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Unformatted text preview: SOLUTIONS FOR PRACTICE PROBLEMS Section 1.1 3. (c) The plane in question is the set of all points of the form ( 8 , 2 , 0) + t ( ( 8 , 2 , 0) (1 , 3 , 0) ) + s ( ( 8 , 2 , 0) (6 , 5 , 0) ) = ( 8 9 t 14 s, 2 t + 7 s, 0) , with t, s R . Section 1.2 17. No , V is not a vector space. For instance, the condition (VS8) (one of the distributive laws) fails. To illustrate this, consider the scalars a = b = 1, and the vector x = (1 , 0). Then ( a + b ) x = 2 x = (1 , 0), yet ax + bx = x + x = (2 , 0). Section 1.3 10. First show that W 1 = { ( a 1 , . . . , a n ) F n : a 1 + . . . + a n = 0 } is a subspace of F n . We have to prove that W 1 is closed under addition and scalar multiplication. More precisely, suppose u = ( u 1 , . . . , u n ) and v = ( v 1 , . . . , v n ) are elements of W 1 , and c is a scalar. We have to show that u + v and cu also belong to W 1 . We have u + v = ( u 1 + v 1 , . . . , u n + v n ), and then ( u 1 + v 1 ) + . . . + ( u n + v n ) = ( u 1 + . . . + u n ) + ( v 1 + . . . + v n ) = 0 + 0 = 0 , hence u + v W 1 . Similarly, u = ( cu 1 , . . . , cu n ), and the equality cu 1 + . . . + cu n = c ( u 1 + . . . + u n ) = c 0 = 0 , which shows cu W 1 ....
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.
 Spring '10
 FUCKHEAD

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