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# HW4_sol - 48 Chapter 3 SECTION 4 3(a Since 2 1 4 3 =2=0 it...

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48 Chapter 3 SECTION 4 3. (a) Since 2 4 1 3 = 2 = 0 it follows that x 1 and x 2 are linearly independent and hence form a basis for R 2 . (b) It follows from Theorem 3.4.1 that any set of more than two vectors in R 2 must be linearly dependent. 5. (a) Since 2 3 2 1 - 1 6 3 4 4 = 0 it follows that x 1 , x 2 , x 3 are linearly dependent. (b) If c 1 x 1 + c 2 x 2 = 0 , then 2 c 1 + 3 c 2 = 0 c 1 - c 2 = 0 3 c 1 + 4 c 2 = 0 and the only solution to this system is c 1 = c 2 = 0. Therefore x 1 and x 2 are linearly independent. 8 (a) Since the dimension of R 3 is 3, it takes at least three vectors to span R 3 . Therefore x 1 and x 2 cannot possibly span R 3 . (b) The matrix X must be nonsingular or satisfy an equivalent condition such as det( X ) = 0. (c) If x 3 = ( a, b, c ) T and X = ( x 1 , x 2 , x 3 ) then det( X ) = 1 3 a 1 - 1 b 1 4 c = 5 a - b - 4 c If one chooses a , b , and c so that 5 a - b - 4 c = 0 then { x 1 , x 2 , x 3 } will be a basis for R 3 . 9. (a) If a 1 and a 2 are linearly independent then they span a 2-dimensional subspace of R 3 . A 2-dimensional subspace of R 3 corresponds to a plane through the origin in 3-space. (b) If b = A x then b = x 1 a 1 + x 2 a 2 so b is in Span( a 1 , a 2 ) and hence the dimension of Span( a 1 , a 2 , b ) is 2. 10. We must find a subset of three vectors that are linearly independent. Clearly x 1 and x 2 are linearly independent, but x 3 = x 2 - x 1

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Section 4 49 so x 1 , x 2 , x 3 are linearly dependent. Consider next the vectors x 1 , x 2 , x 4 . If X = ( x 1 , x 2 , x 4 ) then det( X ) = 1 2 2 2 5 7 2 4 4 = 0 so these three vectors are also linearly dependent. Finally if use x 5 and form the matrix X = ( x 1 , x 2 , x 5 ) then det( X ) = 1 2 1 2 5 1 2 4 0 = - 2 so the vectors x 1 , x 2 , x 5 are linearly independent and hence form a basis for R 3 . 16. dim U = 2. The set { e 1 , e 2 } is a basis for U . dim V = 2. The set { e 2 , e 3 } is a basis for V . dim U V = 1. The set { e 2 } is a basis for U V . dim U + V = 3. The set { e 1 , e 2 , e 3 } is a basis for U + V . 17. Let { u 1 , u 2 } be a basis for U and { v 1 , v 2 } be a basis for V . It follows from Theorem 3.4.1 that u 1 , u 2 , v 1 , v 2 are linearly dependent. Thus there exist scalars c 1 , c 2 , c 3 , c 4 not all zero such that c 1 u 1 + c 2 u 2 + c 3 v 1 + c 4 v 2 = 0 Let x = c 1 u 1 + c 2 u 2 = - c 3 v 1 - c 4 v 2 The vector x is an element of U V . We claim x = 0 , for if x = 0 , then c 1 u 1 + c 2 u 2 = 0 = - c 3 v 1 - c 4 v 2 and by the linear independence of u 1 and u 2 and the linear independence of v 1 and v 2 we would have c 1 = c 2 = c 3 = c 4 = 0 contradicting the definition of the c i ’s. 18. Let U and V be subspaces of R n with the property that U V = { 0 } . If either U = { 0 } or V = { 0 } the result is obvious, so assume that both subspaces are nontrivial with dim U = k > 0 and dim V = r > 0. Let { u 1 , . . ., u k } be a basis for U and let { v 1 , . . ., v r } be a basis for V . The vectors u 1 , . . ., u k , v 1 , . . ., v r span U + V . We claim that these vectors form a basis for U + V and hence that dim U + dim V = k + r . To show this we must show that the vectors are linearly independent. Thus we must show that if c 1 u 1 + · · · + c k u k + c k +1 v 1 + · · · + c k + r v r = 0 (2) then c 1 = c 2 = · · · = c k + r = 0. If we set u = c 1 u 1 + · · · + c k u k and v = c k +1 v 1 + · · · + c k + r v r
50 Chapter 3 then equation (2) becomes u + v = 0 This implies u = - v

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HW4_sol - 48 Chapter 3 SECTION 4 3(a Since 2 1 4 3 =2=0 it...

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