Question bank — Eigenvalues of A — system modes, stability

False — $\det A=\prod {\lambda }_i$, so its sign only tracks the parity (odd/even count) of real negative eigenvalues, not the presence of any positive real part. Counterexample: eigenvalues $\lambda =-1,-2,-3$ give $\det A=-6<0$ yet all lie in the left half-plane, so the system is asymptotically stable.

False — the trace is the sum of eigenvalues, so $\lambda =-10$ and $\lambda =+3$ give a negative trace yet one growing mode. You need the real part of every eigenvalue negative, not just their sum.

False — stability depends only on the real part; $\lambda =-2\pm 50j$ are complex and beautifully stable, just heavily oscillatory as they decay.

False — it means the response stays bounded but need not settle to zero; e.g. a pair $\pm j\omega$ gives a sustained sinusoid that neither grows nor dies.

False — eigenvalues fix growth/decay rates, but eigenvectors (the directions/patterns) depend on the full matrix; different $A$'s can share $\lambda$'s while pointing modes along different directions.

True in effect — replacing $A$ by $A-\alpha I$ shifts every eigenvalue left by $\alpha$ (since eigenvalues become ${\lambda }_i-\alpha$), pushing all modes toward the stable half-plane. This is the idea behind spectral shifting.

False — $\det A=0$ means one eigenvalue is exactly $0$, i.e. a mode sitting on the imaginary axis. If that zero is simple it is only marginally stable (a constant, non-decaying direction), not necessarily unstable.

False — long-term behaviour is set by the eigenvalue with the largest real part (closest to / furthest right of the imaginary axis), because $|e^{\lambda t}|=e^{\sigma t}$ depends only on $\sigma$, not on $|\lambda |$.

Wrong — for a triangular matrix the eigenvalues are the diagonal entries only; off-diagonal terms never enter the characteristic polynomial. A big off-diagonal changes the eigenvectors, not the eigenvalues.

Incomplete — $\det A>0$ only guarantees the two eigenvalues have the same sign (or complex-conjugate). You also need $\text{tr}A<0$; e.g. $\lambda =+1,+6$ gives $\det =6>0$ but both modes grow.

Wrong — a repeated eigenvalue on the axis can produce a term like $t{e}^{0t}=t$ that grows without bound. Axis eigenvalues must be simple/semisimple for marginal stability.

Only valid if $A$ is already diagonal or triangular. For a general $A$ the diagonal entries are not the eigenvalues; you must solve $\det (A-\lambda I)=0$.

No — complex eigenvalues are perfectly normal and physical; they encode oscillation. A real matrix simply produces them in conjugate pairs $\sigma \pm j\omega$.

That is only one mode. The general solution is the superposition ${\sum }_i{c}_i{e}^{{\lambda }_{i}t}{\mathbf{v}}_i$ over all eigenpairs, with $c_i$ fixed by the initial condition.

Because $|e^{(\sigma +j\omega )t}|=e^{\sigma t}|\cos \omega t+j\sin \omega t|=e^{\sigma t}\cdot 1$; the oscillatory factor has magnitude $1$ always, so only $e^{\sigma t}$ can grow or shrink the mode's size.

Along an eigenvector, $A\mathbf{v}=\lambda \mathbf{v}$ replaces matrix multiplication with plain scaling, so the vector ODE $\dot{\mathbf{x}}=A\mathbf{x}$ collapses to the scalar ODE $\dot{s}=\lambda s$ in that direction.

We demand a nonzero eigenvector; $(A-\lambda I)\mathbf{v}=0$ has a nonzero solution only when $A-\lambda I$ is singular, i.e. its determinant vanishes. An invertible matrix would force $\mathbf{v}=0$, which we banned.

It has the least-negative real part, so its mode decays slowest; the slowest-decaying term is the last to fade, and it therefore sets how long the whole system takes to settle.

The entries mix states together (coupling); stability is a property of the decoupled directions. Only after solving $\det (A-\lambda I)=0$ do the true growth/decay rates emerge.

In that basis $A$ becomes diagonal, so the coupled equations become $n$ independent scalar ODEs ${\dot{z}}_i={\lambda }_i{z}_i$, each solvable on its own — the whole point of finding eigenvalues.

Because solving $\det (A-\lambda I)=0$ for degree $\ge 3$ has no clean formula; Routh–Hurwitz tests whether all roots lie in the left half-plane using only the polynomial's coefficients, with no root-finding.

Yes — every eigenvalue has negative real part, so even a "defective" repeat only adds a $t{e}^{-3t}$ factor, and $e^{-3t}$ crushes the polynomial $t$ to zero. Repeats are only dangerous on the axis.

Marginally stable — the zero mode contributes a constant ($e^{0t}=1$) direction that neither grows nor decays, while every other mode dies out. The state settles to a nonzero constant, not to the origin.

Marginally stable — this gives a sustained, undying oscillation of fixed amplitude. Bounded but never settling; a tiny perturbation of the model could tip it either way.

Unstable — the defect produces a $t{e}^{0t}=t$ term that grows without bound, so even though the eigenvalue is "only zero", the response blows up linearly.

All eigenvalues are $0$ and $A=0$ is diagonal (semisimple), so $\dot{\mathbf{x}}=0$ means $\mathbf{x}$ stays at its initial value forever — marginally stable, frozen in place.

No — the real part is $-1<0$, so the envelope $e^{-t}$ still decays to zero; it just rings a great deal on the way down. Oscillation frequency $\omega$ never affects stability. See Damping Ratio and Natural Frequency.