1.3. Linear Combinations
Definition 1.3.1. Let \(A\) be a set equipped with an associative addition operation and an additive identity \(0_A\), let \(n \in \mathbb{N}\), and let
\[a:[n]\longrightarrow A\]be a function. The finite sum of the family \(\lbrace a_i\rbrace_{i\in [n]}\), denoted by
\[\sum_{i=1}^{n} a_i,\]is defined recursively as follows:
\[\sum_{i=1}^{0} a_i = 0_A,\]and, for each \(k \in \mathbb{N}\) with \(k<n\),
\[\sum_{i=1}^{k+1} a_i = \left(\sum_{i=1}^{k} a_i\right) + a_{k+1}.\]Definition 1.3.2. Let \(V\) be a vector space over \(F\), and let \(S \subseteq V\). A vector \(v \in V\) is a linear combination of the vectors in \(S\) if and only if
\[\text{there exist}\quad n\in\mathbb N,\quad a:[n] \to F,\quad u:[n] \to S \quad \text{such that} \quad v = \sum_{i=1}^{n} a_i u_i.\]If \(v\) is a linear combination of the vectors in \(S\) with \(n > 0,\) we say that \(v\) is a linear combination of \(u_1, \dots, u_n,\) and we call \(a_1, \dots, a_n\) the coefficients of the linear combination.
Definition 1.3.3. Let \(S\) be a subset of a vector space \(V\) over a field \(F.\) The span of \(S\) is the set \(\text{span} (S)\) defined by
\[\operatorname{span}(S) = \left\lbrace v\in V : v\text{ is a linear combination of the vectors in } S \right\rbrace.\]Remark 1.3.4. Let \(V\) be a vector space over \(F\). Then
\[\text{span} (\varnothing) = \lbrace 0_V\rbrace.\]Indeed, it is clear that \(0_V\in \text{span}(\varnothing)\). We now suppose \(v\in \text{span} (\varnothing)\). Then there exist functions \(u:[n]\to \varnothing\) and \(a:[n]\to F\) such that \(v = \sum_{i=1}^{n} a_i u_i\) for some \(n\in \mathbb N\). We have \(n=0\), for otherwise \(u\) does not exist. Therefore
\[v = \sum_{i=1}^{0} a_i u_i = 0_V,\]and hence \(\text{span} (\varnothing) = \lbrace 0_V\rbrace\).
Theorem 1.3.4. Let \(S\) be a subset of a vector space \(V\) over a field \(F.\)
- \(\text{span} (S)\) is a subspace of \(V\) that contains \(S.\)
- Every subspace of \(V\) that contains \(S\) contains \(\text{span} (S).\)
Proof. (1) If \(S=\varnothing\), then \(\text{span} (\varnothing) = \lbrace 0_V\rbrace\) so that \(\text{span} (\varnothing)\) is a subspace of \(V\). If \(S\ne \varnothing\) then there is a vector \(z\) in \(S\). Since \(0z=0_V,\)
\[0_V\in \text{span} (S).\]Let \(x,y\in \text{span} (S).\) Then there exist \(u_1,\dots,u_m,v_1,\dots,v_n\in S\) and \(a_1,\dots,a_m,b_1,\dots,b_n\in F\) such that
\[x = a_1u_1 + \cdots + a_mu_m \quad \text{and} \quad y = b_1v_1 + \cdots + b_nv_n.\]Then
\[x+y = a_1u_1 + \cdots + a_mu_m + b_1v_1 + \cdots + b_nv_n\]and, for any \(c\in F,\)
\[cx = (ca_1)u_1 + \cdots + (ca_m)u_m\]are linear combinations of the vectors in \(S\), so that \(x+y\) and \(cx\) are in \(\text{span} (S)\). Thus \(\text{span} (S)\) is a subspace of \(V\). Furthermore, if \(v\in S\) then \(v=1\cdot v \in \text{span}(S)\); so \(\text{span} (S)\) contains \(S\). (2) If \(S=\varnothing\) then \(\text{span} (\varnothing) = \lbrace 0_V\rbrace\), which is contained by any subspace of \(V\). Now let \(S \ne \varnothing\) and let \(W\) be a subspace of \(V\) that contains \(S\). If \(w\in \text{span} (S)\), then
\[w = c_1w_1 + \cdots + c_kw_k\]for some \(w_1,\dots,w_k\in S\) and \(c_1,\dots,c_k\in F.\) Since \(S\subseteq W\), we have \(w_1,\dots,w_k\in W.\) Therefore \(w\in W\) and thus \(\text{span} (S)\subseteq W\). \(\square\)
Definition 1.3.5. Let \(S\) be a subset of a vector space \(V\) over a field \(F.\) \(S\) is said to generate \(V\) if and only if
\[\text{span} (S) = V.\]References
- Friedberg, S. H., Insel, A. J., & Spence, L. E. (2025). Linear algebra (5th ed.). Pearson Education South Asia Pte Ltd.