Page 170 - A Course in Linear Algebra with Applications

154 Chapter Six: Linear Transformations

Example 6.1.3
Define F 3 : R -»• R by F 2(x) = 2x - 1. This function is both
injective and surjective, so it is bijective. (The reader should
supply the proof.)

Composition of functions
Consider two functions F : X —> Y and G : U —» V such
that the image of G is a subset of X. Then it is possible to
combine the functions to produce a new function called the
composite of F and G

FoG : U->Y,

by applying first G and then F; thus the image of an element
x of U is given by the formula

FoG(x) = F(G(x)).

Here it is necessary to know that Im(G') is contained in X,
since otherwise the expression F(G(x)) might be meaningless.

Example 6.1.4
2 2
Consider the functions F : R —> R and G : C —• R defined
by the rules

ft-
a 2
F(( b )) = v V + 6 and G(a + v ^ ) =

Here a and 6 are arbitrary real numbers. Then F o G : C —* R
exists and its effect is described by

2 a 2
F o G(a + V^lb) = F(( 2 b)) = ^/Aa? + 4b .

A basic fact about functional composition is that it sat-
isfies the associative law. First let us agree that two functions