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2.4-2.5s

2.4-2.5s - MATH321 HOMEWORK SOLUTIONS HOMEWORK#5 Section...

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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 , 0) } and it follows that R ( T ) = R 3 by nullity-rank equation. Hence, T is invertible. (d) Here N ( T ) is the space of constant polynomials f ( x ) = c . Hence T is not one-to-one and cannot be inverted. (e) Here N ( T ) consists of matrices of the form 0 0 c - c . Hence T is not oon-to-one and cannot be inverted. (f) Here N ( T ) is trivial so it follows that R ( T ) = M 2 × 2 ( R ) by nullity-rank equation. Hence, T is both one-to-one 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 ) .

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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 3-dimensional 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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