Addition and scalar multiplication of linear maps
Weìve already introduced the space of all linear maps ℱ(V, W) of all function f : V → W between two vector spaces V and W over 𝔽 can be made into a vector space over the same field. Vector addition and multiplication of functions are defined as,
(f + g)(v) = f(v) + g(w), (αf)(v) = αf(v) 3.4.1
where f, g ∈ ℱ(V, W), v ∈ V and α ∈ 𝔽. We are interested in understanding whether this property of linearity is preserved under addition and scalar multiplication of functions.
Theorem 3.4.1 Let f, g : V → W be two linear maps between vector spaces V and W over 𝔽 and α ∈ 𝔽 a scalar. Then the sum f + g and the scalar multiple αf, as defined in Eq. (3.4.1), are linear.
Proof. We need to check the linearity condition (12.1) for f + g and αf, given it is satisfied for f and g.
(f + g)(α1 v1 + α2v2) = f(α1 v1 + α2 v2) + g(α1 v1 + α2v2)
= α1 f(v1) + α2 f(v2) + α1 g(v1) + α2g(v2)
= α1 (f(v1) + g(v1)) + α2 (f v2) + g(v2))
= α1 (f + g)(v1) + α2 (f + g)(v2)
The proof for αf works analogously and is left as Exercise. □
We can reformulate this result more abstractly by saying that the set of linear maps f : V → W forms a vector subspace of ℱ(V, W). This vector space of linear maps from V to W is also denoted by Hom(V, W), which stands for homomorphisms from V to W. For the special case V = W, linear maps V → V are also called (vector space) endomorphisms and the vector space of endomorphisms is denoted by End(V).
Theorem 3.4.2 The space Hom(V, W) of linear maps V → W, with addition and scalar multiplication as defined in Eq. 3.4.1, forms a vector spaces over the same field as V and W. If V and W are finite-dimensional its dimension is given by
dim𝔽 (Hom(V, W)) = dim𝔽 (V) dim𝔽 (W). 3.4.2
Proof. It remains to proof the dimension formula. We choose bases (v1 , ..., vn) and (w1 , ..., wn) of V and W and define the linear maps fij ∈ Hom(V, W) for i = 1, ... , n and j = 1, ..., m by
We want to show that these linear maps form a basis of Hom(V, W). For linear independence, start with the equation Σij λij fij = 0 and act on the vector vk which results in Σij λkj wj = 0. Since (w1, ..., wn) forms a basis, this implies that all λkj = 0. Hence, the fij are linearly independent.
Next, consider the function f = Σi,j aijfij, for aij ∈ 𝔽. Since f(vk) = Σj akj wj and the wj are a basis, it follows that any image vectors f (vk) can be obtained for suitable choices of the aij. From Theorem 3.4.1 this means the fij span Hom(V, W). Since the number of these functions is nm, Eq. (3.4.2) follows. □
Map composition and inverse
Linearity is also preserved under map composition and inversion, as the following proposition illustates.
Proposition 3.4.3 Let f, f1, f2: V → W and g, g1, g2: W → U be linear maps and α1, α2 ∈ 𝔽.
The composition g ◦ f : V → U is linear.
g ◦ (α1 f1 + α2 f2) = α1 (g ◦ f1) + α2 (g ◦ f2).
(α1 g1 + α2 g2) ◦ f = α1 (g1 ◦ f) + α2 (g2 ◦ f).
If f−1: W → V exists it is linear.
Poof. (i) All we need to do is check the linearity condition for g ◦ f given it is satisfied for f and g.
(g ◦ f)(α1 v1 + α2 v2) = g(f(α1 v1 + α2 v2)) = g(α1 f(v1) + α2 f (v2)) = α1 g(f(v1)) + α2 g(f(v2)) = α1 (g ◦ f)(v1) + α2 (g ◦ f)(v2)
(ii) For v ∈ V we have, from the definition of map composition and linearity, that
(g ◦ (α1 f1 + α2 f2 ))(v) = g((α1 f1 + α2 f2)(v)) = g(α1f1 (v) + α2 f2 (v))
= α1 (g ◦ f1 )(v) + α2 (g ◦ f2)(v) = (α1 (g ◦ f1) + α2 (g ◦ f2))(v)
(iii) This works exactly like the proof for part (ii).
(iv) Let f : V → W be an invertible linear maps with inverse f−1 : W → V. Consider two vectors w1, w2 ∈ W. Since f is surjective they can be written as w1 = f (v1) and w2 = f (v2) for two vectors v1, v2 ∈ V, so that v1 = f−1 (w1) and v2 = f−1 (w2). The linearity condition for f−1 is verified by
f−1 (α1 w1 + α2w2) = f−1 (α1 f(v1) + α2 f (v2)) = f−1 (f (α1 v1 + α2 v2))
= α1 v1 + α2 v2 =
α1 f−1 (w1) + α2 f−1(w2) □