Linear Algebra

178 Chapter Three. Maps Between Spacesthat homomorphisms needn’t be onto and needn’t be one-to-one. This meansthat homomorphisms are a more general kind of map, subject to fewer restrictionsthan isomorphisms. We will examine what can happen with homomorphismsthat is prevented by the extra restrictions satisfied by isomorphisms.We first consider the effect of dropping the onto requirement, of not requiringas part of the definition that a homomorphism be onto its codomain. Forinstance, the injection map ι: R 2 → R 3⎛(x↦→ ⎝y) x ⎞y⎠0is not an isomorphism because it is not onto. Of course, being a function, ahomomorphism is onto some set, namely its range; the map ι is onto the xyplanesubset of R 3 .2.1 Lemma Under a homomorphism, the image of any subspace of the domainis a subspace of the codomain. In particular, the image of the entire space, therange 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 a subset of the codomain W . It is nonempty because S isnonempty and thus to show that h(S) is a subspace of W we need only showthat it is closed under linear combinations of two vectors. If h(⃗s 1 ) and h(⃗s 2 ) aremembers of h(S) then c 1·h(⃗s 1 )+c 2·h(⃗s 2 ) = h(c 1·⃗s 1 )+h(c 2·⃗s 2 ) = h(c 1·⃗s 1 +c 2·⃗s 2 )is also a member of h(S) because it is the image of c 1 ·⃗s 1 + c 2 ·⃗s 2 from S. QED2.2 Definition The rangespace of a homomorphism h: V → W isR(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 amatrix.)2.3 Example Recall that the derivative map d/dx: P 3 → P 3 given by a 0 +a 1 x + a 2 x 2 + a 3 x 3 ↦→ a 1 + 2a 2 x + 3a 3 x 2 is linear. The rangespace R(d/dx) isthe set of quadratic polynomials {r + sx + tx 2 ∣ ∣ r, s, t ∈ R}. Thus, the rank ofthis map is three.2.4 Example With this homomorphism h: M 2×2 → P 3( )a b↦→ (a + b + 2d) + 0x + cx 2 + cx 3c dan image vector in the range can have any constant term, must have an xcoefficient 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.