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Unformatted text preview: 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 = ~ } Null( A T ) = { ~x ∈ R m  A T ~x = ~ } [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 [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 . Since Col( A ) = Row( A T ), we get that a basis for Col( A ) is 1 2 , 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....
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This note was uploaded on 09/30/2011 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
 Spring '08
 CELMIN
 Math

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