Linear Algebra Quiz: Stability In Linear Systems
20 questions · exam conditions
0:00
Stability In Linear SystemsQuestion 1 of 20

The origin is classified as a saddle point for the discrete dynamical system x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k, where AA is a 2×22 \times 2 matrix with real entries. Which statement about the eigenvalues of AA must be true?

One eigenvalue λ1\lambda_1 satisfies ∣λ1∣>1|\lambda_1| > 1 and the other eigenvalue λ2\lambda_2 satisfies ∣λ2∣<1|\lambda_2| < 1.
The eigenvalues are real numbers, with one being positive and the other negative.
The determinant of AA must be negative, ensuring that the eigenvalues have opposite signs.
The real parts of the eigenvalues must have opposite signs, one being positive and one being negative.
← Back to quizzes

Linear Algebra Quiz

Linear Algebra Quiz: Stability In Linear Systems

Practice Stability In Linear Systems in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Stability In Linear Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The origin is classified as a saddle point for the discrete dynamical system x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k, where AA is a 2×22 \times 2 matrix with real entries. Which statement about the eigenvalues of AA must be true?

  1. One eigenvalue λ1\lambda_1 satisfies ∣λ1∣>1|\lambda_1| > 1 and the other eigenvalue λ2\lambda_2 satisfies ∣λ2∣<1|\lambda_2| < 1. (correct answer)
  2. The eigenvalues are real numbers, with one being positive and the other negative.
  3. The determinant of AA must be negative, ensuring that the eigenvalues have opposite signs.
  4. The real parts of the eigenvalues must have opposite signs, one being positive and one being negative.
Explanation: 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 λ1=2,λ2=0.5\lambda_1=2, \lambda_2=0.5 produce a saddle point, but both are positive. Eigenvalues λ1=2,λ2=−2\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.

Question 2

A student concludes that the origin is a stable attractor for the discrete system x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k with A=(−0.700.5−1.2)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?

  1. The conclusion is flawed because stability in discrete systems requires all eigenvalues to have a magnitude less than 1, and ∣−1.2∣>1|-1.2| > 1. (correct answer)
  2. The reasoning is correct, but the conclusion is wrong; two negative eigenvalues always mean the origin is a repeller, not an attractor.
  3. The reasoning is flawed because stability is determined by the determinant; since det⁡(A)=0.84<1\det(A)=0.84 < 1, the origin is actually stable.
  4. The reasoning is flawed because stability is determined by the signs of the real parts, which is a concept from continuous systems, not the signs of the eigenvalues themselves.
Explanation: 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: λ1=−0.7\lambda_1 = -0.7 and λ2=−1.2\lambda_2 = -1.2. For the origin to be an attractor, all eigenvalues must satisfy ∣λ∣<1|\lambda| < 1. Here, ∣λ1∣=∣−0.7∣=0.7<1|\lambda_1| = |-0.7| = 0.7 < 1, but ∣λ2∣=∣−1.2∣=1.2>1|\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.

Question 3

The characteristic polynomial of a 2×22 \times 2 matrix AA for a discrete system is p(λ)=λ2−0.5λ+cp(\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 cc?

  1. c=1.1c = 1.1
  2. c=−0.7c = -0.7
  3. c=0.4c = 0.4 (correct answer)
  4. c=−0.5c = -0.5
Explanation: For a 2×22 \times 2 matrix, the characteristic polynomial is λ2−tr(A)λ+det⁡(A)=0\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0. So, tr(A)=0.5\text{tr}(A) = 0.5 and c=det⁡(A)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|D| < 1, (2) 1−T+D>01 - T + D > 0, and (3) 1+T+D>01 + T + D > 0. In our case, T=0.5T=0.5 and D=cD=c. (1) ∣c∣<1|c| < 1, which means −1<c<1-1 < c < 1. This eliminates choice A. (2) 1−0.5+c>0  ⟹  0.5+c>0  ⟹  c>−0.51 - 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  ⟹  1.5+c>0  ⟹  c>−1.51 + 0.5 + c > 0 \implies 1.5 + c > 0 \implies c > -1.5. This condition is weaker than c>−0.5c > -0.5. Combining the conditions, we need −0.5<c<1-0.5 < c < 1. Only choice C, c=0.4c=0.4, falls within this range.

Question 4

For the discrete dynamical system x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k with A=(0.80.301.5)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?

