Module 6 Practice: Eigenvalues and Eigenvectors
10 problems covering eigenvalue computation, eigenvectors, diagonalization, and applications.
Problem 1
Find the eigenvalues of A = [3 1; 0 2].
Solution
A is upper triangular. Eigenvalues are the diagonal entries: lambda = 3 and lambda = 2.
Problem 2
Find the eigenvalues and eigenvectors of A = [2 1; 1 2].
Solution
p(lambda) = lambda^2 - 4*lambda + 3 = (lambda-3)(lambda-1) = 0. Eigenvalues: 3 and 1.
lambda = 3: A - 3I = [-1 1; 1 -1] → [1 -1; 0 0]. v = (1, 1).
lambda = 1: A - I = [1 1; 1 1] → [1 1; 0 0]. v = (-1, 1).
Problem 3
A 3x3 matrix has eigenvalues 1, -2, 4. Find det(A) and trace(A).
Solution
det(A) = 1 * (-2) * 4 = -8. trace(A) = 1 + (-2) + 4 = 3.
Problem 4
Find the characteristic polynomial of A = [1 -1; 2 4].
Solution
p(lambda) = lambda^2 - trace*lambda + det = lambda^2 - 5*lambda + (4+2) = lambda^2 - 5*lambda + 6 = (lambda-2)(lambda-3).
Eigenvalues: lambda = 2 and lambda = 3.
Problem 5
Diagonalize A = [5 4; 1 2] or explain why it cannot be done.
Solution
p(lambda) = lambda^2 - 7*lambda + 6 = (lambda-1)(lambda-6). Eigenvalues: 1 and 6 (distinct, so diagonalizable).
lambda = 1: A - I = [4 4; 1 1] → [1 1; 0 0]. v1 = (-1, 1).
lambda = 6: A - 6I = [-1 4; 1 -4] → [1 -4; 0 0]. v2 = (4, 1).
P = [-1 4; 1 1], D = [1 0; 0 6].
Problem 6
Is A = [2 1; 0 2] diagonalizable?
Solution
Eigenvalue lambda = 2 with algebraic multiplicity 2. A - 2I = [0 1; 0 0]. One free variable, geometric multiplicity = 1 < 2. Not diagonalizable.
Problem 7
If A = PDP^{-1} with D = [3 0; 0 -1], compute A^4.
Solution
A^4 = PD^4P^{-1}. D^4 = [3^4 0; 0 (-1)^4] = [81 0; 0 1]. So A^4 = P[81 0; 0 1]P^{-1}.
Problem 8
Find the steady-state vector for the stochastic matrix A = [0.5 0.3; 0.5 0.7].
Solution
(A - I)q = 0: [-0.5 0.3; 0.5 -0.3]q = 0. So 0.5*q1 = 0.3*q2, q1 = 0.6*q2.
Normalize: 0.6*q2 + q2 = 1.6*q2 = 1. q2 = 5/8, q1 = 3/8. Steady state: (3/8, 5/8) = (0.375, 0.625).
Problem 9
The system x_{k+1} = Ax_k has eigenvalues 0.9 and -0.4. Does the system converge or diverge?
Solution
|0.9| = 0.9 < 1 and |-0.4| = 0.4 < 1. Both eigenvalues have magnitude less than 1, so the system converges to 0.
Problem 10
Find the eigenvalues of A = [0 1 0; 0 0 1; 6 -11 6] given that p(lambda) = -lambda^3 + 6*lambda^2 - 11*lambda + 6.
Solution
Factor: lambda^3 - 6*lambda^2 + 11*lambda - 6 = (lambda-1)(lambda-2)(lambda-3) = 0.
Eigenvalues: lambda = 1, 2, 3. Check: product = 6 = det, sum = 6 = trace.