University:
The University Of Western Ontario
Course:
CALC 1000A | Calculus I
Academic year:

2020
Views:

273

Pages:

2
Author:

Kyuwon J.

Question 1: Eigenvalues and Eigenvectors Find the eigenvalues and eigenvectors of: 3 −1 A= 1 1 Step 1: Eigenvalues Solve the characteristic equation: det(A − λI) = 0 3 − λ −1 A − λI = 1 1−λ det(A − λI) = (3 − λ)(1 − λ) − (−1)(1) = λ2 − 4λ + 4 = (λ − 2)2 Eigenvalue: λ = 2 (repeated). Step 2: Eigenvectors Solve (A − 2I)v = 0: 1 −1 A − 2I = 1 −1 1 −1 x1 0 = 1 −1 x2 0 1 From the first row: x1 − x2 = 0 =⇒ x1 = x2 . Eigenvector: v = . 1 — Question 2: Diagonal Eigenvalues   2 1 0 For B = 0 3 0, verify that the eigenvalues are the diagonal elements. 0 0 4 Solution:   2−λ 1 0 3−λ 0  det(B − λI) = det  0 0 0 4−λ Expand along the diagonal: det(B − λI) = (2 − λ)(3 − λ)(4 − λ) Eigenvalues: λ = 2, 3, 4. The eigenvalues match the diagonal elements. — 1 Question 3: Determinant and Eigenvalues 2 3 Show that the determinant of A = equals the product of its eigenvalues. 1 4 Step 1: Determinant det(A) = 2 3 = (2)(4) − (3)(1) = 8 − 3 = 5 1 4 Step 2: Eigenvalues Solve det(A − λI) = 0: 2−λ 3 A − λI = 1 4−λ det(A − λI) = (2 − λ)(4 − λ) − (3)(1) = λ2 − 6λ + 5 (λ − 5)(λ − 1) = 0 Eigenvalues: λ = 5, 1. Step 3: Verify Determinant and Product of Eigenvalues The product of eigenvalues: λ1 · λ2 = 5 · 1 = 5 The determinant equals the product of the eigenvalues. 2