Stability in Linear Systems - Linear Algebra

1. For a discrete system $\vec{x}_{k+1} = A\vec{x}_k$ with $A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$, and a generic initial vector $\vec{x}_0$ not in an eigenspace, the vector $\vec{x}_k$ will align with a specific direction as $k \to \infty$. What is this long-term direction?
   Answer: Parallel to the vector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. The long-term direction of the system is determined by the eigenvector corresponding to the dominant eigenvalue (the eigenvalue with the largest magnitude). The characteristic equation is $(4-\lambda)(1-\lambda) - (-2)(1) = 0$, which is $\lambda^2 - 5\lambda + 6 = 0$, or $(\lambda-3)(\lambda-2) = 0$. The eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 2$. The dominant eigenvalue is $\lambda_1 = 3$. To find its eigenvector, we solve $(A - 3I)\vec{v} = \vec{0}$: $\begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. This gives the equation $x - 2y = 0$, so the eigenvector is any scalar multiple of $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.
2. For the discrete system $\vec{x}_{k+1} = \begin{pmatrix} 2 & 0 \\ 0 & 0.5 \end{pmatrix} \vec{x}_k$, which of the following sets of initial vectors $\vec{x}_0$ will produce a trajectory that converges to the origin as $k \to \infty$?
   Answer: Any vector lying on the y-axis, of the form $\begin{pmatrix} 0 \\ c \end{pmatrix}$. The matrix is diagonal, so its eigenvalues are the diagonal entries $\lambda_1 = 2$ and $\lambda_2 = 0.5$. The corresponding eigenvectors are the standard basis vectors $\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (the x-axis) and $\vec{v}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ (the y-axis). Since $|\lambda_1| > 1$ and $|\lambda_2| < 1$, the origin is a saddle point. The direction associated with $\lambda_1$ is unstable (repelling), and the direction associated with $\lambda_2$ is stable (attracting). A trajectory converges to the origin only if its initial vector has no component in the unstable direction. For an initial vector $\vec{x}_0 = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$, the trajectory is $\vec{x}_k = \begin{pmatrix} c_1(2^k) \\ c_2(0.5^k) \end{pmatrix}$. This converges to $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ only if $c_1=0$. Therefore, the initial vector must be of the form $\begin{pmatrix} 0 \\ c_2 \end{pmatrix}$, which is any vector on the y-axis.
3. A population model tracks two competing species with transition matrix $P = \begin{pmatrix} 0.7 & 0.1 \\ 0.2 & 1.2 \end{pmatrix}$. After computing eigenvalues $\lambda_1 = 0.6$ and $\lambda_2 = 1.3$, what is the most accurate description of the system's long-term behavior?
   Answer: The system exhibits exponential growth in the direction of the eigenvector for $\lambda_2 = 1.3$ while decay occurs perpendicular to it With eigenvalues 0.6 and 1.3, the dominant eigenvalue λ₂ = 1.3 > 1 causes exponential growth in its eigenvector direction, while components in the λ₁ = 0.6 direction decay. The long-term behavior is dominated by growth along the eigenvector for λ₂. Choice A is wrong because mixed eigenvalues don't lead to finite stabilization. Choice C incorrectly describes oscillation. Choice D incorrectly focuses on individual diagonal entries rather than eigenvalue analysis.
4. Consider the discrete system $\vec{x}_{k+1} = A\vec{x}_k$ where $A = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{pmatrix}$. Which statement best describes the long-term behavior of this system for an arbitrary initial vector $\vec{x}_0$?
   Answer: Trajectories converge to a specific point on the line $y=x$, with the specific point depending on $\vec{x}_0$. First, we find the eigenvalues of $A$. The characteristic equation is $(0.5-\lambda)^2 - 0.25 = 0$, which gives $\lambda^2 - \lambda = 0$, so $\lambda_1 = 1$ and $\lambda_2 = 0$. The eigenvector for $\lambda_1=1$ is found by solving $(A-I)\vec{v}=\vec{0}$, which gives $-0.5x + 0.5y = 0$, so the eigenspace is the line $y=x$, spanned by $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. The eigenvector for $\lambda_2=0$ is spanned by $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Any initial vector can be written as $\vec{x}_0 = c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Then $\vec{x}_k = A^k\vec{x}_0 = c_1(1)^k\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2(0)^k\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. For $k \ge 1$, this becomes $\vec{x}_k = c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Thus, the system converges to the point $c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, which lies on the line $y=x$. The exact point depends on the component of $\vec{x}_0$ in the direction of the first eigenvector.
