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: 1 11 1 where k is a constant. This tells us thatv 1 + v 2v 3 + v 4 = 0. Therefore, they are not linearly independent. No solution exists for 1 1 0 0 1 0 0 1 0 1 1 0 0 0 1 1 c 1 c 2 c 3 c 4 = 1 so they do not span R 4 . Problem 2.3.26 a) True, any vector linearly independent of the basis vectors for S will extend the basis to R 6 . b) False, because the basis vectors may not be in S . Problem 2.3.42 Basis 1 , x, x 2 , x 3 ; basis x1 , x 21 , x 31 when p (1) = 0. 2...
View
Full
Document
This note was uploaded on 03/02/2008 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.
 Spring '08
 Neely

Click to edit the document details