1.3-1.6 - 1 Math 110 Homework 2 Partial Solutions If you...

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1 Math 110 Homework 2 Partial Solutions If you have any questions about these solutions, or about any problem not solved, please ask via email or in office hours, etc. 1.3.19 : Suppose that W 1 W 2 is a subspace. The proof will proceed by contradiction. Suppose that neither W 1 W 2 nor W 2 W 1 . Then there is x W 1 not in W 2 and y W 2 not in W 1 . Since W 1 W 2 is assumed to be a subspace, x + y W 1 W 2 . But, if x + y W 1 , then y = ( x + y ) - x W 1 (since - x W 1 ). But this is impossible. Therefore, we must have x + y W 2 . But this implies that x = ( x + y ) - y W 2 , also a contradiction. Since we arrive at a contradiction in either case, we must have that W 1 W 2 or vice versa. : If W 1 W 2 , then W 1 W 2 = W 2 is a clearly subspace. A similar argument holds if W 2 W 1 . 1.3.23 For part (a), if x W 1 , then x = x +0 W 1 + W 2 , as 0 W 2 since W 2 is a subspace. Thus W 1 W 1 + W 2 . Similarly W 2 W 1 + W 2 . In order to show that W 1 + W 2 is a subspace, we first show that 0 = 0+0 W 1 + W 2 . Second, if x + y, u + v W 1 + W 2 ( x, u W 1 and y, v W 2 ), then ( x + y )+( u + v ) = ( x + u )+( y + v ) W 1 + W 2 . Finally, if x + y W 1 + W 2 and a F , then a ( x + y ) = ax + ay W 1 + W 2 . Thus W 1 + W 2
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.

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1.3-1.6 - 1 Math 110 Homework 2 Partial Solutions If you...

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