5. The origin is classified as a saddle point for the discrete dynamical system $\vec{x}_{k+1} = A\vec{x}_k$, where $A$ is a $2 \times 2$ matrix with real entries. Which statement about the eigenvalues of $A$ must be true?
   Answer: One eigenvalue $\lambda_1$ satisfies $|\lambda_1| > 1$ and the other eigenvalue $\lambda_2$ satisfies $|\lambda_2| < 1$. A saddle point in a discrete dynamical system is characterized by having both attractive and repulsive behavior. This occurs when at least one eigenvalue has a magnitude greater than 1 (causing repulsion along its eigenspace) and at least one eigenvalue has a magnitude less than 1 (causing attraction along its eigenspace). The other options are incorrect because they describe conditions that are neither necessary nor sufficient. For example, eigenvalues $\lambda_1=2, \lambda_2=0.5$ produce a saddle point, but both are positive. Eigenvalues $\lambda_1=2, \lambda_2=-2$ produce a repeller, even though they have opposite signs. The condition on the real parts of eigenvalues applies to continuous systems, not discrete ones.
6. Consider two stable discrete systems, System 1 governed by $A = \begin{pmatrix} 0.5 & 0 \\ 0.1 & 0.6 \end{pmatrix}$ and System 2 governed by $B = \begin{pmatrix} 0.8 & -0.2 \\ 0 & 0.7 \end{pmatrix}$. Which statement correctly compares their long-term behavior?
   Answer: System 1 converges to the origin faster than System 2. When analyzing discrete dynamical systems of the form $\mathbf{x}_{n+1} = M\mathbf{x}_n$, the convergence rate to the origin depends on the spectral radius—the largest absolute value among all eigenvalues of the matrix. The smaller the spectral radius, the faster the system converges. For System 1 with matrix $A = \begin{pmatrix} 0.5 & 0 \\ 0.1 & 0.6 \end{pmatrix}$, we need the eigenvalues. The characteristic polynomial is $(0.5-\lambda)(0.6-\lambda) - 0 = \lambda^2 - 1.1\lambda + 0.3$. Using the quadratic formula: $\lambda = \frac{1.1 \pm \sqrt{1.21-1.2}}{2} = \frac{1.1 \pm 0.1}{2}$, giving eigenvalues 0.6 and 0.5. The spectral radius is 0.6. For System 2 with matrix $B = \begin{pmatrix} 0.8 & -0.2 \\ 0 & 0.7 \end{pmatrix}$ (upper triangular), the eigenvalues are simply the diagonal entries: 0.8 and 0.7. The spectral radius is 0.8. Since System 1 has spectral radius 0.6 < 0.8 (System 2's spectral radius), System 1 converges faster. This confirms answer D. Answer A is wrong because convergence rate depends only on the matrix eigenvalues, not initial conditions. Answer B incorrectly reverses the comparison—System 2's larger spectral radius means slower convergence. Answer C is wrong because the spectral radii are clearly different (0.6 ≠ 0.8). Study tip: For discrete system convergence rates, always compute the spectral radius. Smaller spectral radius = faster convergence to equilibrium.
7. The characteristic polynomial of a $2 \times 2$ matrix $A$ for a discrete system is $p(\lambda) = \lambda^2 - 0.5\lambda + c$. If the origin is a stable attractor for this system, which of the following is a possible value for $c$?
   Answer: $c = 0.4$ For a $2 \times 2$ matrix, the characteristic polynomial is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$. So, $\text{tr}(A) = 0.5$ and $c = \det(A)$. For the origin to be a stable attractor, three conditions on the trace (T) and determinant (D) must be met: (1) $|D| < 1$, (2) $1 - T + D > 0$, and (3) $1 + T + D > 0$. In our case, $T=0.5$ and $D=c$. (1) $|c| < 1$, which means $-1 < c < 1$. This eliminates choice A. (2) $1 - 0.5 + c > 0 \implies 0.5 + c > 0 \implies c > -0.5$. This eliminates choices B and D. (3) $1 + 0.5 + c > 0 \implies 1.5 + c > 0 \implies c > -1.5$. This condition is weaker than $c > -0.5$. Combining the conditions, we need $-0.5 < c < 1$. Only choice C, $c=0.4$, falls within this range.
