Assignment 4 solution

Assignment 4 solution - MATH 223 Linear Algebra Winter 2008...

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MATH 223, Linear Algebra Winter, 2008 Solutions to Assignment 4 1. Let V = Z 4 2 and W 1 = Span 1 0 1 0 , 0 1 0 1 and W 2 = Span 0 1 1 0 , 1 0 0 1 be subspaces of V . Find a basis for W 1 + W 2 and one for W 1 W 2 . Solution: Note that clearly W 1 and W 2 are both 2-dimensional, so the sets that are given to span them are bases for them. Considering the matrix 1 0 | 0 1 0 1 | 1 0 1 0 | 1 0 0 1 | 0 1 , the column space of which is W 1 + W 2 , and the fact that it row-reduces very quickly to 1 0 | 0 1 0 1 | 0 1 0 0 | 1 1 0 0 | 0 0 , it’s immediate that a basis for W 1 + W 2 is 1 0 1 0 , 0 1 0 1 , 0 1 1 0 . For the inter- section, the RREF tells us that that the fourth column is the sum of the first three, so the sum of the first two equals the sum of the last two, which is a nonzero vector in the intersection. Also Lunch In Chinatown tells us that the intersection has dimension 1, so a basis for W 1 W 2 is just 1 1 1 1 . 2. Let V = C 5 and W 1 = Span 1 i 1 i 1 , 2 i - 1 1 + i 0 2 , 4 3 i 3 - i 2 i 2 - 2 i , 2 + 2 i - 1 + 2 i 3 + i 1 + 2 i 4 and
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This note was uploaded on 04/29/2008 for the course MATH 223 taught by Professor Loveys during the Fall '07 term at McGill.

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Assignment 4 solution - MATH 223 Linear Algebra Winter 2008...

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