This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: MATH 235 Solutions Assignment 4b 1. Let A be an n n matrix. Also, let , , be three distinct eigenvalues of A having corresponding eigenvectors x , y , z , respectively. Consider the vector v = x + y + z . Can v be an eigenvector of A corresponding to an eigenvalue (possibly different from , , ? Explain. Solution : Given A x = x, A y = y , A z = z and v = x + y + z , then A v = A x + A y + A z = x + y z . (1) Assume v is an eigenvector of A corresponding to the eigenvalue . Then A v = v . From (1), it follows that x + y + z = parenleftBig x + y + z parenrightBig ( - ) x +( - ) y +( - ) z = . (2) Since the eigenvectors x, y , z belong to distinct eigenvalues, the eigenvectors x, y , z are linearly independent. Thus, (2) gives that - = - = - = 0 = =...
View Full Document