8. For the discrete dynamical system $\vec{x}_{k+1} = A\vec{x}_k$ with $A = \begin{pmatrix} 0.8 & 0.3 \\ 0 & 1.5 \end{pmatrix}$, the origin is a saddle point. Trajectories are attracted to the origin along one direction and repelled from it along another. Along which direction are the trajectories attracted toward the origin?
   Answer: Along the direction of the vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. When analyzing stability in discrete dynamical systems, you need to examine the eigenvalues and eigenvectors of the matrix. The eigenvalues determine whether trajectories grow or decay, while the eigenvectors show the directions along which this behavior occurs. For matrix $A = \begin{pmatrix} 0.8 & 0.3 \\ 0 & 1.5 \end{pmatrix}$, since it's upper triangular, the eigenvalues are simply the diagonal entries: $\lambda_1 = 0.8$ and $\lambda_2 = 1.5$. Since $|\lambda_1| < 1$, trajectories decay (are attracted to the origin) along its eigenvector direction. Since $|\lambda_2| > 1$, trajectories grow (are repelled from the origin) along its eigenvector direction. To find the eigenvector for $\lambda_1 = 0.8$, solve $(A - 0.8I)\vec{v} = \vec{0}$: $\begin{pmatrix} 0 & 0.3 \\ 0 & 0.7 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ This gives $0.3y = 0$ and $0.7y = 0$, so $y = 0$. The eigenvector is $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, making D correct. Choice A, $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$, isn't an eigenvector of this matrix. Choice B, $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$, is actually the eigenvector for $\lambda_2 = 1.5$, which corresponds to repulsion, not attraction. Choice C, $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, also isn't an eigenvector. Remember: for discrete systems, attraction occurs along eigenvectors with eigenvalues having absolute value less than 1, while repulsion occurs when the absolute value exceeds 1.
9. A mechanical system satisfies $\mathbf{x}_{k+1} = M\mathbf{x}_k$ where $M$ has characteristic polynomial $p(\lambda) = \lambda^2 - 0.6\lambda + 0.25$. What can be determined about the stability without computing eigenvectors?
   Answer: The system is stable since both roots satisfy $|\lambda| < 1$, as verified by $p(1) > 0$ and $p(-1) > 0$ For stability, we need |λ| < 1 for all roots. Since p(λ) = λ² - 0.6λ + 0.25, we have p(1) = 1 - 0.6 + 0.25 = 0.65 > 0 and p(-1) = 1 + 0.6 + 0.25 = 1.85 > 0. With the leading coefficient positive, this means no roots have |λ| ≥ 1, so the system is stable. Choice B misinterprets p(0). Choice C incorrectly concludes marginal stability from complex roots. Choice D is wrong because the tests are conclusive.
10. Two towns, Northville and Southtown, have populations that migrate between them. Each year, 10% of Northville's population moves to Southtown, and 5% of Southtown's population moves to Northville. If this pattern continues, what is the long-term stable distribution of the total population between the two towns?
    Answer: The population of Northville will stabilize at half the population of Southtown. Let $n_k$ and $s_k$ be the populations of Northville and Southtown in year $k$. The system is described by: $n_{k+1} = 0.90n_k + 0.05s_k$ and $s_{k+1} = 0.10n_k + 0.95s_k$. The transition matrix is $A = \begin{pmatrix} 0.90 & 0.05 \\ 0.10 & 0.95 \end{pmatrix}$. The long-term stable distribution is the eigenvector corresponding to the eigenvalue $\lambda=1$. We solve $(A-I)\vec{v} = \vec{0}$, which is $\begin{pmatrix} -0.10 & 0.05 \\ 0.10 & -0.05 \end{pmatrix} \begin{pmatrix} n \\ s \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. Both equations simplify to $0.10n = 0.05s$, or $s = 2n$. This means in the long term, the population of Southtown will be twice the population of Northville, or equivalently, Northville's population will be half of Southtown's.
