ko standard basis mein carry karta hai; $P$ ka skew use standard basis mein ek non-rotation-looking matrix mein smear kar deta hai.

Verify: $A$ ke eigenvalues $2e^{\pm i60°}=1\pm i\sqrt{3}$ hone chahiye. Invariants se check karo: $\mathrm{tr}A=(1+\sqrt{3})+(1-\sqrt{3})=2=2\mathrm{Re}\lambda$ ✓; $\det A=r^2=4$ ✓ (VERIFY dekho). Numerically $\mathrm{tr}=2,\det =4$ dete hain $\lambda =1\pm i\sqrt{3}$. ✓

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Active Recall

Neeche, har line vault reveal format Prompt ::: Answer use karti hai — ::: se pehle sab kuch question hai, baad mein sab kuch hidden answer hai.