Midsoln.pdf - Math 235 Midterm Solutions Total 50 2 marks 1 Short Answer Problems a Is P2(R isomorphic to M22(R Discuss why or why not[2 mark(s Solution

# Midsoln.pdf - Math 235 Midterm Solutions Total 50 2 marks 1...

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Math 235 Midterm Solutions Total: 50 + 2 marks 1. Short Answer Problems a) Is P 2 ( R ) isomorphic to M 2 × 2 ( R )? Discuss why or why not. [ 2 mark(s) ] Solution: Since dim( P 2 ) = 3 and dim( M 2 × 2 ( R )) = 4, they are not isomorphic. b) Let B = { 1 - x, 1 + x } be a basis for P 1 ( R ) and let L : P 1 ( R ) P 1 ( R ) be a linear mapping. If [ L ] B = 1 3 - 2 1 and [ p ( x )] B = 2 1 , then express L ( p ( x )) in the form a + bx . [ 3 mark(s) ] Solution: We have [ L ( p ( x ))] B = [ L ] B [ p ( x )] B = 1 3 - 2 1 2 1 = 5 - 3 . Therefore, L ( p ( x )) = 5(1 - x ) - 3(1 + x ) = 2 - 8 x . c) If T : M 4 × 4 ( R ) M 3 × 4 ( R ) is linear and rank( T ) = 6, then prove that nullity( T ) = 10. [ 2 mark(s) ] Solution: By the Rank-Nullity Theorem we have nullity( T ) = dim M 4 × 4 ( R ) - rank( T ) = 16 - 6 = 10 as required. d) Suppose V is an n -dimensional vector space, and h , i : V × V R is a function. If there exists a vector ~ y V such that h ~ y, ~x i < 0 for all ~x V , prove that h , i is not an inner-product. [ 3 mark(s) ] Solution: Assuming the hypothesis is true, we get h ~ y, ~ y i < 0. However, an inner product must satisfy h ~ y, ~ y i ≥ 0, so h , i is not an inner-product. 2. Suppose L : P 1 ( R ) P 1 ( R ) is a linear operator, such that L (1) = 1 + 4 x and L (1 + x ) = 7 + 3 x . (a) Determine the value of L (2 + 3 x ). [ 3 mark(s) ] Solution: We find 2 + 3 x = ( - 1)(1) + (3)(1 + x ), so using linearity of L , get L (2 + 3 x ) = ( - 1) L (1) + 3 L (1 + x ) = ( - 1)(1 + 4 x ) + 3(7 + 3 x ) = 20 + 5 x. 1
(b) For the basis B = { 1 , 1 + x } of P 1 ( R ), we have [1 + 4 x ] B = - 3 4 and [7 + 3 x ] B = 4 3 . Is [ L ] B an orthogonal matrix? [ 3 mark(s) ] Solution: We have [ L ] B = - 3 4 4 3 . The columns of [ L ] B are not unit vectors, and thus cannot form an orthonormal basis for R 2 . Therefore [ L ] B is not an orthogonal matrix. 3. Let L : M 2 × 2 ( R ) M 2 × 2 ( R ) be given by L ( A ) = 1 2 3 4 A T . Find the matrix for L with respect to the standard basis B of M 2 × 2 ( R ), where [ 6 mark(s) ] B = 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 Solution: Evaluate L 1 0 0 0 = 1 2 3 4 1 0 0 0 = 1 0 3 0 = (1) 1 0 0 0 + (0) 0 1 0 0 + (3) 0 0 1 0 + (0) 0 0 0 1 , L 0 1 0 0 = 1 2 3 4 0 0 1 0 = 2 0 4 0 = (2) 1 0 0 0 + (0) 0 1 0 0 + (4) 0 0 1 0 + (0) 0 0 0 1 , L 0 0 1 0 = 1 2 3 4 0 1 0 0 = 0 1 0 3 = (0) 1 0 0 0 + (1) 0 1 0 0 + (0) 0 0 1 0 + (3) 0 0 0 1 , L 0 0 0 1 = 1 2 3 4 0 0 0 1 = 0 2 0 4 = (0) 1 0 0 0 + (2) 0 1 0 0 + (0) 0 0 1 0 + (4) 0 0 0 1 . [ L ] B = 1 2 0 0 0 0 1 2 3 4 0 0 0 0 3 4 . 4. Let A = 1 0 1 0 3 6 1 6 13 . The reduced row echelon form of A is the matrix 1 0 1 0 1 2 0 0 0 .