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235_tt1_w10_soln

# 235_tt1_w10_soln - Math 235 Term Test 1 Solutions 1 Short...

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Math 235 Term Test 1 Solutions 1. Short Answer Problems [3] a) Let A = 1 0 0 - 1 0 0 1 1 0 0 0 0 . Write a basis for the rowspace, columnspace and nullspace of A . Solution: A basis for the row space is { (1 , 0 , 0 , - 1) , (0 , 0 , 1 , 1) } . A basis for the column space is 1 0 0 , 0 1 0 . A basis for the nullspace is 0 1 0 0 , 1 0 - 1 1 [2] b) Let B = { v 1 , . . . , v n } be orthonormal in an inner product space V and let v = a 1 v 1 + · · · + a n v n . Prove that a i = < v, v i > . Solution: Taking the inner product of both sides with v i to get < v, v i > = < a 1 v 1 + · · · + a n v n , v i > = a 1 < v 1 , v i > + · · · + a n < v n , v i > = a i since B is orthonormal. [1] c) Determine if < p, q > = p (0) q (0) + p (1) q (1) is an inner product for P 2 . Solution: It is not an inner product since if p ( x ) = x 2 - x then < p, p > = 0(0) + 0(0) = 0 but p ( x ) = 0. [2] d) State the Rank-Nullity Theorem. Solution: Suppose that V is an n -dimensional vector space and that L : V W is a linear mapping into a vector space W . Then rank( L ) + nul( L ) = n. [2] e) Find the rank and nullity of the linear mapping T : P 2 M (2 , 2) defined by T ( a + bx + cx 2 ) = c b 0 c . Solution: The range of T is given by { c b 0 c | b, c arbitrary scalars } . A basis for that is { 1 0 0 1 , 0 1 0 0 } . So the rank of L is two. Using (a), the nullity is equal to 1, noting that the domain P 2 has dimension three. 1

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2 [3] 2. Let L : M (2 , 2) M (2 , 2) be given by L ( A ) = 1 2 3 4 A T . Find the matrix for L relative to the standard basis B of M (2 , 2) , where B = 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 . Solution: L ( 1 0 0 0 ) = 1 2 3 4 1 0 0 0 = 1 0 3 0 . By inspection, its coordinate vector relative to the basis B is 1 0 3 0 which will form the first column of [ L ] B . Similar computation according to the definition of L yield3 L ( 0 1 0 0 ) = 1 2 3 4 0 0 1 0 = 2 0 4 0 , L ( 0 0 1 0 ) = 1 2 3 4 0 1 0 0 = 0 1 0 3 , L ( 0 0 0 1 ) = 1 2 3 4 0 0 0 1 = 0 2 0 4 . Their coordinates vectors relative to the basis can be obtained by inspection, yielding [ L ] B = 1 2 0 0 0 0 1 2 3 4 0 0 0 0 3 4 .
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