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Unformatted text preview: MATH321 – HOMEWORK SOLUTIONS HOMEWORK #5 Section 2.4: Problems 1, 2, 3, 4, 5, 6, 7, 13, 14, 22 Section 2.5: Problems 1, 2(a)(b), 3(a), 4, 6, 7, 9, 10 Krzysztof Galicki Problem 2.4.1 (See Answers to Selected Exercises ). Problem 2.4.2 (a) T : R 2→ R 3 cannot be invertible. However, N ( T ) = { (0 , 0) } is trivial so that T can be inverted on R ( T ) which is a plane in R 3 . (b) T : R 2→ R 3 cannot be invertible. N ( T ) = { (0 , 0) } is trivial so that T can be inverted on R ( T ) which is a plane in R 3 . (c) Here N ( T ) = { (0 , , 0) } and it follows that R ( T ) = R 3 by nullityrank equation. Hence, T is invertible. (d) Here N ( T ) is the space of constant polynomials f ( x ) = c . Hence T is not onetoone and cannot be inverted. (e) Here N ( T ) consists of matrices of the form c c . Hence T is not oontoone and cannot be inverted. (f) Here N ( T ) is trivial so it follows that R ( T ) = M 2 × 2 ( R ) by nullityrank equation. Hence, T is both onetoone and onto. T is invertible. Problem 2.4.3 (a) F 3 and P 3 ( F ) are of dimension 3 and 4, respectively so thay cannot be isomorphic. (b) F 4 and P 3 ( F ) are isomorphic. One could take T : P 3 ( F )→ F 4 to be T ( a + bx + cx 2 + dx 3 ) = ( a, b, c, d ) . Clearly, T is invertible. (c) Take T a b c d = a + bx + cx 2 + dx 3 . This map is linear and invertible. (d) V is 3dimensional so it cannot be isomorphic to R 4 . Problem 2.4.4 Let A , B ∈ M n × n ( F ) be invertible with inverses A 1 and B 1 . Consider the product C = B 1 A 1 . We have C ( AB ) = B 1 A 1 AB = B 1 B = I n and, similarly, λ ( AB ) C = ABB 1 A 1 = AA 1 = I n ....
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.
 Winter '07
 Liu
 Math, Linear Algebra, Algebra

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