Complex eigenvalues — rotation-scaling interpretation: Exercises

4.5.32 · D4 · Maths › Linear Algebra (Full) › Complex eigenvalues — rotation-scaling interpretation

Yeh ek self-testing sheet hai parent topic ke liye. Har problem ko pehle khud try karo — phir uska solution kholo. Difficulty dheere-dheere badh'ti hai: "kya tum ise pehchaan sakte ho" se lekar "kya tum poori picture bana sakte ho" tak. Neeche use kiya gaya har notation parent note mein pehle se explain hai — agar koi symbol unfamiliar lage, pehle wahan jaao.

Kuch reminders jinhe tum baar-baar use karoge (sab parent se liye gaye hain):

Level 1 — Recognition

Exercise 1.1

Har matrix ke liye, bina solve kiye decide karo ki eigenvalues real hain ya complex. Sirf trace, determinant, aur discriminant ka sign use karo.

$M_{1} = (21 - 3 2), M_{2} = (3410), M_{3} = (52 - 2 5) .$

Recall Solution 1.1

KYA compute karte hain: discriminant $D = (tr A)^{2} - 4 det A$ . KYUN: sirf iska sign bata deta hai real vs complex (negative number ke square root se $i$ banta hai).

$M_{1}$ : $tr = 4, det = (2) (2) - (- 3) (1) = 4 + 3 = 7.$ $D = 16 - 28 = - 12 < 0 \Rightarrow$ complex. $M_{2}$ : $tr = 3, det = (3) (0) - (1) (4) = - 4.$ $D = 9 - 4 (- 4) = 9 + 16 = 25 > 0 \Rightarrow$ real (actually $λ = \frac{3 \pm 5}{2} = 4, - 1$ ). $M_{3}$ : $tr = 10, det = 25 - (- 2) (2) = 25 + 4 = 29.$ $D = 100 - 116 = - 16 < 0 \Rightarrow$ complex.

Exercise 1.2

Ek student claim karta hai ki $A = (01 - 1 0)$ "ke koi eigenvalues nahi hain." Sach ya jhooth, aur kyun?

Recall Solution 1.2

Jhooth. Iske eigenvalues $\pm i$ hain — woh complex hain, absent nahi. KYUN confusion hoti hai: real line $R$ par koi real eigenvector nahi hai (yeh $9 0^{\circ}$ rotation hai, har arrow ghoomta hai, toh koi bhi nonzero real vector sirf stretch nahi hota). Lekin $C$ par characteristic equation $λ^{2} + 1 = 0$ ke phir bhi do roots hain. "Koi real eigenvectors nahi" $\neq =$ "koi eigenvalues nahi."

Level 2 — Application

Exercise 2.1

$A = (21 - 3 2)$ ke eigenvalues nikalo, phir scale factor $r$ aur rotation angle $θ$ batao (convention $λ = α - i β$ use karo).

Recall Solution 2.1

Step 1 — eigenvalues. $tr = 4, det = 7$ , toh $λ = \frac{4 \pm 16 - 28}{2} = \frac{4 \pm - 12}{2} = \frac{4 \pm 2 3 i}{2} = 2 \pm 3 i .$ KYUN $- 12 = 23 i$ : $- 12 = (- 1) (4) (3)$ , aur $- 1 = i$ .

Step 2 — convention chuno. $λ = α - i β = 2 - 3 i$ lo, toh $α = 2, β = 3$ .

Step 3 — scale. $r = ∣ λ ∣ = α^{2} + β^{2} = 4 + 3 = 7 \approx 2.646$ .

Step 4 — angle. $θ = atan2 (β, α) = atan2 (3, 2) = arctan (\frac{3}{2}) \approx 40. 9^{\circ}$ . Reading: $A$ ka har application $\approx 40. 9^{\circ}$ spin karta hai aur $7$ se magnify karta hai — ek bahar ki taraf spiral, kyunki $r > 1$ .

Exercise 2.2

$A = (0.6 0.8 - 0.8 0.6)$ ke liye $λ$ , $r$ , $θ$ nikalo. Yeh geometrically kaun si special motion hai?

