Linear Algebra Solutions 37

# Linear Algebra Solutions 37 - R and u 6∈ S v ∈ S and v...

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Hence Y 1 ,...,Y s are linearly dependent. 17. Let R and S be subspaces of F n , with R S . We Frst prove dim R dim S. Let X 1 ,...,X r be a basis for R . Now by Theorem 3.5.2, because X 1 ,...,X r form a linearly independent family lying in S , this family can be extended to a basis X 1 ,...,X r ,...,X s for S . Then dim S = s r = dim R. Next suppose that dim R = dim S . Let X 1 ,...,X r be a basis for R . Then because X 1 ,...,X r form a linearly independent family in S and S is a sub- space whose dimension is r , it follows from Theorem 3.4.3 that X 1 ,...,X r form a basis for S . Then S = h X 1 ,...,X r i = R. 18. Suppose that R and S are subspaces of F n with the property that R S is also a subspace of F n . We have to prove that R S or S R . We argue by contradiction: Suppose that R 6⊆ S and S 6⊆ R . Then there exist vectors u and v such that u
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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...
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