11. The trajectories of a 2D discrete dynamical system $\vec{x}_{k+1} = A\vec{x}_k$ are observed to spiral outward, away from the origin. Which of the following could be the eigenvalues of the matrix $A$?
    Answer: $\lambda = 1.1 \pm 0.4i$ The behavior of trajectories is determined by the eigenvalues of $A$. A 'spiral' motion indicates that the eigenvalues are a complex conjugate pair. Motion 'outward, away from the origin' indicates that the origin is a repeller, which for a discrete system means the magnitude of the eigenvalues must be greater than 1. Let's check the magnitudes for the complex options: (A) $|0.5 \pm 0.5i| = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.5} < 1$. This would be a spiral attractor. (C) $|1.1 \pm 0.4i| = \sqrt{1.1^2 + 0.4^2} = \sqrt{1.21 + 0.16} = \sqrt{1.37} > 1$. This corresponds to a spiral repeller. (D) $|\pm i| = \sqrt{0^2 + 1^2} = 1$. This would be a center, with trajectories orbiting but not moving outward.
12. A student concludes that the origin is a stable attractor for the discrete system $\vec{x}_{k+1} = A\vec{x}_k$ with $A = \begin{pmatrix} -0.7 & 0 \\ 0.5 & -1.2 \end{pmatrix}$, reasoning that both eigenvalues are negative. Why is this reasoning and conclusion flawed?
    Answer: The conclusion is flawed because stability in discrete systems requires all eigenvalues to have a magnitude less than 1, and $|-1.2| > 1$. The student's reasoning is incorrect because the stability of a discrete dynamical system depends on the magnitude of the eigenvalues, not their sign. The matrix is lower triangular, so the eigenvalues are the diagonal entries: $\lambda_1 = -0.7$ and $\lambda_2 = -1.2$. For the origin to be an attractor, all eigenvalues must satisfy $|\lambda| < 1$. Here, $|\lambda_1| = |-0.7| = 0.7 < 1$, but $|\lambda_2| = |-1.2| = 1.2 > 1$. Since one eigenvalue has a magnitude greater than 1, the origin is unstable (specifically, a saddle point). The student incorrectly applied a rule related to the sign of eigenvalues, which is not the correct criterion.
13. Consider the linear system $\frac{d\mathbf{x}}{dt} = A\mathbf{x}$ where $A = \begin{pmatrix} -2 & 1 \\ 4 & -3 \end{pmatrix}$. The eigenvalues of $A$ are $\lambda_1 = -1 + i$ and $\lambda_2 = -1 - i$. What type of stability does this equilibrium point exhibit?
    Answer: Asymptotically stable with oscillatory approach since both eigenvalues have negative real parts and nonzero imaginary parts Since both eigenvalues have negative real parts (-1), the equilibrium is asymptotically stable. The nonzero imaginary parts (±i) cause oscillatory behavior as solutions spiral inward to the origin. Choice B is wrong because negative real parts ensure stability, not instability. Choice C confuses the continuous case with discrete systems. Choice D incorrectly focuses on magnitude and describes monotonic behavior.
14. Consider a continuous linear system $\dot{\mathbf{x}} = C\mathbf{x}$ where $C$ has eigenvalues $\alpha = 0$ and $\beta = -0.5$. If the eigenvector for $\alpha = 0$ is $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, what characterizes the long-term behavior of solutions?
    Answer: Solutions approach the line spanned by $\mathbf{v}_1$ with the approach rate determined by the $\beta = -0.5$ eigenvalue When analyzing linear dynamical systems $\dot{\mathbf{x}} = C\mathbf{x}$, the eigenvalues of matrix $C$ completely determine the system's long-term behavior. Each eigenvalue governs how solutions behave along its corresponding eigenvector direction. With eigenvalues $\alpha = 0$ and $\beta = -0.5$, the general solution is $\mathbf{x}(t) = c_1 e^{0 \cdot t}\mathbf{v}_1 + c_2 e^{-0.5t}\mathbf{v}_2 = c_1\mathbf{v}_1 + c_2 e^{-0.5t}\mathbf{v}_2$. As $t \to \infty$, the exponential term $e^{-0.5t} \to 0$, so solutions approach $c_1\mathbf{v}_1$—a point on the line spanned by $\mathbf{v}_1$. The rate of approach is governed by how quickly $e^{-0.5t}$ decays, which depends on $\beta = -0.5$. This confirms answer D. A is wrong because solutions don't converge to the origin unless $c_1 = 0$. The zero eigenvalue means the $\mathbf{v}_1$ component persists forever—solutions approach the line spanned by $\mathbf{v}_1$, not the origin. B incorrectly suggests oscillations. Oscillations require complex eigenvalues with imaginary parts. Here, both eigenvalues are real, producing exponential behavior without oscillation. C mischaracterizes zero eigenvalues as causing instability. A zero eigenvalue creates neutral stability along that direction—neither growth nor decay. Instability requires positive real parts in eigenvalues. Study tip: Remember that eigenvalues with negative real parts decay to zero, while zero eigenvalues persist unchanged. The long-term behavior is dominated by eigenvalues with the largest real parts.
