Linear Algebra

Section II. Homomorphisms 185
an image vector in the range can have any constant term, must have an x
coefficient of zero, and must have the same coefficient of x 2 as of x 3 . That is,
the rangespace is R(h) ={r +0x + sx 2 + sx 3 � � r, s ∈ R} and so the rank is two.
The prior result shows that, in passing from the definition of isomorphism to
the more general definition of homomorphism, omitting the ‘onto’ requirement
doesn’t make an essential difference. Any homomorphism is onto its rangespace.
However, omitting the ‘one-to-one’ condition does make a difference. A
homomorphism may have many elements of the domain map to a single element
in the range. The general picture is below. There is a homomorphism and its
domain, codomain, and range. The homomorphism is many-to-one, and two
elements of the range are shown that are each the image of more than one
member of the domain.
domain V
.
.
codomain W
�
R(h)
(Recall that for a map h: V → W , the set of elements of the domain that are
mapped to �w in the codomain {�v ∈ V � � h(�v) = �w} is the inverse image of �w. It
is denoted h −1 ( �w); this notation is used even if h has no inverse function, that
is, even if h is not one-to-one.)
2.5 Example Consider the projection π : R 3 → R 2
⎛ ⎞
x
⎝y⎠
z
π
↦−→
� �
x
y
which is a homomorphism but is not one-to-one. Picturing R 2 as the xy-plane
inside of R 3 allows us to see π(�v) as the “shadow” of �v in the plane. In these
terms, the preservation of addition property says that
�v1 above (x1,y1) plus �v2 above (x2,y2) equals �v1 + �v2 above (x1 + x2,y1 + y2).
Briefly, the shadow of a sum equals the sum of the shadows. (Preservation of
scalar multiplication has a similar interpretation.)