Recall Solution 2.2

$tr = 1.2, det = (0.6) (0.6) - (- 0.8) (0.8) = 0.36 + 0.64 = 1.$ $λ = \frac{1.2 \pm 1.44 - 4}{2} = \frac{1.2 \pm - 2.56}{2} = 0.6 \pm 0.8 i .$ $λ = 0.6 - 0.8 i$ lo: $r = 0.36 + 0.64 = 1 = 1$ , aur $θ = atan2 (0.8, 0.6) \approx 53.1 3^{\circ}$ . Motion: $r = 1$ matlab koi scaling nahi — ek pure rotation $\approx 53.1 3^{\circ}$ se. Yeh exactly $(cos θ sin θ - sin θ cos θ)$ hai jahan $cos θ = 0.6, sin θ = 0.8$ . Figure dekho.

Figure — Complex eigenvalues — rotation-scaling interpretation

Level 3 — Analysis

Exercise 3.1

Ek real $2 \times 2$ matrix ke eigenvalues $λ = α \pm i β$ hain. Diya hai ki matrix ka $tr A = 1$ aur $det A = 1$ hai, $α, β, r, θ$ nikalo. Kya iterated map $A^{n}$ (discrete) grow karega ya bounded rahega? Kya flow $\overset{v}{˙} = A v$ (continuous) grow karega ya decay karega?

Recall Solution 3.1

Trace aur determinant se $λ$ recover karo. Conjugate pair ke liye, $tr A = λ + \overset{ˉ}{λ} = 2 α$ aur $det A = λ \overset{ˉ}{λ} = ∣ λ ∣^{2} = r^{2}$ .

$2 α = 1 \Rightarrow α = \frac{1}{2} .$
$r^{2} = det A = 1 \Rightarrow r = 1.$
$β$ : $r^{2} = α^{2} + β^{2}$ se, $1 = \frac{1}{4} + β^{2} \Rightarrow β = 3 /2 \approx 0.866.$
$θ = atan2 (β, α) = atan2 (0.866, 0.5) = 6 0^{\circ} .$

Discrete $A^{n}$ : growth per step $r = 1$ hai → na grow karta hai na shrink — bounded rehta hai. KYUN orbits ellipses hote hain: $P$ -frame mein (columns $x, y$ ) motion circles par pure rotation hai; wapas standard basis mein $P$ lagate hain, jo har circle ko ek tilted ellipse mein shear karta hai. Toh $∣ A^{n} v ∣$ na grow karta hai na decay, kyunki $r^{n} = 1^{n} = 1$ , aur har orbit ek fixed ellipse trace karta hai.

Continuous $\overset{v}{˙} = A v$ : growth set hoti hai $Re λ = α = \frac{1}{2} > 0$ se → solutions $e^{α t}$ ki tarah grow karti hain — ek bahar ki taraf spiral. KYUN difference hai: discrete iteration $λ$ se multiply karta hai har step ( $∣ λ ∣ = r$ matter karta hai); ODE exponentiate karta hai, $e^{λ t}$ , jiska magnitude $e^{α t}$ hai ( $Re λ$ matter karta hai). Dekho Linear Dynamical Systems aur Phase Portraits.

Exercise 3.2

Discriminant aur Trace and Determinant use karke, $(tr, det)$ plane mein real aur complex eigenvalues ke beech ki poori boundary explain karo. Kin curve par dono regimes milte hain?

Recall Solution 3.2

Real vs complex decide hota hai $D = (tr A)^{2} - 4 det A$ se. Likho $T = tr A, Δ = det A$ .

$D > 0 ⟺ Δ < \frac{T ^{2}}{4}$ : do real eigenvalues.
$D < 0 ⟺ Δ > \frac{T ^{2}}{4}$ : complex conjugate pair.
$D = 0 ⟺ Δ = \frac{T ^{2}}{4}$ : ek repeated real eigenvalue — boundary parabola.

Toh is plane mein jahan horizontal axis $T$ hai aur vertical axis $Δ$ hai, parabola $Δ = T^{2} /4$ dono worlds ko alag karta hai; uske upar (trace ke relative bada determinant) rotation-scaling milta hai. Figure dekho.

