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

2024
Views:

401

Pages:

3
Author:

Kyuwon J.

Practice Problems on Eigenvectors and Characteristic Polynomials Problem 1: Eigenvalues and Eigenvectors of a 2x2 Matrix Let: 3 A= 0 1 2 Find: 1. The eigenvalues of A, 2. The eigenvectors corresponding to each eigenvalue. Solution Step 1: Compute the Characteristic Polynomial The characteristic equation is: det(A − λI) = 0 3−λ 1 A − λI = 0 2−λ det(A − λI) = (3 − λ)(2 − λ) − (0)(1) = (3 − λ)(2 − λ) p(λ) = λ2 − 5λ + 6 Step 2: Solve for Eigenvalues Solve: λ2 − 5λ + 6 = 0 Factoring: (λ − 3)(λ − 2) = 0 The eigenvalues are: λ1 = 3, Step 3: Find Eigenvectors For λ1 = 3: λ2 = 2 0 A − 3I = 0 Solve: 0 0 1 −1 1 x 0 = −1 y 0 From the first row: y=0 1 Thus, v1 = . 0 For λ2 = 2: A − 2I = Solve: 1 0 1 0 1 0 1 x 0 = 0 y 0 From the first row: x+y =0 ⇒ 1 Thus, v2 = . −1 Final Answer: • Eigenvalues: λ1 = 3, λ2 = 2 1 1 • Eigenvectors: v1 = , v2 = 0 −1 — 1 y = −x Problem 2: Eigenvalues and Eigenvectors of a 3x3 Matrix Let:  4 B = 0 0  0 0 2 1 4 0 Find: 1. The eigenvalues of B, 2. The eigenvectors corresponding to each eigenvalue. Solution Step 1: Compute the Characteristic Polynomial The characteristic equation is: det(B − λI) = 0   4−λ 1 0 4−λ 0  B − λI =  0 0 0 2−λ det(B − λI) = (4 − λ)2 (2 − λ) The characteristic polynomial is: p(λ) = (4 − λ)2 (2 − λ) Step 2: Solve for Eigenvalues The eigenvalues are: λ1 = 4 (with multiplicity 2), Step 3: Find Eigenvectors For λ1 = 4: Solve:  0 B − 4I = 0 0  0 0 0 1 0 0 λ2 = 2  0 0 −2 1 0 0     0 x 0 0  y  = 0 0 z −2 From the first row: y=0 From the third row: z=0   1 Thus, v1 = 0. 0 For λ2 = 2: Solve:  2 B − 2I = 0 0  2 0 0 1 2 0 1 2 0  0 0 0     0 x 0 0 y  = 0 0 z 0 From the first row: 2x + y = 0  ⇒  1 Thus, v2 = −2. 0 Final Answer: 2 y = −2x • Eigenvalues: λ1 = 4 (multiplicity 2), λ2 = 2     1 1 • Eigenvectors: v1 = 0 , v2 = −2 0 0 — Problem 3: Application of Eigenvectors Consider a matrix C = 1 1 1 . Find the eigenvalues and eigenvectors. Interpret the results in terms of 1 the transformation Cv. Solution Step 1: Compute the Characteristic Polynomial 1−λ C − λI = 1 1 1−λ det(C − λI) = (1 − λ)2 − 1 = λ2 − 2λ p(λ) = λ(λ − 2) Step 2: Solve for Eigenvalues The eigenvalues are: λ1 = 0, λ2 = 2 C − 0I = 1 1 Step 3: Find Eigenvectors For λ1 = 0: Solve: 1 1 1 1 1 x 0 = 1 y 0 ⇒ x+y =0 y = −x 1 Thus, v1 = . −1 For λ2 = 2: C − 2I = −1 1 Solve: x=y 1 Thus, v2 = . 1 Final Answer: • Eigenvalues: λ1 = 0, λ2 = 2 1 1 • Eigenvectors: v1 = , v2 = −1 1 3 1 −1