Page 608 - Elementary_Linear_Algebra_with_Applications_Anton__9_edition

Proof We already know that (a), (b), and (c) are equivalent, so we can complete the proof by proving the equivalence of (c)
and (d).

Suppose that and . It follows from the Dimension Theorem (Theorem 8.2.3) that

By definition, is the dimension of the range of T, so the range of T has dimension n. It now follows from Theorem

5.4.7 that the range of T is V , since the two spaces have the same dimension.

Suppose that and . It follows from these relationships that , or, equivalently,

. Thus it follows from the Dimension Theorem (Theorem 8.2.3) that

EXAMPLE 5 A Transformation That Is Not One-To-One
Let be multiplication by

Determine whether is one-to-one. , since the

Solution

As noted in Example 1, the given problem is equivalent to determining whether A is invertible. But
first two rows of A are proportional, and consequently, A is not invertible. Thus is not one-to-one.

Inverse Linear Transformations

In Section 4.3 we defined the inverse of a one-to-one matrix operator to be , and we
showed that if is the image of a vector under , then
to general linear transformations. maps back into . We shall now extend these ideas

Recall that if is a linear transformation, then the range of T, denoted by , is the subspace of W consisting of

all images under T of vectors in V. If T is one-to-one, then each vector in is the image of a unique vector in V. This

uniqueness allows us to define a new function, called the inverse of T and denoted by , that maps back into (Figure

8.3.1).

Figure 8.3.1 back into .
The inverse of T maps