Math201mt2.pdf - B U Department of Mathematics Math 201 Matrix Theory Fall 2002 Second Midterm This archive is a property of Bo\u02d8 gazi\u00b8ci University

Math201mt2.pdf - B U Department of Mathematics Math 201...

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B U Department of Mathematics Math 201 Matrix Theory Fall 2002 Second Midterm This archive is a property of Bo˘ gazi¸ ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. (a) Find the equation (do not solve) for the coefficients C, D, E in b = C + Dt + Et 2 , the parabola which best fits the four points:( t, b ) = (0 , 0) , (1 , 1) , (1 , 3) and (2 , 2). Solution: We have C = 0 , C + D + E = 1 , C + D + E = 3 , C + 2 D + 4 E = 2 . Let A = 1 0 0 1 1 1 1 1 1 1 2 4 , x = C D E , b = 0 1 3 2 . We have the system Ax = b which is inconsistent. We need to look for its least square solution and solve the system: A T A ¯ x = A T b, i.e. 4 4 6 4 6 10 6 10 18 ¯ C ¯ D ¯ E = 6 8 12 (b) (Fill in the blanks) In solving this problem you are projecting the vector ( b ) onto the subspace spanned by ( the columns of A )
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2. Let A = 1 1 2 - 1 - 2 4 (a) Find orthonormal vectors e 1 , e 2 and e 3 so that e 1 and e 2 form a basis for the column space of A . (b) Find the projection matrix P which projects onto the left nullspace of A . Solution: (a) Let a = 1 2 - 2 , b = 1 - 1 4 , e 1 = a a = 1 3 1 2 - 2 b = b - ( e T 1 b ) e 1 , e 2 = b b = 1 3 2 1 2 . Then e 3 must be in the left nullspace N ( A T ): A T y = 1 2 - 2 1 - 1 4 y 1 y 2 y 3 = 0 0 y 1 = - 2 y 3 , y 2 = 2 y 3 , and y 3 is free. dim N ( A T ) = 1 A basis for N ( A T ) is : c = - 2 2 1 . Then e 3 = c c = 1 3 - 2 2 1 . Now { e 1 , e 2 , e 3 } is now an ON basis for R 3 . (b) We have P = e 3 ( e T 3 e 3 ) - 1 e T 3 = e 3 e T 3 = 1 9 - 2 2 1 ( - 2 2 1 ) = 1 9 4 - 4 - 2 - 4 4 2 - 2 2 1
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3. (a) Let P 2 be the vector space of polynomials of degree less than or equal to 2. Suppose L : P 2 R 3 is the linear transformation defined by L [ p ( t )] = p (1) p (0) p ( - 1) , p ( t ) P 2 . Find the matrix representation of L relative to the standard bases of P 2 and R 3 . Solution: We know that p 1 = 1 , p 2 = t, p 3 = t 2 is the standard basis of P 2 and e T 1 = ( 1 0 0 ) , e T 2 = ( 0 1 0 ) , e T 3 = ( 0 0 1 ) is the standard basis of R 3 . Then L [ p 1 ] = ( 1 1 1 ) = e 1 + e 2 + e 3 , L [ p 2 ] = ( 1 0 - 1 ) = e 1 - e 2 , L [ p 3 ] = ( 1 0 1 ) = e 1 + e 3 . Since L [ p j ] = 3 i =1 a ij e i , where ( a ij ) is the matrix representation, we have A = ( a ij ) = 1 1 1 1 0 0 1 - 1 1 . (b) Let P be an n × n matrix satisfying P 2 = P and let λ = 1 be real. Prove that the matrix: I - λP is invertible and ( I - λP ) - 1 = I + λ 1 - λ P Solution: Consider the nullspace of I - λP : ( I - λP ) x = 0 λPx = x. Apply P : λP 2 x = Px = λPx ( λ - 1) Px = 0 Px = 0 x = 0 . This means that the nullspace is trivial: dim N ( I - λP ) = 0; I - λP is of rank n ( I - λP ) - 1 exists. Then ( I - λP )( I - λP ) - 1 = ( I - λP )[ I + λ 1 - λ P ] = I - λP - λ 2 1 - λ P 2 + λP 1 - λ = I + λP ( - λ 1 - λ - 1 + 1 1 - λ ) = I Therefore, ( I - λP ) - 1 = I + λ 1 - λ P.
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4. (a) Let A = 1 0 1 0 0 1 0 1 1 0 - 1 0 0 - 1 0 1 . How many of the 24 terms in detA are nonzero? Justify your answer and find detA .
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