This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ) =...
View Full
Document
 Spring '10
 SHALOM
 Math, Linear Algebra, Algebra

Click to edit the document details