235_midterm_s11_soln

235_midterm_s11_soln - Math 235 Midterm Solutions 1 Short...

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Math 235 Midterm Solutions 1. Short Answer Problems [1] a) State the definition of an orthonormal basis for an inner product space V . Solution: A basis { ~v 1 , . . . ,~v n } is an orthonormal basis for V if it is orthogonal and each vector is a unit vector. [2] b) Let P and Q be orthogonal matrices. Prove that PQ is an orthogonal matrix. Solution: Since P and Q are orthogonal we have PP T = I and QQ T = I . Hence, PQ ( PQ ) T = PQQ T P T = PIP T = PP T = I Thus, ( PQ ) T = ( PQ ) - 1 so PQ is orthogonal. [4] c) Define the four fundamental subspaces of a matrix A . Solution: Let A be an m × n matrix. Col( A ) = { A~x | ~x R n } Row( A ) = { A T ~x | ~x R m } Null( A ) = { ~x R n | A~x = ~ 0 } Null( A T ) = { ~x R m | A T ~x = ~ 0 } [2] d) Let A be a 42 × 37 matrix with rank 20. Determine the dimension of the four fundamental subspaces of A . Solution: dim Col ( A ) = rank ( A ) = 20. dim Row ( A ) = rank ( A ) = 20. dim Null ( A ) = 37 - rank ( A ) = 17. dim Null ( A T ) = 42 - rank ( A ) = 22. 1
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[3] 2. Let A = 1 2 - 1 2 2 4 - 2 4 - 1 3 1 1 . Find a basis for the columnspace of A and for the left nullspace of A . Solution: Row reducing A T we get 1 2 - 1 2 4 3 - 1 - 2 1 2 4 1 1 2 0 0 0 1 0 0 0 0 0 0 . Thus, a basis for Null( A T ) is - 2 1 0 . Since Col( A ) = Row( A T ), we get that a basis for Col( A ) is 1 2 0 , 0 0 1 . NOTE: If you row reduce A , you find that a basis for Col( A ) is 1 2 - 1 , 2 4 3 . [3] 3. Let L : V W and M : W U be linear mappings. Prove that M L is a linear mapping. Solution: Let ~x, ~ y V and s, t R . Then ( M L )( s~x + t~ y ) = M [ L ( s~x + t~ y )] = M [ sL ( ~x ) + tL ( ~ y )] since L is linear = sM ( L ( ~x )) + tM ( L ( ~ y )) since M is linear = s ( M L )( ~x ) + t ( M L )( ~ y ) Thus, M L is linear. 2
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4. Define a linear mapping L : P 3 R 2 by L ( p ( x )) = p (0) p (1) . [2] a) Find a basis for the kernel of L .
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