  1. Along the direction of the vector (1−1)\begin{pmatrix} 1 \\ -1 \end{pmatrix}.
  2. Along the direction of the vector (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix}.
  3. Along the direction of the vector (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}.
  4. Along the direction of the vector (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix}. (correct answer)
Explanation: 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=(0.80.301.5)A = \begin{pmatrix} 0.8 & 0.3 \\ 0 & 1.5 \end{pmatrix} , since it's upper triangular, the eigenvalues are simply the diagonal entries: λ1=0.8\lambda_1 = 0.8 and λ2=1.5\lambda_2 = 1.5. Since ∣λ1∣<1|\lambda_1| < 1, trajectories decay (are attracted to the origin) along its eigenvector direction. Since ∣λ2∣>1|\lambda_2| > 1, trajectories grow (are repelled from the origin) along its eigenvector direction. To find the eigenvector for λ1=0.8\lambda_1 = 0.8, solve (A−0.8I)v⃗=0⃗(A - 0.8I)\vec{v} = \vec{0}: (00.300.7)(xy)=(00)\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=00.3y = 0 and 0.7y=00.7y = 0, so y=0y = 0. The eigenvector is $$ \begin{pmatrix} 1 \ 0 \end{pmatrix} Choice A, $$ \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.

Question 5

Consider the discrete system x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k where A=(0.50.50.50.5)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 x⃗0\vec{x}_0?

  1. All non-zero trajectories are repelled from the origin and go to infinity.
  2. All trajectories converge to the origin, making it a global attractor.
  3. Trajectories converge to a specific point on the line y=xy=x, with the specific point depending on x⃗0\vec{x}_0. (correct answer)
  4. Trajectories oscillate between two points and never converge.
Explanation: First, we find the eigenvalues of AA. The characteristic equation is (0.5−λ)2−0.25=0(0.5-\lambda)^2 - 0.25 = 0, which gives λ2−λ=0\lambda^2 - \lambda = 0, so λ1=1\lambda_1 = 1 and λ2=0\lambda_2 = 0. The eigenvector for λ1=1\lambda_1=1 is found by solving (A−I)v⃗=0⃗(A-I)\vec{v}=\vec{0}, which gives −0.5x+0.5y=0-0.5x + 0.5y = 0, so the eigenspace is the line y=xy=x, spanned by (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}. The eigenvector for λ2=0\lambda_2=0 is spanned by (1−1)\begin{pmatrix} 1 \\ -1 \end{pmatrix}. Any initial vector can be written as x⃗0=c1(11)+c2(1−1)\vec{x}_0 = c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix}. Then x⃗k=Akx⃗0=c1(1)k(11)+c2(0)k(1−1)\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≥1k \ge 1, this becomes x⃗k=c1(11)\vec{x}_k = c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix}. Thus, the system converges to the point c1(11)c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix}, which lies on the line y=xy=x. The exact point depends on the component of x⃗0\vec{x}_0 in the direction of the first eigenvector.

Question 6

For the discrete system x⃗k+1=(2000.5)x⃗k\vec{x}_{k+1} = \begin{pmatrix} 2 & 0 \\ 0 & 0.5 \end{pmatrix} \vec{x}_k, which of the following sets of initial vectors x⃗0\vec{x}_0 will produce a trajectory that converges to the origin as k→∞k \to \infty?

  1. Only the zero vector x⃗0=(00)\vec{x}_0 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.
  2. Any vector lying on the x-axis, of the form (c0)\begin{pmatrix} c \\ 0 \end{pmatrix}.
  3. Any vector lying on the y-axis, of the form (0c)\begin{pmatrix} 0 \\ c \end{pmatrix}. (correct answer)
  4. Any vector x⃗0\vec{x}_0 for which the sum of its components is zero.
Explanation: The matrix is diagonal, so its eigenvalues are the diagonal entries λ1=2\lambda_1 = 2 and λ2=0.5\lambda_2 = 0.5. The corresponding eigenvectors are the standard basis vectors v⃗1=(10)\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} (the x-axis) and v⃗2=(01)\vec{v}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} (the y-axis). Since ∣λ1∣>1|\lambda_1| > 1 and ∣λ2∣<1|\lambda_2| < 1, the origin is a saddle point. The direction associated with λ1\lambda_1 is unstable (repelling), and the direction associated with λ2\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 x⃗0=(c1c2)\vec{x}_0 = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}, the trajectory is x⃗k=(c1(2k)c2(0.5k))\vec{x}_k = \begin{pmatrix} c_1(2^k) \\ c_2(0.5^k) \end{pmatrix}. This converges to (00)\begin{pmatrix} 0 \\ 0 \end{pmatrix} only if c1=0c_1=0. Therefore, the initial vector must be of the form (0c2)\begin{pmatrix} 0 \\ c_2 \end{pmatrix}, which is any vector on the y-axis.

