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Unformatted text preview: MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 4 SOLUTIONS 1. Let W 1 , W 2 be subspaces of V such that V = W 1 W 2 . Let W be a subspace of V . Show that if W 1 W or W 2 W , then W = ( W W 1 ) ( W W 2 ) . Is this still true if we omit the condition W 1 W or W 2 W ? Solution. Without loss of generality, we will assume that W 1 W . So W W 1 = W 1 and we just need to show that W = W 1 ( W W 2 ) . By Problem 2 (b) in Homework 2 , the smallest subspace containing both W 1 and W W 2 is W 1 + ( W W 2 ); since W 1 W and W W 2 W , it follows that W 1 + ( W W 2 ) W. (1.1) To show the reverse inclusion, let w W . Then w V . Since V = W 1 W 2 , there exist w 1 W 1 and w 2 W 2 such that w = w 1 + w 2 . Observe that w 2 = w w 1 W W 2 since w 2 W 2 and w w 1 W . Hence w = w 1 + w 2 W 1 + ( W W 2 ). Since this is true for arbitrary w W , we conclude that W W 1 + ( W W 2 ) . (1.2) By (1.1) and (1.2), W = W 1 + ( W W 2 ) . This is a direct sum since ( W W 1 ) ( W W 2 ) W 1 W 2 = { } . The statement is false without the condition W 1 W or W 2 W . Here is a counterexample. Let W = { ( x,y ) R 2  x = y } , W 1 = { ( x,y ) R 2  y = 0 } , and W 2 = { ( x,y ) R 2  x = 0 } . Then W 1 W 2 = R 2 since every ( x,y ) R 2 may be written uniquely as ( x,y ) = ( x, 0) + (0 ,y ). However, W W 1 = { (0 , 0) } = W W 2 and so ( W W 1 ) ( W W 2 ) = { (0 , 0) } 6 = W. 2. For the following vector spaces V , find the coordinate representation of the respective elements. (a) V = P 2 = { ax 2 + bx + c  a,b,c R } . Find [ p ( x )] B where p ( x ) = 2 x 2 5 x + 6 , B = [1 ,x 1 , ( x 1) 2 ] . Solution. Since p ( x ) = 2 x 2 5 x + 6 = 3 ( x 1) + 2( x 1) 2 , so [ p ( x )] B = 3 1 2 . Date : March 15, 2008 (Version 1.1). 1 (b) V = R 2 2 . Find [ A ] B where A = 2 3 4 7 , B = 1 1 1 1 , 1 1 , 1 1 , 1 0 0 0 . Solution. Let 2 3 4 7 = 1 1 1 1 +  1 1 + 1 1 + 1 0 0 0 . Then solving + + = 2 ,   = 3 , + = 4 , = 7 , gives = 7 , = 11 , = 21 , = 30. So [ A ] B =  7 11 21 30 ....
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra

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