University:
The University Of Western Ontario
Course:
MATH 3020A | Introduction to Abstract Algebra
Academic year:

2024
Views:

241

Pages:

3
Author:

Kyuwon J.

Practice Exam: Eigenvectors and Characteristic Polynomials Instructions Answer the following questions. Show all your work for full credit. — Question 1: Eigenvalues of a 2x2 Matrix Let: 4 A= 2 1 3 1. Find the eigenvalues of A. 2. Find the eigenvectors corresponding to each eigenvalue. — Solution to Question 1 Part (a): Find the Eigenvalues The characteristic equation is: det(A − λI) = 0 4−λ 1 A − λI = 2 3−λ det(A − λI) = (4 − λ)(3 − λ) − (2)(1) = λ2 − 7λ + 10 Solve: λ2 − 7λ + 10 = 0 (λ − 5)(λ − 2) = 0 λ1 = 5, Part (b): Find the Eigenvectors For λ1 = 5: A − 5I = Solve: −1 2 1 −2 λ2 = 2 −1 2 1 −2 x 0 = y 0 From the first row: −x + y = 0 ⇒ y=x Eigenvector: 1 v1 = 1 For λ2 = 2: Solve: 2 A − 2I = 2 2 2 1 1 1 x 0 = 1 y 0 From the first row: 2x + y = 0 ⇒ y = −2x Eigenvector: 1 v2 = −2 — 1 Question 2: Eigenvalues of a 3x3 Matrix Let:  3 B = 0 1  1 0 3 0 3 0 1. Compute the characteristic polynomial of B. 2. Find all eigenvalues of B. 3. Determine the eigenvector corresponding to λ = 4. — Solution to Question 2 Part (a): Compute the Characteristic Polynomial   3−λ 0 1 3−λ 0  B − λI =  0 1 0 3−λ det(B − λI) = (3 − λ) 3−λ 0 0 0 − (1) 1 3−λ = (3 − λ)((3 − λ)2 ) − 0 = (3 − λ)3 p(λ) = (3 − λ)3 Part (b): Find All Eigenvalues The eigenvalue is: λ = 3 (with multiplicity 3) Part (c): Determine the Eigenvector for λ = 3  0 0 B − 3I = 0 0 1 0 Solve:  0 0 1 0 0 0     0 x 1 0 y  = 0 z 0 0 From the first and third rows: z = 0, x=0 Eigenvector:   0 v = 1 0 — Question 3: Application of Eigenvectors Consider the transformation matrix:  1 0 0 2 C= 1 1. Compute the eigenvalues of C. 2. Find a basis of eigenvectors for C. 3. Describe the geometric effect of C on any vector. — 2 1 2 0 3−λ Solution to Question 3 Part (a): Compute the Eigenvalues C − λI = 2−λ 1 1 2−λ det(C − λI) = (2 − λ)2 − 1 = λ2 − 4λ + 3 (λ − 3)(λ − 1) = 0 λ1 = 3, λ2 = 1 Part (b): Find a Basis of Eigenvectors For λ1 = 3: −1 C − 3I = 1 1 −1 Solve: x=y Eigenvector: 1 v1 = 1 For λ2 = 1: 1 C −I = 1 1 1 Solve: x = −y Eigenvector: 1 −1 v2 = Part (c): Geometric Effect of C 1 1 The transformation scales vectors along v1 = by 3 and along v2 = by 1. 1 −1 3