Question 7

For a discrete system x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k with A=(4−211)A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}, and a generic initial vector x⃗0\vec{x}_0 not in an eigenspace, the vector x⃗k\vec{x}_k will align with a specific direction as k→∞k \to \infty. What is this long-term direction?

  1. Parallel to the vector (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}.
  2. Parallel to the vector (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}. (correct answer)
  3. Parallel to the vector (1−2)\begin{pmatrix} 1 \\ -2 \end{pmatrix}.
  4. Parallel to the vector (41)\begin{pmatrix} 4 \\ 1 \end{pmatrix}.
Explanation: 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−λ)(1−λ)−(−2)(1)=0(4-\lambda)(1-\lambda) - (-2)(1) = 0, which is λ2−5λ+6=0\lambda^2 - 5\lambda + 6 = 0, or (λ−3)(λ−2)=0(\lambda-3)(\lambda-2) = 0. The eigenvalues are λ1=3\lambda_1 = 3 and λ2=2\lambda_2 = 2. The dominant eigenvalue is λ1=3\lambda_1 = 3. To find its eigenvector, we solve (A−3I)v⃗=0⃗(A - 3I)\vec{v} = \vec{0}: (1−21−2)(xy)=(00)\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=0x - 2y = 0, so the eigenvector is any scalar multiple of (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}.

Question 8

Consider two stable discrete systems, System 1 governed by A=(0.500.10.6)A = \begin{pmatrix} 0.5 & 0 \\ 0.1 & 0.6 \end{pmatrix} and System 2 governed by B=(0.8−0.200.7)B = \begin{pmatrix} 0.8 & -0.2 \\ 0 & 0.7 \end{pmatrix}. Which statement correctly compares their long-term behavior?

  1. The long-term behavior cannot be compared without knowing the initial vectors.
  2. System 2 converges to the origin faster than System 1.
  3. Both systems converge to the origin at the same rate.
  4. System 1 converges to the origin faster than System 2. (correct answer)
Explanation: When analyzing discrete dynamical systems of the form xn+1=Mxn\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=(0.500.10.6)A = \begin{pmatrix} 0.5 & 0 \\ 0.1 & 0.6 \end{pmatrix} , we need the eigenvalues. The characteristic polynomial is (0.5−λ)(0.6−λ)−0=λ2−1.1λ+0.3(0.5-\lambda)(0.6-\lambda) - 0 = \lambda^2 - 1.1\lambda + 0.3. Using the quadratic formula: λ=1.1±1.21−1.22=1.1±0.12\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} SinceSystem1hasspectralradius0.6<0.8(System2′sspectralradius),System1convergesfaster.ThisconfirmsanswerD.AnswerAiswrongbecauseconvergenceratedependsonlyonthematrixeigenvalues,notinitialconditions.AnswerBincorrectlyreversesthecomparison—System2′slargerspectralradiusmeansslowerconvergence.AnswerCiswrongbecausethespectralradiiareclearlydifferent(0.6≠0.8).∗∗Studytip:∗∗Fordiscretesystemconvergencerates,alwayscomputethespectralradius.Smallerspectralradius=fasterconvergencetoequilibrium. 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.

Question 9

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?

  1. The populations of the two towns will become equal.
  2. The population of Northville will stabilize at half the population of Southtown. (correct answer)
  3. The population of Southtown will stabilize at half the population of Northville.
  4. All residents will eventually move to Northville, the more attractive town.
Explanation: Let nkn_k and sks_k be the populations of Northville and Southtown in year kk. The system is described by: nk+1=0.90nk+0.05skn_{k+1} = 0.90n_k + 0.05s_k and sk+1=0.10nk+0.95sks_{k+1} = 0.10n_k + 0.95s_k. The transition matrix is A=(0.900.050.100.95)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 λ=1\lambda=1. We solve (A−I)v⃗=0⃗(A-I)\vec{v} = \vec{0}, which is (−0.100.050.10−0.05)(ns)=(00)\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=0.05s0.10n = 0.05s, or s=2ns = 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.