15. For the discrete system $\mathbf{y}_{n+1} = B\mathbf{y}_n$, suppose $B$ has eigenvalues $\mu_1 = -0.8$ and $\mu_2 = 0.6$. Starting from a generic initial condition, which statement best describes the solution's behavior?
    Answer: The solution exhibits damped oscillations approaching the origin due to the negative eigenvalue with largest absolute value When analyzing discrete linear systems, the eigenvalues of the coefficient matrix determine the long-term behavior of solutions. For the system $\mathbf{y}_{n+1} = B\mathbf{y}_n$, you need to examine both the magnitude and sign of each eigenvalue to predict how solutions evolve. With eigenvalues $\mu_1 = -0.8$ and $\mu_2 = 0.6$, the solution has the general form $\mathbf{y}_n = c_1(-0.8)^n\mathbf{v}_1 + c_2(0.6)^n\mathbf{v}_2$, where $\mathbf{v}_1$ and $\mathbf{v}_2$ are the corresponding eigenvectors. Since $|-0.8| = 0.8 > |0.6| = 0.6$, the first term dominates the behavior. The negative eigenvalue $(-0.8)^n$ creates alternating signs each iteration (oscillation), while the magnitude less than 1 ensures decay over time (damping). Both terms eventually approach zero, so the solution exhibits damped oscillations approaching the origin. Choice A incorrectly ignores the oscillatory nature caused by the negative eigenvalue. Choice B wrongly suggests increasing amplitude—both eigenvalues have magnitude less than 1, so all components decay. Choice D misunderstands the dynamics: the solution doesn't approach a steady state along $\mathbf{v}_2$ because that component also decays to zero, and the alternating signs come from the $\mu_1$ term, not $\mu_2$. Study tip: For discrete systems, eigenvalue magnitude determines stability ($|\mu| < 1$ means stable), while eigenvalue sign determines oscillation (negative eigenvalues cause alternating behavior). Always identify the dominant eigenvalue—the one with largest absolute value—to predict primary solution characteristics.
16. For the continuous system $\frac{d\mathbf{u}}{dt} = Q\mathbf{u}$, matrix $Q$ has eigenvalues $\sigma_1 = -0.1$ and $\sigma_2 = -0.001$. If the initial condition has equal components along both eigenvectors, approximately how long does it take for the solution to be within 1% of its final steady state?
    Answer: About 4606 time units, determined by $\ln(100)/0.001$ since the slower eigenvalue determines when steady state is reached When analyzing continuous linear systems, the long-term behavior depends critically on understanding how different eigenvalues control the decay to steady state. For the system $\frac{d\mathbf{u}}{dt} = Q\mathbf{u}$ with negative eigenvalues, the general solution is $\mathbf{u}(t) = c_1 e^{\sigma_1 t}\mathbf{v}_1 + c_2 e^{\sigma_2 t}\mathbf{v}_2$ where $\mathbf{v}_1$ and $\mathbf{v}_2$ are the corresponding eigenvectors. Since both eigenvalues are negative, the system approaches zero as $t \to \infty$. With $\sigma_1 = -0.1$ and $\sigma_2 = -0.001$, the term $e^{-0.1t}$ decays much faster than $e^{-0.001t}$. After sufficient time, the $e^{-0.1t}$ term becomes negligible, and the solution is dominated by $c_2 e^{-0.001t}\mathbf{v}_2$. To reach within 1% of steady state (zero), we need the remaining component to satisfy $|c_2 e^{-0.001t}| = 0.01|c_2|$, giving us $e^{-0.001t} = 0.01$. Solving: $t = \frac{\ln(100)}{0.001} \approx 4606$ time units. Option A incorrectly uses the faster eigenvalue, missing that the slower mode determines convergence time. Option B doubles the fast eigenvalue calculation but still ignores the dominant slow mode. Option C attempts to average the eigenvalues, but the slowest mode alone controls the final approach to steady state. Key takeaway: In multi-mode systems, the eigenvalue closest to zero (smallest magnitude) determines how long it takes to reach steady state—the "bottleneck" principle governs convergence time.
