Linear Algebra

162 Chapter 3. Maps Between Spaces
This computation shows that f preserves addition.
f � (a1 cos θ + a2 sin θ)+(b1cos θ + b2 sin θ) �
= f � (a1 + b1)cosθ +(a2 + b2)sinθ �
� �
a1 + b1
=
a2 + b2
� � � �
a1 b1
= +
a2
b2
= f(a1 cos θ + a2 sin θ)+f(b1 cos θ + b2 sin θ)
A similar computation shows that f preserves scalar multiplication.
f � r · (a1 cos θ + a2 sin θ) � = f( ra1 cos θ + ra2 sin θ )
� �
ra1
=
ra2
� �
a1
= r ·
a2
= r · f(a1 cos θ + a2 sin θ)
With that, conditions (1) and (2) are verified, so we know that f is an
isomorphism, and we can say that the spaces are isomorphic G ∼ = R 2 .
1.5 Example Let V be the space {c1x + c2y + c3z � � c1,c2,c3 ∈ R} of linear
combinations of three variables x, y, andz, under the natural addition and
scalar multiplication operations. Then V is isomorphic to P2, the space of
quadratic polynomials.
To show this we will produce an isomorphism map. There is more than one
possibility; for instance, here are four.
c1x + c2y + c3z
f1
↦−→ c1 + c2x + c3x2 f2
↦−→ c2 + c3x + c1x2 f3
↦−→ −c1 − c2x − c3x2 f4
↦−→ c1 +(c1 + c2)x +(c1 + c3)x2 Although the first map is the more natural correspondence, below we shall
verify that the second one is an isomorphism, to underline that there are many
isomorphisms other than the obvious one that just carries the coefficients over
(showing that f1 is an isomorphism is Exercise 12).
To show that f2 is one-to-one, we will prove that if f2(c1x + c2y + c3z) =
f2(d1x + d2y + d3z) then c1x + c2y + c3z = d1x + d2y + d3z. The assumption
that f2(c1x+c2y +c3z) =f2(d1x+d2y +d3z) gives, by the definition of f2, that
c2 + c3x + c1x 2 = d2 + d3 + d1x 2 . Equal polynomials have equal coefficients, so
c2 = d2, c3 = d3, andc1 = d1. Thusf2(c1x + c2y + c3z) =f2(d1x + d2y + d3z)
implies that c1x + c2y + c3z = d1x + d2y + d3z and therefore f2 is one-to-one.