Question 10

A population model tracks two competing species with transition matrix P=(0.70.10.21.2)P = \begin{pmatrix} 0.7 & 0.1 \\ 0.2 & 1.2 \end{pmatrix} . After computing eigenvalues λ1=0.6\lambda_1 = 0.6 and λ2=1.3\lambda_2 = 1.3, what is the most accurate description of the system's long-term behavior?

  1. Both populations stabilize at finite levels since one eigenvalue is less than 1 and one is greater than 1
  2. The system exhibits exponential growth in the direction of the eigenvector for λ2=1.3\lambda_2 = 1.3 while decay occurs perpendicular to it (correct answer)
  3. The populations oscillate between growth and decay since the eigenvalues have different signs relative to unity
  4. Population 2 grows exponentially while population 1 decays since P22>1P_{22} > 1 and P11<1P_{11} < 1
Explanation: 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.

Question 11

For the discrete system yn+1=Byn\mathbf{y}_{n+1} = B\mathbf{y}_n, suppose BB has eigenvalues μ1=−0.8\mu_1 = -0.8 and μ2=0.6\mu_2 = 0.6. Starting from a generic initial condition, which statement best describes the solution's behavior?

  1. The solution converges monotonically to the origin since both eigenvalues have absolute value less than 1
  2. The solution oscillates with increasing amplitude since one eigenvalue is negative and one is positive
  3. The solution exhibits damped oscillations approaching the origin due to the negative eigenvalue with largest absolute value (correct answer)
  4. The solution approaches a steady state along the eigenvector for μ2=0.6\mu_2 = 0.6 with alternating signs each iteration
Explanation: When analyzing discrete linear systems, the eigenvalues of the coefficient matrix determine the long-term behavior of solutions. For the system yn+1=Byn\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 μ1=−0.8\mu_1 = -0.8 and μ2=0.6\mu_2 = 0.6, the solution has the general form yn=c1(−0.8)nv1+c2(0.6)nv2\mathbf{y}_n = c_1(-0.8)^n\mathbf{v}_1 + c_2(0.6)^n\mathbf{v}_2, where v1\mathbf{v}_1 and v2\mathbf{v}_2 are the corresponding eigenvectors. Since ∣−0.8∣=0.8>∣0.6∣=0.6|-0.8| = 0.8 > |0.6| = 0.6, the first term dominates the behavior. The negative eigenvalue (−0.8)n(-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 v2\mathbf{v}_2 because that component also decays to zero, and the alternating signs come from the μ1\mu_1 term, not μ2\mu_2. Study tip: For discrete systems, eigenvalue magnitude determines stability (∣μ∣<1|\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.

Question 12

Consider a continuous linear system x˙=Cx\dot{\mathbf{x}} = C\mathbf{x} where CC has eigenvalues α=0\alpha = 0 and β=−0.5\beta = -0.5. If the eigenvector for α=0\alpha = 0 is $$\mathbf{v}_1 = \begin{pmatrix} 1 \ 2 \end{pmatrix}

  1. All solutions converge to the origin since the largest eigenvalue is zero, which indicates marginal stability
  2. Solutions oscillate around the line spanned by v1\mathbf{v}_1 with decreasing amplitude governed by β=−0.5\beta = -0.5
  3. The system is unstable because having a zero eigenvalue creates a degenerate case with unbounded growth
  4. Solutions approach the line spanned by v1\mathbf{v}_1 with the approach rate determined by the β=−0.5\beta = -0.5 eigenvalue (correct answer)
Explanation: When analyzing linear dynamical systems x˙=Cx\dot{\mathbf{x}} = C\mathbf{x}, the eigenvalues of matrix CC completely determine the system's long-term behavior. Each eigenvalue governs how solutions behave along its corresponding eigenvector direction. With eigenvalues α=0\alpha = 0 and β=−0.5\beta = -0.5, the general solution is x(t)=c1e0⋅tv1+c2e−0.5tv2=c1v1+c2e−0.5tv2\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→∞t \to \infty, the exponential term e−0.5t→0e^{-0.5t} \to 0, so solutions approach c1v1c_1\mathbf{v}_1—a point on the line spanned by v1\mathbf{v}_1. The rate of approach is governed by how quickly e−0.5te^{-0.5t} decays, which depends on β=−0.5\beta = -0.5. This confirms answer D. A is wrong because solutions don't converge to the origin unless c1=0c_1 = 0. The zero eigenvalue means the v1\mathbf{v}_1 component persists forever—solutions approach the line spanned by v1\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.

