Page 263 - Advanced engineering mathematics

7.10 Linear Transformations 243

S is onto, since every vector in R is the image of a vector in R under S. For example,
2
3
√ √ √
< −3, 97 >= S(−3, 97,0).But S is not one-to-one. For example, S(−3, 97,22) also
√
equals < −3, 97 >.
There is a convenient test to tell whether a linear transformation is one-to-one. We know
that every linear transformation maps the zero vector to the zero vector. The transformation is
one-to-one when this is the only vector mapping to the zero vector.

THEOREM 7.18
m
n
Let T : R → R be a linear transformation. Then T is one-to-one if and only if T (u)=O m occurs
only if u = O n .
Proof Suppose first that T is one-to-one. If T (u) = O m , then

T (u) = T (O n ) = O m
so the assumption that T is one-to-one requires that u = O n .
Conversely, suppose T (u) = O m occurs only if u = O n . To show that T is one-to-one,
n
suppose, for some u and v in R ,
T (u) = T (v).
By the linearity of T ,

T (u − v) = O m .
By assumption, this implies that
u − v = O n .
But then u = v,so T is one-to-one.

To illustrate, S in Example 7.35 is not one-to-one, because nonzero vectors map to the zero
vector. In Example 7.34, T is not one-to-one for the same reason.

EXAMPLE 7.36

Let T : R → R be defined by
4
7
T (x, y, z,w) =< x − y + 2z + 8w, y − z, x − w, y + 4w,5x + 5y − z,0,0 >.
To see if T is one-to-one, examine whether nonzero vectors can map to the zero vector. Suppose

T (x, y, z,w) = O 7 =< 0,0,0,0,0,0,0 >.
Then
< x + y + z + w, y − z, x − w, x − y + z − w,5x + 5y − z,0,0 >=< 0,0,0,0,0,0,0 >.

Looking at the second and third components of both sides of this equation, we must have y − z =
0 and x − w = 0, so y = z and x = w. From the first components,
x + y + z + w = 2x + 2y = 0,

so y =−x. From the fifth component,
5x + 5y − z = 5x − 5x − z = 0

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October 14, 2010 14:23 THM/NEIL Page-243 27410_07_ch07_p187-246