# ass5 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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MATH 223, Linear Algebra Fall, 2007 Solutions to Assignment 5 1. W 1 = span 1 0 5 - 2 , 3 1 15 4 , 1 - 1 5 - 12 and W 2 = span 5 0 25 - 9 , 13 2 65 - 5 , - 11 - 4 - 55 - 17 are subspaces of R 4 . Find a basis for each of W 1 , W 2 , W 1 + W 2 and W 1 W 2 . Solution: We start by row-reducing the big matrix 1 3 1 | 5 13 - 11 0 1 - 1 | 0 2 - 4 5 15 5 | 25 65 - 55 - 2 4 - 12 | - 9 - 5 - 17 , getting 1 0 4 | 0 2 - 4 0 1 - 1 | 0 2 - 4 0 0 0 | 1 1 1 0 0 0 | 0 0 0 . It is evident from the left half of this that the ﬁrst two columns are independent, but not all three. A basis for W 1 is then 1 0 5 - 2 , 3 1 15 4 . Only slightly less ob- vious is that two of the three last columns are independent, but not all three; a basis for W 2 is 5 0 25 - 9 , 13 2 65 - 5 . Of the columns of the big row-reduced matrix, we can choose only three independent ones, most obviously the ﬁrst, second and fourth. A basis for W 1 + W 2 is 1 0 5 - 2 , 3 1 15 4 5 0 25 - 9 . Finally, if we label the columns C 1 ,..., C 6 , then from the row-reduced matrix, it’s clear that (say) C 5 = 2 C 1 + 2 C 2 + C 4 ; i.e C 5 - C 4 = 2 C 1 + 2 C 2 . This is true for the original big matrix and that vector C 5 - C 4 is in the intersection. Since 2 + 2 = 3 + 1, the intersection has dimension 1, and a basis for W 1 W 2 is 8 2 40 4 . 2. Let V = M 3 ( R ) be the real vector space of 3 × 3 matrices with real 1

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entries. Let A = 3 5 2 1 0 - 1 7 5 - 2 . Now let T : V -→ V be deﬁned by T ( X ) = AXA T for any X V . (a) Show that
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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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ass5 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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