Question 13

A mechanical system satisfies xk+1=Mxk\mathbf{x}_{k+1} = M\mathbf{x}_k where MM has characteristic polynomial p(λ)=λ2−0.6λ+0.25p(\lambda) = \lambda^2 - 0.6\lambda + 0.25. What can be determined about the stability without computing eigenvectors?

  1. The system is stable since both roots satisfy ∣λ∣<1|\lambda| < 1, as verified by p(1)>0p(1) > 0 and p(−1)>0p(-1) > 0 (correct answer)
  2. The system is unstable since p(0)=0.25>0p(0) = 0.25 > 0, indicating both eigenvalues are positive and potentially greater than 1
  3. The system is marginally stable since the discriminant Δ=0.36−1<0\Delta = 0.36 - 1 < 0 gives complex eigenvalues on the unit circle
  4. Stability cannot be determined without explicitly solving p(λ)=0p(\lambda) = 0 since the coefficient tests are inconclusive
Explanation: 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.

Question 14

A discrete dynamical system is given by xn+1=Axn\mathbf{x}_{n+1} = A\mathbf{x}_n where A=(0.80.30.20.7)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}

  1. The system converges to the zero vector since both diagonal entries are less than 1
  2. The system converges to a steady state determined by the eigenvector corresponding to eigenvalue 1 (correct answer)
  3. The system oscillates indefinitely since the matrix has complex eigenvalues with modulus less than 1
  4. The system grows without bound since the sum of each column exceeds 1
Explanation: 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.

Question 15

For the continuous system dudt=Qu\frac{d\mathbf{u}}{dt} = Q\mathbf{u}, matrix QQ has eigenvalues σ1=−0.1\sigma_1 = -0.1 and σ2=−0.001\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?

  1. About 46 time units, determined by ln⁡(100)/0.1\ln(100)/0.1 since the faster eigenvalue dominates the approach
  2. About 92 time units, determined by 2ln⁡(100)/0.12\ln(100)/0.1 since both eigenvalue effects must be considered simultaneously
  3. About 230 time units, determined by the average ln⁡(100)/0.0505\ln(100)/0.0505 since both eigenvalues contribute equally
  4. About 4606 time units, determined by ln⁡(100)/0.001\ln(100)/0.001 since the slower eigenvalue determines when steady state is reached (correct answer)
Explanation: 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 dudt=Qu\frac{d\mathbf{u}}{dt} = Q\mathbf{u} with negative eigenvalues, the general solution is u(t)=c1eσ1tv1+c2eσ2tv2\mathbf{u}(t) = c_1 e^{\sigma_1 t}\mathbf{v}_1 + c_2 e^{\sigma_2 t}\mathbf{v}_2 where v1\mathbf{v}_1 and v2\mathbf{v}_2 are the corresponding eigenvectors. Since both eigenvalues are negative, the system approaches zero as t→∞t \to \infty. With σ1=−0.1\sigma_1 = -0.1 and σ2=−0.001\sigma_2 = -0.001, the term e−0.1te^{-0.1t} decays much faster than e−0.001te^{-0.001t}. After sufficient time, the e−0.1te^{-0.1t} term becomes negligible, and the solution is dominated by c2e−0.001tv2c_2 e^{-0.001t}\mathbf{v}_2. To reach within 1% of steady state (zero), we need the remaining component to satisfy ∣c2e−0.001t∣=0.01∣c2∣|c_2 e^{-0.001t}| = 0.01|c_2|, giving us e−0.001t=0.01e^{-0.001t} = 0.01. Solving: t=ln⁡(100)0.001≈4606t = \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.

Question 16

Two similar dynamical systems have matrices A1A_1 with eigenvalues {0.9,0.7}\{0.9, 0.7\} and A2A_2 with eigenvalues {0.95,0.3}\{0.95, 0.3\}. Both systems start from the same initial condition x0\mathbf{x}_0. How do their convergence properties compare?

  1. System 1 converges faster overall since both eigenvalues are closer to each other, reducing oscillatory transients
  2. System 2 converges faster since its smallest eigenvalue 0.30.3 is much smaller than system 1's smallest eigenvalue 0.70.7
  3. System 1 converges faster since its dominant eigenvalue 0.90.9 is smaller than system 2's dominant eigenvalue 0.950.95 (correct answer)
  4. The convergence rates are essentially equal since both systems have all eigenvalues less than 1, ensuring stability
Explanation: 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.

