BASIC LINEAR ALGEBRA AN EXERCISE APPROACH Gabriel Nagy Kansas State University c Gabriel Nagy CHAPTER 1 Vector spaces and linear maps In this chapter we introduce the basic algebraic notions of vector spaces and linear maps. 1. Vector spaces Suppose k is a field. (Although the theory works for arbitrary fields, we will eventually focus only on the choices k = R, C.) Definition. A k-vector space is an abelian group (V, +), equipped with an external operation1 k × V 3 (λ, v) 7−→ λv ∈ V, called scalar multiplication, with the following properties: • λ · (v + w) = (λ · v) + (λ · w), for all λ ∈ k, v, w ∈ V . • (λ + µ) · v = (λv) + (µv), for all λ, µ ∈ k, v ∈ V . • (λ · µ)v = λ · (µ · v), for all λ, µ ∈ k, v ∈ V . • 1 · v = v, for all v ∈ V . The elements of a vector space are sometimes called vectors. Examples. The field k itself is a k-vector space, with its own multiplication as scalar multiplication. A trivial group (with one element) is always a k-vector space (with the only possible scalar multiplication). 1 Suppose V is a k-vector space. Prove that 0 · v = 0, for all v ∈ V. (The zero in the left-hand side is the field k. The zero in the right-hand side is the neutral element in the abelian group V .) Use the above fact to conclude that for any v ∈ V , the vector −v (the inverse of v in the abelian group V ) can also be described by −v = (−1) · v. 2 Fix a field k, a non-empty set I, and a family (Vi)i∈I of k-vector spaces. Q Consider the product i∈I Vk, equipped with the operations: • (vi)i∈I + (wi)i∈I = (vi + wi)i∈I ; • λ · (vi)i∈I = (λvi)i∈I . 1 When convenient, the symbol · may be omitted. 1 2 1. VECTOR SPACES AND LINEAR MAPS Q Prove that i∈I Vk is a k-vector space. This structure is called the k-vector space direct product of the family (Vi)i∈I . Definition. Suppose V is a k-vector space. A subset X ⊂ V is called a k-linear subspace, if • Whenever x, y ∈ X, we have x + y ∈ X. • Whenever x, y ∈ Xand λ ∈ k, we have λx ∈ X. 3 If X is a k-linear subspace of the k-vector space V , then X itself is a k-vector space, when equipped with the operations “inherited” from V . Prove than any linear subspace of V contains the zero vector 0 ∈ V 4 Let (Vi)i∈I be a family of k-vector spaces (indexed by a non-empty set I). For Q an element v = (vi)i∈I ∈ i∈I Vi let us define the set bvc = {i ∈ I : vi 6= 0}. Prove that the set Y {v ∈ Vi : bvc is finite } i∈I Q is a linear subspace of Vi. This space is called the k-vector space direct sum i∈I L of the family (Vi)i∈I , and is denoted by i∈I Vi. Definition. Suppose we have a family (Vi)i∈I of vector spaces. For a fixed L index j ∈ I, define the map εj : Vj → i∈I Vi as follows. For a vector v ∈ Vj we construct εj(v) = (wi)i∈I , where v if i = j w = i 0 if i 6= j. L We call the maps εj : Vj → i∈I Vi, j ∈ I, the standard inclusions. The maps Y πj : Vi 3 (vi)i∈I 7−→ vj ∈ Vj i∈I are called the coordinate maps. 5 Let (Vi)i∈I be a family of vector spaces. Prove that the standard inclusions εi, i ∈ I. are injective. In fact prove that πi ◦ εi = IdV . Prove that any element L i v ∈ i∈I Vi is given as X v = (εi ◦ πi)(v). i∈bvc P In other words, if v = (vi)i∈I , then v = i∈bvc εi(vi). 6 Suppose (Xj)j∈J is a family of k-linear subspaces of V . Prove that the inter- T section j∈J Xj is again a k-linear subspace of V . 1. VECTOR SPACES 3 Definition. Let V be a k-vector space, and let M ⊂ V be an arbitrary subset of V . Consider the family F = {X : X k-linear subspace of V , and X ⊃ M}. The set \ Spank(M) = X, X∈F which is a linear subspace of V by the preceding exercise, is called the k-linear span of M in V . Convention. Spank(∅) = {0}. Example. The linear span of a singleton is described as Spank({v}) = kv(= {λv : λ ∈ k}). 7 Prove that if M and N are subsets of a k-vector space V , with M ⊂ N, then we also have the inclusion Spank(M) ⊂ Spank(N). Give an example where M ( N, but their spans coincide. 