Level 4 — Synthesis

Exercise 4.1

$A = (12 - 1 3)$ ke liye complex eigenvalue $λ = α - i β$ nikalo, uska eigenvector $v = x + i y$ nikalo, $P = (x y)$ banao, aur verify karo ki $C = P^{- 1} A P = (α β - β α)$ .

Recall Solution 4.1

Step 1 — eigenvalues. $tr = 4, det = (1) (3) - (- 1) (2) = 3 + 2 = 5.$ $λ = \frac{4 \pm 16 - 20}{2} = \frac{4 \pm 2 i}{2} = 2 \pm i .$ Convention: $λ = α - i β = 2 - i$ , toh $α = 2, β = 1$ .

Step 2 — eigenvector. $(A - λ I) v = 0$ solve karo $λ = 2 - i$ ke saath: $A - λ I = (1 - (2 - i) 2 - 1 3 - (2 - i)) = (- 1 + i 2 - 1 1 + i) .$ Row 2 clean hai: $2 v_{1} + (1 + i) v_{2} = 0 \Rightarrow v_{1} = - \frac{1 + i}{2} v_{2}$ . $v_{2} = 2$ lo: $v = (- (1 + i) 2) = x (- 1 2) + i y (- 1 0) .$ Row 1 check: $(- 1 + i) (- (1 + i)) - 1 \cdot 2 = (- 1 + i) (- 1 - i) - 2 = (1 + i - i - i^{2}) - 2 = (1 + 1) - 2 = 0.$ ✓

Step 3 — $P$ aur $C$ banao. $P = (x y) = (- 1 2 - 1 0), det P = (- 1) (0) - (- 1) (2) = 2, P^{- 1} = \frac{1}{2} (0 - 2 1 - 1) .$ $C = P^{- 1} A P = \frac{1}{2} (0 - 2 1 - 1) (12 - 1 3) (- 1 2 - 1 0) = (21 - 1 2) .$ Yeh $(α β - β α)$ ke barabar hai $α = 2, β = 1$ ke saath. ✓ Yeh padha jaata hai: rotate karo $θ = atan2 (1, 2) \approx 26.5 7^{\circ}$ se, scale $r = 5 \approx 2.236$ .

Level 5 — Mastery

Exercise 5.1

Ek real $2 \times 2$ matrix $A$ design karo jiska kisi bhi vector par action ho: har step mein exactly $3 0^{\circ}$ rotate karo aur factor $0.9$ se shrink karo, standard basis mein. Phir uske eigenvalues $α \pm i β$ form mein batao aur confirm karo ki $n \to \infty$ par $A^{n} v$ ka long-run behaviour kya hoga.

Recall Solution 5.1

KYA chahiye: standard basis mein rotation-scaling ka matlab hai $P = I$ , toh $A = C$ directly. $C = r (cos θ sin θ - sin θ cos θ)$ use karo $r = 0.9, θ = 3 0^{\circ}$ ke saath. $cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660, sin 3 0^{\circ} = \frac{1}{2} = 0.5.$ $A = 0.9 (0.8660 0.5 - 0.5 0.8660) = (0.7794 0.45 - 0.45 0.7794) .$ Eigenvalues: construction se $α = r cos θ = 0.9 (0.8660) = 0.7794$ , $β = r sin θ = 0.9 (0.5) = 0.45$ , toh $λ = 0.7794 \pm 0.45 i, ∣ λ ∣ = r = 0.9, ar g λ = 3 0^{\circ} .$ Check: $det A = r^{2} = 0.81$ , $tr A = 2 r cos θ = 1.5588$ ; discriminant $= 1.558 8^{2} - 4 (0.81) = 2.4299 - 3.24 = - 0.8101 < 0$ ✓ complex. Long run: har step length ko $r = 0.9 < 1$ se multiply karta hai, toh $∣ A^{n} v ∣ = 0. 9^{n} ∣ v ∣ \to 0$ — ek andar ki taraf spiral jo origin par collapse ho jaata hai (discrete sense mein ek stable spiral). Figure dekho.