Question 17

Consider the linear system dxdt=Ax\frac{d\mathbf{x}}{dt} = A\mathbf{x} where A=(−214−3)A = \begin{pmatrix} -2 & 1 \\ 4 & -3 \end{pmatrix} . The eigenvalues of AA are λ1=−1+i\lambda_1 = -1 + i and λ2=−1−i\lambda_2 = -1 - i. What type of stability does this equilibrium point exhibit?

  1. Asymptotically stable with oscillatory approach since both eigenvalues have negative real parts and nonzero imaginary parts (correct answer)
  2. Unstable since the eigenvalues are complex, causing the solution to spiral outward indefinitely
  3. Neutrally stable since the real parts are negative but the imaginary parts cause persistent oscillations
  4. Asymptotically stable with monotonic approach since the eigenvalues have magnitude less than 1
Explanation: 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.

Question 18

Consider the discrete dynamical system defined by x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k, where A=(1.30.4−0.50.5)A = \begin{pmatrix} 1.3 & 0.4 \\ -0.5 & 0.5 \end{pmatrix}. How is the origin classified for this system?

  1. An attractor (sink), because the eigenvalues are complex with magnitude less than 1. (correct answer)
  2. A repeller (source), because the trace of the matrix is greater than 1.
  3. A saddle point, because one eigenvalue has a magnitude greater than 1 and one is less than 1.
  4. A center, because the eigenvalues are complex, leading to purely rotational trajectories.
Explanation: To classify the origin, we must find the eigenvalues of AA by solving the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0. This gives (1.3−λ)(0.5−λ)−(0.4)(−0.5)=0(1.3-\lambda)(0.5-\lambda) - (0.4)(-0.5) = 0, which simplifies to λ2−1.8λ+0.85=0\lambda^2 - 1.8\lambda + 0.85 = 0. Using the quadratic formula, the eigenvalues are λ=1.8±(−1.8)2−4(1)(0.85)2=0.9±0.2i\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±0.2i∣=0.92+0.22=0.81+0.04=0.85|0.9 \pm 0.2i| = \sqrt{0.9^2 + 0.2^2} = \sqrt{0.81 + 0.04} = \sqrt{0.85}. Since 0.85<1\sqrt{0.85} < 1, both eigenvalues have a magnitude less than 1, so the origin is an attractor (sink).

Question 19

Consider the system x⃗k+1=Ax⃗k\vec{x}_{k+1} = A\vec{x}_k where A=(k0.300.9)A = \begin{pmatrix} k & 0.3 \\ 0 & 0.9 \end{pmatrix}. For what range of values of kk is the origin an attractor?

  1. k<1k < 1
  2. 0≤k<10 \le k < 1
  3. −0.9<k<0.9-0.9 < k < 0.9
  4. −1<k<1-1 < k < 1 (correct answer)
Explanation: For the origin to be an attractor, all eigenvalues of the matrix AA must have a magnitude less than 1. Since AA is an upper triangular matrix, its eigenvalues are its diagonal entries: λ1=k\lambda_1 = k and λ2=0.9\lambda_2 = 0.9. We need to satisfy ∣λi∣<1|\lambda_i| < 1 for both eigenvalues. For λ2\lambda_2, we have ∣0.9∣=0.9<1|0.9| = 0.9 < 1, which is satisfied. For λ1\lambda_1, we need ∣k∣<1|k| < 1. This inequality is equivalent to −1<k<1-1 < k < 1.

Question 20

The trajectories of a 2D discrete dynamical system x⃗k+1=Ax⃗k\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 AA?

  1. λ=0.5±0.5i\lambda = 0.5 \pm 0.5i
  2. λ1=1.2\lambda_1 = 1.2, λ2=0.8\lambda_2 = 0.8
  3. λ=1.1±0.4i\lambda = 1.1 \pm 0.4i (correct answer)
  4. λ=±i\lambda = \pm i
Explanation: The behavior of trajectories is determined by the eigenvalues of AA. 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±0.5i∣=0.52+0.52=0.5<1|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±0.4i∣=1.12+0.42=1.21+0.16=1.37>1|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) ∣±i∣=02+12=1|\pm i| = \sqrt{0^2 + 1^2} = 1. This would be a center, with trajectories orbiting but not moving outward.