M 340L HW 10 Solutions

Thus x0 c1 v1 c2 v2 but v1 v2 are eigenvectors

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nd x0 = 3v1 + 4v2 . 1 2k 6. xk = 6(2k )v1 + 2 v2 1 2k v2 Explanation: Since v1 , v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus x0 = c1 v1 + c2 v2 . But v1 , v2 are eigenvectors corresponding to 1 respective eigenvalues 2, 2 so To compute c1 , c2 we apply row reduction to the augmented matrix x1 = Ax0 = 3Av1 + 4Av2 −1 1 +2 = 6v1 + 2v2 = 6 1 1 [ v1 v2 x0 ] = . Consequently, 1 5 ∼ 0 1 1 −1 11 03 , 1 −2 for then c1 = 3, c2 = −2 and x0 = 3v1 − 2v2 . x1 = 4 8 But v1 , v2 are eigenvectors corresponding to 1 respective eigenvalues 2, 2 so set . . 1 2k xk = 3(2k )v1 − 2 003 10.0 points When A is a 2 × 2 matrix with eigenvalues 1 2, 2 and corresponding eigenvectors 1 , v1 = 1 −1 , v2 = 1 determine the solution {xk } of the difference equation xk+1 = Axk , x0 = 5 . 1 v2 for k ≥ 0. Then x0 is the given initial value and Axk = 3 2k Av1 − 2 = 3 2k+1 v1 − 2 1 2k Av2 1 2k+1 v2 = xk+1 , so {xk } solve the difference equation. 004 10.0 points cruz (fmc326)...
View Full Document

This document was uploaded on 03/16/2014 for the course M 340L at University of Texas.

Ask a homework question - tutors are online