Unformatted text preview: R and u 6∈ S, v ∈ S and v 6∈ R. Consider the vector u + v . As we are assuming R ∪ S is a subspace, R ∪ S is closed under addition. Hence u + v ∈ R ∪ S and so u + v ∈ R or u + v ∈ S . However if u + v ∈ R , then v = ( u + v )u ∈ R , which is a contradiction; similarly if u + v ∈ S . Hence we have derived a contradiction on the asumption that R 6⊆ S and S 6⊆ R . Consequently at least one of these must be false. In other words R ⊆ S or S ⊆ R . 19. Let X 1 ,...,X r be a basis for S . (i) ±irst let Y 1 = a 11 X 1 + ··· + a 1 r X r . . . (2) Y r = a r 1 X 1 + ··· + a rr X r , 40...
View
Full Document
 Fall '10
 Dreibelbis
 Linear Algebra, Linear Independence, Vector Space, basis, XR, X1

Click to edit the document details