8 Let V be a k-vector space, and M be a subset of V . For an element v ∈ V , prove that the following are equivalent: (i) v ∈ Spank(M); (ii) there exists an integer n ≥ 1, elements x1, . , xn ∈ M, and scalars 2 λ1, . , λn ∈ k such that v = λ1x1 + ··· + λnxn. Hint: First prove that the set of elements satisfying property (ii) is a linear subspace. Second, prove that the linear span of M contains all elements satisfying (ii). Notation. Suppose V is a vector space, and A1,...,An are subsets of V . We define A1 + ··· + A2 = {a1 + ··· + an : ak ∈ Ak, k = 1, . , n}. 9 Let V be a k-vector space. Suppose A1,...,An are k-homogeneous, in the sense that for every k = 1, . , n we have the equality: Ak = {λx : λ ∈ k, x ∈ Ak}. Prove the equality Span(A1 ∪ · · · ∪ An) = Span(A1) + ··· + Span(An). 10 Let V be a k-vector space, and (Xj)j∈J be a family of linear subspaces of V . For an element v ∈ V , prove that the following are equivalent: S (i) v ∈ Spank j∈J Xj); S (ii) there exists an integer n ≥ 1 and x1, . , xn ∈ j∈J Xj, such that v = x1 + ··· + xn. 2 From now on we will use the usual convention which gives the scalar multiplication prece- dence over addition. 4 1. VECTOR SPACES AND LINEAR MAPS Comment. If X1,...,Xn are linear subspaces of the vector space V , then using the notation preceding Exercise ??, and the above result, we get Span(X1 ∪ · · · ∪ Xn) = X1 + ··· + Xn. 11 In general, a union of linear subspaces is not a linear subspace. Give an 2 example of two linear subspaces X1,X2 ⊂ R , such that X1 ∪ X2 is not a linear subspace. 12 Prove that the union of a directed family of linear subspaces is a linear subspace. That is, if (Xj)j∈J is a family of k-linear subspaces of the k-vector space V , with the property • For any j, k ∈ J there exists some ` ∈ J such that Xj ⊂ X` ⊃ Xk, S then j∈J Xj is again a k-linear subspace of V . Hint: Use the preceding exercise. Definition. Suppose V is a k-vector space. A set M ⊂ V is said to be k- linearly independent, if (compare with Exercise ??) for every strict subset P ( M, one has the strict inclusion Spank(P ) ( Spank(M). 13 Let M be a subset in the k-vector space V . Prove that the following are equivalent: (i) M is linearly independent. (ii) If n ≥ 1 is an integer, if x1, . , xn ∈ M are different elements, and if λ1, . , λn ∈ k satisfy λ1x1 + ··· + λnxn = 0, then λ1 = ··· = λn = 0. Hint: To prove (i) ⇒ (ii) show that if one has a relation as in (ii), then one of the x’s can be eliminated, without changing the linear Span. 14 Prove that a linearly independent set M cannot contain the zero element. Prove that a subset of a linearly independent set is again linearly independent. 15 Prove that the union of a directed family of linearly independent sets is again a linearly independent set. That is, if (Xj)j∈J is a family of k-linearly independent subsets of the k-vector space V , with the property • For any j, k ∈ J there exists some ` ∈ J such that Xj ⊂ X` ⊃ Xk, S then j∈J Xj is again a k-linearly independent subset of V . 16 Suppose V is a vector space, and x1, x2,... is a sequence (finite or infinite) of different non-zero vectors in V . Prove that the following are equivalent: (i) The set M = {x1, x2,... } is linearly independent. (ii) The sequence of susbspaces Wk = Span({x1, . , xk}) is strictly increasing, in the sense that we have strict inclusions W1 ( W2 ( .... 1. VECTOR SPACES 5 Hint: The implication (i) ⇒ (ii) is clear from the definition. Conversely, if M were not linearly independent, there exist scalars λ1, . , λn ∈ k such that λ1x1 + ··· + λnxn = 0, and at least one of the λ’s non-zero. If we take k = max{j : 1 ≤ j ≤ n and λj 6= 0}, then we get xk ∈ Span{x1, . , xk−1, which proves that Wk = Wk−1. Definition. Let V be a k-vector space. A subset B ⊂ V is called a k-linear basis for V , if: • Spank(B) = V ; • B is k-linearly independent. 17 Prove the following: Theorem 1.1.1. Let V be a k-vector space, and let P ⊂ M ⊂ V be subsets, with P linearly independent, and Spank(M) = V . Then there exists a linear basis B such that P ⊂ B ⊂ M. Sketch of proof: Consider the set B = {B ⊂ V : B linearly independent, and P ⊂ B ⊂ M}, equipped with the inclusion as the order relation.