2.4 - [ UT ] . 13. Isomorphism is reexive: V V as witnessed...

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Selected answers to assignment 6: 2.4 2. (a), (b), (d), (e) noninvertible; dimensions wrong. (c) invertible; standard basis mapped to (3 , 0 , 3) , (0 , 1 , 4) , ( - 2 , 0 , 0), which is a basis. (f) invertible; standard basis mapped to ± 1 1 0 0 ² , ± 1 0 0 0 ² , ± 0 0 1 1 ² , ± 0 0 0 1 ² , which is a basis. 3. (b), (c) are isomorphic pairs; (a), (d) are not. All by dimension. 4. Multiply AB by B - 1 A - 1 ; replace the inverse pairs by I n from the inside-out. 5. The transpose of a product is the product of the transposes in reverse order. Apply that to AA - 1 or the reverse. 6. Since A is invertible there is some A - 1 . Then A - 1 AB = A - 1 0, which simplifies to B = 0. 10. (a) Since I n is invertible and A , B are square, exercise 9 applies and shows A and B are invertible. (b) Since A and B are invertible we may multiply the equality AB = I n on the left by A - 1 or on the right by B - 1 ; simplification gives B = A - 1 and A = B - 1 . (c) If V and W are n -dimensional vector spaces and T : V W , U : W V are such that UT = I V , then T and U are invertible and are in fact each others’ inverses. Proof by applying (a) and (b) to
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Unformatted text preview: [ UT ] . 13. Isomorphism is reexive: V V as witnessed by I V . Symmetric: if V W is witnessed by T , then T-1 witnesses W V . Transitive: if V W and W Z , shown by T and U respectively, then V Z is shown by UT . 16. B-1 ( cA + D ) B = cB-1 AB + B-1 DB by a few applications of Theorem 2.12, p. 89. Use exercise 6 twice to argue that if B-1 AB = 0, A must be 0. Hence the null space is zero and is one-to-one. Since the vector space is nite-dimensional that suces to show is also onto and hence an isomorphism. 20. Since is an isomorphism, R ( L A ) = R ( L A ). Since is an isomorphism, by #17 R ( T ) has equal rank to R ( T ). Commutativity of Figure 2.2 then shows rank( T ) = rank( L A ). The nullity argument is similar. 1...
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