Exercise 5.2

Do students ek hi $A$ (complex eigenvalues wali) ko $C$ par aur $R$ par diagonalize karte hain. $C$ par ek likhta hai $A = S D S^{- 1}$ jahan $D = diag (λ, \overset{ˉ}{λ})$ ; $R$ par doosra likhta hai $A = P C P^{- 1}$ rotation-scaling block ke saath. Argue karo ki yeh dono ek hi fact hain do alag languages mein, aur woh explicit bridge do jo ek ko doosre mein convert karta hai.

Recall Solution 5.2

Dono similarity transforms hain jo $A$ ko sabse simple coordinates mein le jaate hain. $C$ par, Diagonalization ek diagonal $D = (λ 0 0 \overset{ˉ}{λ})$ deta hai — lekin change-of-basis matrix $S = (v \overset{v}{ˉ})$ ke columns complex hain (eigenvectors $v, \overset{v}{ˉ}$ ). Real block $C = (α β - β α)$ mein $P = (x y)$ ke columns real hain ( $v = x + i y$ ).

Explicit bridge. Real aur complex frames ek fixed complex matrix se related hain: $W = \frac{1}{2} (11 - i i), W^{- 1} = (1 i 1 - i),$ jo sirf "real part, imaginary part" ko " $λ$ -component, $\overset{ˉ}{λ}$ -component" mein repackage karta hai. Direct multiply karne par: $W D W^{- 1} = \frac{1}{2} (11 - i i) (λ 0 0 \overset{ˉ}{λ}) (1 i 1 - i) = (α β - β α) = C,$ $λ = α - i β$ , $\overset{ˉ}{λ} = α + i β$ use karke. KYUN yeh kaam karta hai: $\frac{λ + λ ˉ}{2} = α$ diagonal par land karta hai aur $\frac{λ - λ ˉ}{2 i} = - β$ (apne conjugate ke saath) off-diagonal fill karta hai — exactly block entries. Toh $C = W D W^{- 1}$ : $C$ aur $D$ ke same eigenvalues $α \pm i β$ hain (iska characteristic equation $t^{2} - 2 α t + (α^{2} + β^{2}) = 0$ hai), aur $S = P W$ . Dekho Complex Numbers - Polar Form — wahan samjhaya gaya hai ki $λ = r e^{i θ}$ kyun dono views ko interchangeable banata hai.

Real form ka faayda: $P$ aur $C$ real hain, toh tum $A^{n} = P C^{n} P^{- 1}$ iterate kar sakte ho aur geometry padhh sakte ho (spin $n θ$ , scale $r^{n}$ ) bina real arithmetic chode — koi complex bookkeeping nahi, aur phase portrait seedha nikal aata hai.

Active Recall

Kaun sa ek number real-vs-complex decide karta hai, aur kaunsa sign matlab complex?

Discriminant

(tr A)^{2} - 4 det A

; negative value matlab complex conjugate eigenvalues.

Sirf

tr A

aur

det A

se, conjugate pair ke liye

α

aur

r

kaise milte hain?

α = \frac{1}{2} tr A

aur

r = ∣ λ ∣ = det A

(kyunki

tr A = 2 α

aur

det A = r^{2}

Discrete map

A^{n}

vs continuous ODE

\overset{v}{˙} = A v

— dono mein growth kaun si quantity govern karti hai?

Discrete mein

r = ∣ λ ∣

use hota hai (grow if

r > 1

); continuous mein

Re λ = α

(grow if

α > 0

Standard-basis motion kab ek true circle-spiral hoti hai naa ki tilted-ellipse spiral?

Jab

P = I

ho, yaani

A

already apne rotation-scaling block

C

ke barabar ho; warna

P

circles ko ellipses mein shear karta hai.

Complex-eigenvalue wali

A

kis real normal form se similar hai, aur kis matrix ke zariye?

A = P C P^{- 1}

jahan

C = (α β - β α)

aur

P = (x y)

eigenvector

v = x + i y

λ = α - i β

ke liye.