Define the span of vectors \(v_1, \dots, v_k\) in \(\mathbb{R}^n\).
Prove that if \(w\) lies in the span of \(v_1, \dots, v_k\), then the span of \(v_1, \dots, v_k, w\) is equal to the span of \(v_1, \dots, v_k\).
<1>2. Suppose \(w\in\operatorname{span}(v_1,\dots,v_k)\). Then \[ \operatorname{span}(v_1,\dots,v_k) \subseteq \operatorname{span}(v_1,\dots,v_k,w). \]
<1>3. Conversely, \[ \operatorname{span}(v_1,\dots,v_k,w) \subseteq \operatorname{span}(v_1,\dots,v_k). \]
<1>4. Therefore \[ \operatorname{span}(v_1,\dots,v_k,w) = \operatorname{span}(v_1,\dots,v_k). \]
Solution
1. The span of \(v_1,\dots,v_k\) is\[ \operatorname{span}(v_1,\dots,v_k) =\left\{\sum_{i=1}^k a_i v_i : a_1,\dots,a_k\in\mathbb R\right\}. \]
Every linear combination of \(v_1,\dots,v_k\) is also a linear combination of \(v_1,\dots,v_k,w\), by taking the coefficient of \(w\) to be \(0\).
Because \(w\) lies in the first span, there are \(b_1,\dots,b_k\in\mathbb R\) such that \[ w=\sum_{i=1}^k b_i v_i. \] Hence every element of the enlarged span has the form \[ \sum_{i=1}^k a_i v_i+c w =\sum_{i=1}^k (a_i+c b_i)v_i, \] which belongs to \(\operatorname{span}(v_1,\dots,v_k)\).
This follows from the two inclusions in 2 and 3.