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assign7_soln

# assign7_soln - Math 136 Assignment 7 Solutions 1 Prove that...

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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 non-empty. 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 non-empty. Thus, we can apply the Subspace Test. Let a 1 a 2 a 3 a 4 , b 1 b 2 b 3 b 4 S 2 . Then, a 1 - a 4 = a 2 and b 1 - b 4 = b 2 . Hence, a 1 a 2 a 3 a 4 + b 1 b 2 b 3 b 4 = a 1 + b 1 a 2 + b 2 a 3 + b 3 a 4 + b 4 S 2 since ( a 1 + b 1 ) - ( a 4 + b 4 ) = a 1 - a 4 + b 1 - b 4 = a 2 + b 2 .

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