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Unformatted text preview: Math 136 Assignment 7 Solutions 1. Given that the set B = 1 1 1 , 1 0 1 1 , 0 1 1 2 is a basis for the subspace of M (2 , 2) which it spans. Find the Bcoordinates of ~u = 1 3 2 3 and ~v = 1 0 3 7 . Solution: We need to find c 1 , c 2 , c 3 and d 1 , d 2 , d 3 such that c 1 1 1 1 + c 2 1 0 1 1 + c 3 0 1 1 2 = 1 3 2 3 d 1 1 1 1 + d 2 1 0 1 1 + d 3 0 1 1 2 = 1 0 3 7 Row reducing the corresponding doubly augmented matrix gives 1 1 0 1 1 1 0 1 3 1 1 2 3 1 1 2 3 7 1 0 0 2 2 0 1 0 3 1 0 0 1 1 2 Hence, for the first system we have c 1 = 2, c 2 = 3, and c 3 = 1, and for the second system we have d 1 = 2, d 2 = 1, and d 3 = 2. Thus, [ ~u ] B =  2 3 1 and [ ~v ]  mB =  2 1 2 2. Find a basis and determine the dimension of the following vector spaces. a) S = { xp ( x )  p ( x ) P 3 } . Solution: Every vector in S has the form x ( d + cx + bx 2 + ax 3 ) = dx + cx 2 + bx 3 + ax 4 for some a, b, c, d R . Thus, S = span { x, x 2 , x 3 , x 4 } . Moreover, since these are vectors from the standard basis for P 4 , we know that they are linearly independent. Hence, { x, x 2 , x 3 , x 4 } is a basis for S and thus dim S = 4. b) K = { A M (2 , 2)  A T = A } . (Note: This is called the subspace of skewsymmetric matrices.) Solution: Let A = a b c d K . Then, since A T = A , we have a c b d = a b c d = a b c d Thus, we have a = a , c = b , b = c , and d = d . Therefore, a = 0 = d and b = c ....
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 Spring '08
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 Math, Linear Algebra, Algebra

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