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HW6solutions_3

# HW6solutions_3 - MATH 115 SOLUTION SET 6 ANSWERS TO...

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MATH 115 SOLUTION SET 6 ANSWERS TO SELECTED PROBLEMS § 2.4: 3. Only the pairs in b) and c) are isomorphic. / 4. Since A and B are invertible, A - 1 and B - 1 exist. We have ( AB )( B - 1 A - 1 ) = I and ( B - 1 A - 1 ) AB = I , so AB is invertible, with inverse equal to B - 1 A - 1 . / 5. Since A is invertible, A - 1 exists. We have A t ( A - 1 ) t = ( A - 1 A ) t = I and ( A - 1 ) t A t = ( AA - 1 ) t = I , so A t is invertible and its inverse is ( A - 1 ) t . / 6. Since A is invertible, A - 1 exists. If AB = O , then A - 1 AB = A - 1 O , so IB = O , so B = O . / 12. Let β = { v 1 , . . . , v n } be a basis for a vector space V . We have φ β : V F n is a linear transformation between vector spaces of the same dimension. To show that φ β is an isomorphism it is enough to show that it is one-to-one. Assume that φ β ( v ) = (0 , 0 , . . . , 0) is the zero vector. This means that v = 0 v 1 + · · · + 0 v n = 0 must itself be the zero vector. Since N ( φ β ) = { 0 } , φ β is one-to-one as required. / 13. This required proving that is symmetric, reflexive and transitive. Let’s just do transitive: We assume A B , B C and prove A C . We have isomorphisms T : A B and U : B C . Then UT : A C is invertible (its inverse is T - 1 U - 1 ) and therefore A C . The other properties are no more difficult to establish.

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HW6solutions_3 - MATH 115 SOLUTION SET 6 ANSWERS TO...

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