17. Consider the system $\vec{x}_{k+1} = A\vec{x}_k$ where $A = \begin{pmatrix} k & 0.3 \\ 0 & 0.9 \end{pmatrix}$. For what range of values of $k$ is the origin an attractor?
    Answer: $-1 < k < 1$ For the origin to be an attractor, all eigenvalues of the matrix $A$ must have a magnitude less than 1. Since $A$ is an upper triangular matrix, its eigenvalues are its diagonal entries: $\lambda_1 = k$ and $\lambda_2 = 0.9$. We need to satisfy $|\lambda_i| < 1$ for both eigenvalues. For $\lambda_2$, we have $|0.9| = 0.9 < 1$, which is satisfied. For $\lambda_1$, we need $|k| < 1$. This inequality is equivalent to $-1 < k < 1$.
18. A discrete dynamical system is given by $\mathbf{x}_{n+1} = A\mathbf{x}_n$ where $A = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}$. If the system starts with initial condition $\mathbf{x}_0 = \begin{pmatrix} 100 \\ 50 \end{pmatrix}$, what can be concluded about the long-term behavior?
    Answer: The system converges to a steady state determined by the eigenvector corresponding to eigenvalue 1 For this matrix, the eigenvalues are λ₁ = 1 and λ₂ = 0.5. Since one eigenvalue equals 1 and the other has absolute value less than 1, the system converges to a steady state in the direction of the eigenvector for λ = 1. Choice A is wrong because stability depends on eigenvalues, not diagonal entries. Choice C is wrong because the eigenvalues are real. Choice D is wrong because column sums don't determine stability.
19. Consider the discrete dynamical system defined by $\vec{x}_{k+1} = A\vec{x}_k$, where $A = \begin{pmatrix} 1.3 & 0.4 \\ -0.5 & 0.5 \end{pmatrix}$. How is the origin classified for this system?
    Answer: An attractor (sink), because the eigenvalues are complex with magnitude less than 1. To classify the origin, we must find the eigenvalues of $A$ by solving the characteristic equation $\det(A - \lambda I) = 0$. This gives $(1.3-\lambda)(0.5-\lambda) - (0.4)(-0.5) = 0$, which simplifies to $\lambda^2 - 1.8\lambda + 0.85 = 0$. Using the quadratic formula, the eigenvalues are $\lambda = \frac{1.8 \pm \sqrt{(-1.8)^2 - 4(1)(0.85)}}{2} = 0.9 \pm 0.2i$. The stability of a discrete system is determined by the magnitude of its eigenvalues. The magnitude is $|0.9 \pm 0.2i| = \sqrt{0.9^2 + 0.2^2} = \sqrt{0.81 + 0.04} = \sqrt{0.85}$. Since $\sqrt{0.85} < 1$, both eigenvalues have a magnitude less than 1, so the origin is an attractor (sink).
20. Two similar dynamical systems have matrices $A_1$ with eigenvalues $\{0.9, 0.7\}$ and $A_2$ with eigenvalues $\{0.95, 0.3\}$. Both systems start from the same initial condition $\mathbf{x}_0$. How do their convergence properties compare?
    Answer: System 1 converges faster since its dominant eigenvalue $0.9$ is smaller than system 2's dominant eigenvalue $0.95$ The convergence rate is determined by the largest eigenvalue in absolute value (the dominant eigenvalue). System 1 has dominant eigenvalue 0.9, while system 2 has dominant eigenvalue 0.95. Since 0.9 < 0.95, system 1 converges faster. Choice A incorrectly focuses on eigenvalue spacing. Choice B incorrectly emphasizes the smallest eigenvalue. Choice D ignores the significant difference in dominant eigenvalues.

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