assign6_soln

# assign6_soln - Math 136 Assignment 6 Solutions 1 Determine...

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Math 136 Assignment 6 Solutions 1. Determine, with proof, which of the following are subspaces of the given vector space. Find a basis for each subspace. a) A = { ( x 1 ,x 2 3 ) R 3 | 2 x 1 + x 3 =0 1 + x 2 - x 3 } of R 3 . Solution: A is the solution space of a homogeneous system of linear equations 2 x 1 + x 3 = 0 1 + x 2 - x 3 = 0, so it is a spanning set and hence a subspace of R 3 by theorem (i.e. question 4 below). To ±nd a basis, we need to solve the homogeneous system. Row-reducing the coe²cient matrix gives ± 2 0 1 11 - 1 ² ± 1 0 1 / 2 01 - 3 / 2 ² . Hence, x 3 is a free variable and so we get x 1 x 2 x 3 = - 1 2 x 3 3 2 x 3 x 3 = t - 1 / 2 3 / 2 1 , where x 3 = t R . Thus, A = span - 1 / 2 3 / 2 1 , which is clearly linearly independent and hence a basis for A . b) B = { ax 2 + bx + c P 2 | b 2 - 4 ac } of P 2 . Solution: Observe that x 2 +2 x +1 B and x 2 - 2 x B , but ( x 2 x +1)+( x 2 - 2 x +1) = 2 x 2 ±∈ B . Hence, B is not closed under addition. Thus, it is not a vector space, so it is not a subspace of P 2 . c) C = { ax 2 + bx + c P 2 | a - c = b } of P 2 . Solution: By de±nition C is a subset of P 2 , so we use the subspace test. Observe that 0 x 2 +0 x C since 0 - 0 = 0. So " 0 C . If ax 2 + bx + c and dx 2 + ex + f are both in C , then a - c = b and d - f = e . Then, ( ax 2 + bx + c )+( dx 2 + ex + f ) = ( a + d ) x 2 +( b + e ) x c + f ) , and ( a + d ) - ( c + f )= a - c + d - f = b + e. So, ( ax 2 + bx + c dx 2 + ex + f ) C , thus C is closed under addition. Similarly, k ( ax 2 + bx + c ) = ( ka ) x 2 kb ) x kc ) , 1

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2 and ( ka ) - ( kc )= k ( a - c kb . Thus k ( ax 2 + bx + c ) C , so C is also closed under scalar multiplication. Hence C is a subspace of P 2 . To Fnd a basis, we Frst observe that if ax 2 + bx + c C , then we a - c = b so ax 2 + bx + c = ax 2 +( a - c ) x + c = a ( x 2 + x )+ c ( - x + 1) . Hence C = span { x 2 + x, - x +1 } . It is clear that neither is a scalar multiple of the other, so the set is linearly independent and so a basis for C is { x 2 + x, - x } . d) D = ±² ab cd ³ M (2 , 2) | ad - bc =0 ´ of M (2 , 2). Solution: Observe that ² 12 24 ³ and ² - 2 - 2 - 2 - 2 ³ are in D , but ² ³ + ² - 2 - 2 - 2 - 2 ³ = ² - 10 02 ³ ±∈ D since ( - 1)(2) - (0)(0) ± = 0.
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assign6_soln - Math 136 Assignment 6 Solutions 1 Determine...

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