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Unformatted text preview: Math 136 Assignment 7 Solutions 1. Prove that each of the following sets is a subspace of the given vector space. Find a basis for the subspace and hence determine the dimension of the subspace. a) S 1 = { ax 2 + bx + c  a + b + c = 0 } of P 4 Solution: By definition S 1 contains 2nd degree polynomials which are polynomials of degree less than 4 and hence S 1 is a subset of P 4 . Moreover, taking a = b = c = 0 we get 0+0+0 = 0 and hence 0 ∈ S 1 . Thus S 1 is nonempty. Hence, we can apply the Subspace Test. Let a + bx + cx 2 ,d + ex + fx 2 ∈ S 1 . Then we have a + b + c = 0 and d + e + f = 0. Hence, ( a + bx + cx 2 ) + ( d + ex + fx 2 ) = ( a + d ) + ( b + e ) x + ( c + f ) x 2 ∈ S 1 since ( a + d ) + ( b + e ) + ( c + f ) = a + b + c + d + e + f = 0 + 0 = 0. Similarly, t ( a + bx + cx 2 ) = ( ta )+( tb ) x +( tc ) x 2 ∈ S 1 since ta + tb + tc = t ( a + b + c ) = t (0) = 0. Thus, by the subspace test S 1 is a subspace of P 4 . To find a basis for S 1 , we first find a spanning set. Since a + b + c = 0 we have c = a b . Thus, every polynomial in S 1 has the form a + bx + ( a b ) x 2 = a (1 x 2 ) + b ( x x 2 ) Hence, span { 1 x 2 ,x x 2 } = S 1 . Since neither 1 x 2 nor x x 2 is a scalar multiple of the other, { 1 x 2 ,x x 2 } is also linearly independent. Thus, { 1 x 2 ,x x 2 } is a basis for S 1 . Hence, dim S 1 = 2. b) S 2 = a 1 a 2 a 3 a 4  a 1 a 4 = a 2 of M 2 × 2 ( R ). Solution: By definition S 2 is a subset of M 2 × 2 ( R ). Moreover, the zero matrix O 2 , 2 ∈ S 2 since 0 0 = 0. Hence, S 2 is nonempty. Thus, we can apply the Subspace Test....
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This note was uploaded on 07/13/2011 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.
 Winter '08
 All
 Math, Vector Space, Sets

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