Linear Algebra Solutions 27

Linear Algebra Solutions 27 - (d) Let S be the set of...

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Section 3.6 1. (a) Let S be the set of vectors [ x, y ] satisfying x = 2 y . Then S is a vector subspace of R 2 . For (i) [0 , 0] S as x = 2 y holds with x = 0 and y = 0. (ii) S is closed under addition. For let [ x 1 , y 1 ] and [ x 2 , y 2 ] belong to S . Then x 1 = 2 y 1 and x 2 = 2 y 2 . Hence x 1 + x 2 = 2 y 1 + 2 y 2 = 2( y 1 + y 2 ) and hence [ x 1 + x 2 , y 1 + y 2 ] = [ x 1 , y 1 ] + [ x 2 , y 2 ] belongs to S . (iii) S is closed under scalar multiplication. For let [ x, y ] S and t R . Then x = 2 y and hence tx = 2( ty ). Consequently [ tx, ty ] = t [ x, y ] S. (b) Let S be the set of vectors [ x, y ] satisfying x = 2 y and 2 x = y . Then S is a subspace of R 2 . This can be proved in the same way as (a), or alternatively we see that x = 2 y and 2 x = y imply x = 4 x and hence x = 0 = y . Hence S = { [0 , 0] } , the set consisting of the zero vector. This is always a subspace. (c) Let S be the set of vectors [ x, y ] satisfying x = 2 y + 1. Then S doesn’t contain the zero vector and consequently fails to be a vector subspace.
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Unformatted text preview: (d) Let S be the set of vectors [ x, y ] satisfying xy = 0. Then S is not closed under addition of vectors. For example [1 , 0] S and [0 , 1] S , but [1 , 0] + [0 , 1] = [1 , 1] 6 S . (e) Let S be the set of vectors [ x, y ] satisfying x 0 and y 0. Then S is not closed under scalar multiplication. For example [1 , 0] S and-1 R , but (-1)[1 , 0] = [-1 , 0] 6 S . 2. Let X, Y, Z be vectors in R n . Then by Lemma 3.2.1 h X + Y, X + Z, Y + Z i h X, Y, Z i , as each of X + Y, X + Z, Y + Z is a linear combination of X, Y, Z . 30...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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