Spanning Sets and Linear Independence

Given a general vector space V over the field 𝔽. Given that we have two operations, vector addition and scalar multiplication, at our disposal the most general algebraic expression is of the form

$${\alpha }_1{\mathbf{v}}_1+\dots +{\alpha }_k{\mathbf{v}}_k=\sum _{i=1}{\alpha }_i{\mathbf{v}}_i\text{\hspace{1em}1.3.1}$$

where v₁, ..., v_k ∈ V and α₁, ..., α_k ∈ 𝔽 are k vector and scalars, respectively. An expression of the form (1.3.1) is called a linear combination of the vectors v₁, ..., v_k. The set of all linear combinations of given vectors v₁, ..., v_k is called the span of these vectors, and it is written as

$$\text{Span}({\mathbf{v}}_1+\dots +{\mathbf{v}}_k):=\{\sum _{i=1}^k{\alpha }_i{\mathbf{v}}_i\}$$

Proposition 1.3.1 For any vectors v₁, ..., v_k ∈ V, the span, Span(v₁, ..., v_k), is a vector subspace of V. Proof All we need to do is to verify that the span satisfies the conditions in Def. 6.2. Consider two vectors u, w ∈ Span(v₁, ..., v_k) in the span. By definition of the span, this means they can be written as linear combinations

$$\mathbf{u}=\sum _{i=1}^k{\alpha }_i{\mathbf{v}}_i,\mathbf{w}=\sum _{i=1}^k\beta {\mathbf{w}}_i$$

for suitable scalars α_i, β_i ∈ 𝔽. Their sum and the scalar multiple of u with α ∈ 𝔽 are then given by

$$\mathbf{u}+\mathbf{w}=\sum _{i=1}^k({\alpha }_i+{\beta }_i){\mathbf{v}}_i,\alpha \mathbf{u}=\sum _{i=1}^k(\alpha {\alpha }_i){\mathbf{v}}_i$$

and are, hence, both contained in the span. This shows that the two conditions of Definition 1.5.1 are indeed satisfied.  □

This result means that the span provides us with a way of generating vector subspaces. The span has a straightforward geometric interpretation, at least for coordinate vectors with real entries. The span of a single vector v ∈ ℝⁿ consists of all scalar multiples of this vector and, hence, can be thought of as the line through 0 which contains v. The span of two vectors u, v ∈ ℝⁿ (which are not multiples of each other) represents the plane through 0 which contains both vectors. More generally, spans of column vectors are lines, planes and their higher-dimensional analogues through the ’origin’ 0. We will be more precise about this later but for now just present an example.

Definition 1.3.3 If every element of a vector space V over a field 𝔽 is a linear combination of v₁, ..., v_n we say that the set {v₁, ..., v_n} spans V over F (or generates V).  □

Note that V may be infinite or finite, but any given linear combination of vectors from S involves only a finite number of vectors.

Example 1.3.4 Let A the subset of ℝ³ given by A = {(1,0,0), (0,1,0)}. The span of A is the xy-plane. □

Example 1.3.5 Let P the vector space of polynomials of any degree. Then no finite set of polynomials spans P. If you suppose that p₁, p₂, ..., p_n are polynomials. Let p_k be the polynomial of largest degree in this set and let n = deg p_k. Then it is clear that the polynomial p(x) = xⁿ⁺¹ cannot be written as a linear combination of p₁, p₂, ..., p_n.  □

Linear Independence

Definition 1.3.4 Let V be a vector space. A nonempty set S of vectors in V is linearly independent if for any distinct vectors v₁, v₂, ..., v_n in S, and scalars λ_i

That is, S is linearly independent if the only linear combination of vectors from S that is equal to 0 is the trivial linear combination, all of whose coefficients are 0. If S is not linearly independent, it is said to be linearly dependent.

It is clear that a linearly independent set of vectors cannot contain the zero vector, since then 1 ⋅ 0 = 0 violates the condition of linear independence.

Let V be a vector space over a field F. The preceding examples show that linear independence and spanning do not imply each other; a subset of V may have one, both, or neither of these properties. A subset that has both properties is given a special name

Definition 1.3.6 A subset {v₁, ..., v_n} of a vector space V over a field F is said to be a basis of V If it spans V and is linearly Independent over F.  □

When the vectors u and v are linearly independent, it means that they are not scalar multiples of each other: i.e. they have different directions. Arbitrary linear independent vectors u and v in ℝ² can be illustrated as shown in Fig. 1:

Linearly independent vectors u and v have different directions.

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