Addition and scalar multiplication of linear maps

Weìve already introduced the space of all linear maps ℱ(V, W) of all function f : VW 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), vV 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 : VW 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 : VW 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 VV 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 VW, 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

f i j ( v k ) = { w k for  k = i 0 for  k i

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: VW and g, g1, g2: WU be linear maps and α1, α2 ∈ 𝔽.

  1. The composition gf : VU is linear.

  2. g ◦ (α1 f1 + α2 f2) = α1 (gf1) + α2 (gf2).

  3. 1 g1 + α2 g2) ◦ f = α1 (g1f) + α2 (g2f).

  4. If f−1: WV exists it is linear.

Poof. (i) All we need to do is check the linearity condition for gf given it is satisfied for f and g.

(gf)(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 (gf)(v1) + α2 (gf)(v2)

(ii) For vV 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 (gf1 )(v) + α2 (gf2)(v) = (α1 (gf1) + α2 (gf2))(v)

(iii) This works exactly like the proof for part (ii).

(iv) Let f : VW be an invertible linear maps with inverse f−1 : WV. Consider two vectors w1, w2W. Since f is surjective they can be written as w1 = f (v1) and w2 = f (v2) for two vectors v1, v2V, 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−11 f(v1) + α2 f (v2)) = f−1 (f (α1 v1 + α2 v2)) = α1 v1 + α2 v2 =
α1 f−1 (w1) + α2 f−1(w2)  □

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