Linear Algebra

184 Chapter 3. Maps Between Spaces
Show that F is linear.
(b) Does the converse hold—is any linear map from V to R 2 made up of two
linear component maps to R 1 ?
(c) Generalize.
3.II.2 Rangespace and Nullspace
The difference between homomorphisms and isomorphisms is that while both
kinds of map preserve structure, homomorphisms needn’t be onto and needn’t
be one-to-one. Put another way, homomorphisms are a more general kind of
map; they are subject to fewer conditions than isomorphisms. In this subsection,
we will look at what can happen with homomorphisms that the extra conditions
rule out happening with isomorphisms.
We first consider the effect of dropping the onto requirement. Of course,
any function is onto some set, its range. The next result says that when the
function is a homomorphism, then this set is a vector space.
2.1 Lemma Under a homomorphism, the image of any subspace of the domain
is a subspace of the codomain. In particular, the image of the entire space, the
range of the homomorphism, is a subspace of the codomain.
Proof. Let h: V → W be linear and let S be a subspace of the domain V .
The image h(S) is nonempty because S is nonempty. Thus, to show that h(S)
is a subspace of the codomain W , we need only show that it is closed under
linear combinations of two vectors. If h(�s1) andh(�s2) are members of h(S) then
c1 · h(�s1)+c2 · h(�s2) =h(c1 ·�s1)+h(c2 ·�s2) =h(c1 ·�s1 + c2 ·�s2) isalsoamember
of h(S) because it is the image of c1 · �s1 + c2 · �s2 from S. QED
2.2 Definition The rangespace of h: V → W is
R(h) ={h(�v) � � �v ∈ V }
sometimes denoted h(V ). The dimension of the rangespace is the map’s rank .
(We shall soon see the connection between the rank of a map and the rank of a
matrix.)
2.3 Example Recall that the derivative map d/dx: P3 →P3 given by a0 +
a1x + a2x 2 + a3x 3 ↦→ a1 +2a2x +3a3x 2 is linear. The rangespace R(d/dx) is
the set of quadratic polynomials {r + sx + tx 2 � � r, s, t ∈ R}. Thus, the rank of
this map is three.
2.4 Example With this homomorphism h: M2×2 →P3
� �
a b
c d
↦→ (a + b +2d)+0x + cx 2 + cx 3