Linear Algebra

156 Chapter Three. Maps Between SpacesThis computation shows that f preserves addition.f ( (a 1 cos θ + a 2 sin θ) + (b 1 cos θ + b 2 sin θ) )= f ( (a 1 + b 1 ) cos θ + (a 2 + b 2 ) sin θ )( )a1 + b=1a 2 + b 2( ) ( )a1 b1= +a 2 b 2= f(a 1 cos θ + a 2 sin θ) + f(b 1 cos θ + b 2 sin θ)A similar computation shows that f preserves scalar multiplication.f ( r · (a 1 cos θ + a 2 sin θ) ) = f( ra 1 cos θ + ra 2 sin θ )( )ra1=ra 2( )a1= r ·a 2= r · f(a 1 cos θ + a 2 sin θ)With that, conditions (1) and (2) are verified, so we know that f is anisomorphism and we can say that the spaces are isomorphic G ∼ = R 2 .1.5 Example Let V be the space {c 1 x + c 2 y + c 3 z ∣ c 1 , c 2 , c 3 ∈ R} of linearcombinations of three variables x, y, and z, under the natural addition andscalar multiplication operations. Then V is isomorphic to P 2 , the space ofquadratic polynomials.To show this we will produce an isomorphism map. There is more than onepossibility; for instance, here are four.c 1 x + c 2 y + c 3 zf 1↦−→ c1 + c 2 x + c 3 x 2f 2↦−→ c2 + c 3 x + c 1 x 2f 3↦−→ −c1 − c 2 x − c 3 x 2f 4↦−→ c1 + (c 1 + c 2 )x + (c 1 + c 3 )x 2The first map is the more natural correspondence in that it just carries thecoefficients over. However, below we shall verify that the second one is an isomorphism,to underline that there are isomorphisms other than just the obviousone (showing that f 1 is an isomorphism is Exercise 12).To show that f 2 is one-to-one, we will prove that if f 2 (c 1 x + c 2 y + c 3 z) =f 2 (d 1 x + d 2 y + d 3 z) then c 1 x + c 2 y + c 3 z = d 1 x + d 2 y + d 3 z. The assumptionthat f 2 (c 1 x+c 2 y +c 3 z) = f 2 (d 1 x+d 2 y +d 3 z) gives, by the definition of f 2 , thatc 2 + c 3 x + c 1 x 2 = d 2 + d 3 x + d 1 x 2 . Equal polynomials have equal coefficients, soc 2 = d 2 , c 3 = d 3 , and c 1 = d 1 . Thus f 2 (c 1 x + c 2 y + c 3 z) = f 2 (d 1 x + d 2 y + d 3 z)implies that c 1 x + c 2 y + c 3 z = d 1 x + d 2 y + d 3 z and therefore f 2 is one-to-one.