Question bank — Eigenvalues of A — system modes, stability

False — $\det A=\prod {\lambda }_i$, isliye uska sign sirf real negative eigenvalues ki parity (odd/even count) track karta hai, kisi bhi positive real part ki presence nahi. Counterexample: eigenvalues $\lambda =-1,-2,-3$ se $\det A=-6<0$ milta hai phir bhi sab left half-plane mein hain, toh system asymptotically stable hai.

False — trace eigenvalues ka sum hai, isliye $\lambda =-10$ aur $\lambda =+3$ se negative trace milta hai phir bhi ek growing mode hota hai. Tumhe har eigenvalue ka real part negative chahiye, sirf unka sum nahi.

False — stability sirf real part pe depend karti hai; $\lambda =-2\pm 50j$ complex hain aur khoobsurti se stable hain, bas decay hote waqt zyada oscillatory hain.

False — iska matlab hai response bounded rehta hai lekin necessarily zero pe settle nahi hota; jaise ek pair $\pm j\omega$ ek sustained sinusoid deta hai jo na badhta hai na khatam hota hai.

False — eigenvalues growth/decay rates fix karte hain, lekin eigenvectors (directions/patterns) poori matrix pe depend karte hain; alag $A$'s same $\lambda$'s share kar sakte hain jabki modes alag directions mein point karte hain.

True in effect — $A$ ko $A-\alpha I$ se replace karna har eigenvalue ko left mein $\alpha$ shift karta hai (kyunki eigenvalues ${\lambda }_i-\alpha$ ho jaate hain), saare modes ko stable half-plane ki taraf push karta hai. Ye spectral shifting ka idea hai.

False — $\det A=0$ ka matlab ek eigenvalue exactly $0$ hai, yaani ek mode imaginary axis pe baitha hai. Agar woh zero simple hai toh yeh sirf marginally stable hai (ek constant, non-decaying direction), zaruri nahi ki unstable ho.

False — long-term behaviour largest real part wale eigenvalue se set hoti hai (imaginary axis ke sabse zyada right mein), kyunki $|e^{\lambda t}|=e^{\sigma t}$ sirf $\sigma$ pe depend karta hai, $|\lambda |$ pe nahi.

Galat — ek triangular matrix ke eigenvalues sirf diagonal entries hoti hain; off-diagonal terms kabhi characteristic polynomial mein enter nahi karte. Ek bada off-diagonal eigenvectors ko change karta hai, eigenvalues ko nahi.

Incomplete — $\det A>0$ sirf guarantee karta hai ki do eigenvalues ka same sign hai (ya complex-conjugate). Tumhe $\text{tr}A<0$ bhi chahiye; jaise $\lambda =+1,+6$ se $\det =6>0$ milta hai lekin dono modes badhte hain.

Galat — axis pe ek repeated eigenvalue $t{e}^{0t}=t$ jaisa term produce kar sakta hai jo unbounded badhta hai. Marginal stability ke liye axis eigenvalues simple/semisimple hone chahiye.

Ye sirf tab valid hai jab $A$ already diagonal ya triangular ho. Ek general $A$ ke liye diagonal entries eigenvalues nahi hoti; tumhe $\det (A-\lambda I)=0$ solve karna hoga.

Nahi — complex eigenvalues bilkul normal aur physical hain; ye oscillation encode karte hain. Ek real matrix inhe simply conjugate pairs $\sigma \pm j\omega$ mein produce karta hai.

Ye sirf ek mode hai. General solution saare eigenpairs pe superposition hai ${\sum }_i{c}_i{e}^{{\lambda }_{i}t}{\mathbf{v}}_i$, jahan $c_i$ initial condition se fix hote hain.

Kyunki $|e^{(\sigma +j\omega )t}|=e^{\sigma t}|\cos \omega t+j\sin \omega t|=e^{\sigma t}\cdot 1$; oscillatory factor ka magnitude hamesha $1$ hota hai, isliye sirf $e^{\sigma t}$ mode ki size ko badha ya ghata sakta hai.

Ek eigenvector ke along, $A\mathbf{v}=\lambda \mathbf{v}$ matrix multiplication ko plain scaling se replace kar deta hai, isliye vector ODE $\dot{\mathbf{x}}=A\mathbf{x}$ us direction mein scalar ODE $\dot{s}=\lambda s$ mein collapse ho jaata hai.

Hum ek nonzero eigenvector maangte hain; $(A-\lambda I)\mathbf{v}=0$ ka nonzero solution tab hi hota hai jab $A-\lambda I$ singular ho, yaani uska determinant zero ho. Ek invertible matrix $\mathbf{v}=0$ force kar deta, jo allowed nahi hai.

Uska real part sabse kam negative hai, isliye uska mode sabse dheere decay karta hai; sabse dheere decay karne wala term sabse aakhir mein fade hota hai, aur isliye woh decide karta hai ki poore system ko settle hone mein kitna time lagega.

Entries states ko aapas mein mix karti hain (coupling); stability decoupled directions ki property hai. Sirf $\det (A-\lambda I)=0$ solve karne ke baad hi true growth/decay rates saamne aate hain.

Us basis mein $A$ diagonal ho jaata hai, isliye coupled equations $n$ independent scalar ODEs ${\dot{z}}_i={\lambda }_i{z}_i$ ban jaate hain, har ek apne aap solve ho sakta hai — eigenvalues dhundhne ka yehi to poora point hai.

Kyunki degree $\ge 3$ ke liye $\det (A-\lambda I)=0$ solve karne ka koi clean formula nahi hai; Routh–Hurwitz test karta hai ki kya saare roots left half-plane mein hain, sirf polynomial ke coefficients use karke, bina kisi root-finding ke.

Haan — har eigenvalue ka real part negative hai, isliye ek "defective" repeat sirf $t{e}^{-3t}$ factor add karta hai, aur $e^{-3t}$ polynomial $t$ ko zero kar deta hai. Repeats sirf axis pe khatarnak hote hain.

Marginally stable — zero mode ek constant ($e^{0t}=1$) direction contribute karta hai jo na badhta hai na decay karta hai, jabki baaki saare modes khatam ho jaate hain. State ek nonzero constant pe settle hoti hai, origin pe nahi.

Marginally stable — ye ek sustained, undying oscillation deta hai fixed amplitude ke saath. Bounded hai lekin kabhi settle nahi hota; model ki chhoti si perturbation isse kisi bhi taraf tip kar sakti hai.

Unstable — defect ek $t{e}^{0t}=t$ term produce karta hai jo unbounded badhta hai, isliye chahe eigenvalue "sirf zero" ho, response linearly blow up karta hai.

Saare eigenvalues $0$ hain aur $A=0$ diagonal hai (semisimple), isliye $\dot{\mathbf{x}}=0$ ka matlab hai $\mathbf{x}$ apni initial value pe hamesha ke liye ruk jaata hai — marginally stable, apni jagah frozen.

Nahi — real part $-1<0$ hai, isliye envelope $e^{-t}$ phir bhi zero tak decay karta hai; bas neeche aate waqt bahut zyada ring karta hai. Oscillation frequency $\omega$ stability ko kabhi affect nahi karta. Dekho Damping Ratio and Natural Frequency.