Question

Do #: 1.13 a, b, c, 1.15 a, b, c, d, 1.19 & 1.22

1.13 For the map f: P1 + R2 given by at barn (9-b) Find the image of each of these elements of the domain. (a) 3-2x (b) 2 + 2x (c) x Show that this map is an isomorphism. 1.14 Show that the natural map f, from Example 1.5 is an isomorphism. ✓ 1.15 Decide whether each map is an isomorphism (if it is an isomorphism then prove it and if it isn't then state a condition that it fails to satisfy). (a) f: M2x2 + R given by le ) - ad – bc (b) f: M22Rgiven by 69-(0) 'a+b+c+d a + b + c a + b Section I. Isomorphisms (c) f: M2 → Ps given by 2)+c+(d+c)x+ (b+a)x? + ax? (d) f: M22 - Ps given by ( .) -e+(d+c)x+ (b + a +1)x² + ax? 1.16 Show that the map f:R' R' given by f(x) = is one-to-one and onto. Is it an isomorphism? 1.17 Refer to Example 1.1. Produce two more isomorphisms (of course, you must also verify that they satisfy the conditions in the definition of isomorphism). 1.18 Refer to Example 1.2. Produce two more isomorphisms (and verify that they satinfy the conditions) 1.19 Show that, although R is not itself a subspace of R', it is isomorphic to the xy-plane subspace of R'. 1.20 Find two isomorphisms between Ris and Me 1.21 For what kis M inomorphic to R? 1.22 For what kissomorphic to R"? 1.23 Prove that the

Answer 1

1.13 Consider the mapping f : R → R2 given by (a) Consider the element of the domain: 3-2x (3-2x) = f (3+(-2)x) -1312) Thus, the image of 3-2x is (6) Consider the element of the domain 2+2x ${2+2x)=(223) Thus, the image of 2+2x is (0) Consider the element of the domain: f (x)= f (0+1x) Thus, the image of x is

Consider the mapping f: R → R2 given by a+bx(2) Leta, + bx, 4, +bx € R, CER f((4, +5x)+(az+bx)) = f((a +az) +(+b)x) _lay+a)-(4+b)) 7 th -674-676) = ${{4} +6x)) + f ((az +byx)) cf ((a+bx)) - (ca-co) = | cb = f((ca+cbx)) = f (c(a+bx)) Thus, f is homomorphism. f is one-to-one: Letf(( 47 +x)) = f((az +b,x)) (a-bi) _ (arba) By by = - = a; - b, and b = By an = a, and h = b a+bx= d; +byx Thus, f is one-to-one. f is onto: y=4x'=x= For each (9) e Rº there exists (a+b) + bx € Ry such that f(a+b) +6x)= ((+6)=0)-(6)*** Thus, is onto Therefore, fis an isomorphism.