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Unformatted text preview: Math 115A Homework 3 Solutions Brett Hemenway April 28, 2006 1. Let S 1 ⊂ S 2 ⊂ V . (a) Suppose S 1 is linearly dependent. This means we can find v 1 , . . . , v n ∈ S 1 and a 1 , . . . , a n ∈ F not all zero such that a 1 v 1 + . . . + a n v n = 0 Since { v 1 , . . . , v n } ⊂ S 1 ⊂ S 2 , v 1 , . . . , v n are dependent vectors in S 2 . Thus S 2 is dependent. (b) This is the contrapositive of part (a), so it is equivalent to part (a). We present a proof anyway, in case the proof for part (a) was unclear. Let v 1 , . . . , v n ∈ S 1 , and suppose a 1 v 1 + . . . + a n v n = 0 If we can show that a i = 0 for 1 ≤ i ≤ n , then we will have shown that v 1 , . . . , v n are independent. Since S 1 ⊂ S 2 , we have that v 1 , . . . , v n ∈ S 2 . But S 2 is independent by hypothesis, which implies that a 1 = a 2 = . . . = a n = 0. (A) If S 1 is independent, then S 2 need not be independent, consider S 1 = 1 , S 2 = 1 , ( 2 , ) (B) If S 2 is linearly dependent, then S 1 need not be dependent, and the same example from part (A) works. 1 2. (a) T is linear because T a x 1 x 2 x 3 + b y 1 y 2 y 3 = ( ax 1 + by 1 ) ( ax 2 + by 2 ) 2 ax 3 + 2 by 3 ) = a x 1 x 2 2 x 3 + b y 1 y 2 2 y 3 = aT x 1 x 2 x 3 + bT y 1 y 2 y 3 ker( T ) =...
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This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.
 Spring '10
 SHALOM
 Math, Linear Algebra, Algebra

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