Define what it means for vectors \(v_1, \ldots, v_n\) in a vector space \(V\) to be linearly independent.
Suppose that vectors \(v_1, \ldots, v_n\) in a vector space \(V\) span \(V\) and that no proper subset of \(\{v_1, \ldots, v_n\}\) spans \(V\). Prove that \(v_1, \ldots, v_n\) are linearly independent.
Solution
(a).
1. The vectors \(v_1, \ldots, v_n\) are linearly independent if the only scalars \(c_1, \ldots, c_n\) with \(c_1 v_1 + \cdots + c_n v_n = 0\) are \(c_1 = \cdots = c_n = 0\).
definition of linear independence.
(b).
1. Suppose for contradiction that \(v_1, \ldots, v_n\) are linearly dependent.
assume the conclusion fails.
2. Then there is a nontrivial relation \(c_1 v_1 + \cdots + c_n v_n = 0\) with some \(c_j \neq 0\).
1 (a).
3. Hence \(v_j = -\sum_{i \neq j} \frac{c_i}{c_j} v_i\), so \(v_j\) is a linear combination of the other vectors.
2, solving for \(v_j\).
4. Therefore \(\{v_1, \ldots, v_n\} \setminus \{v_j\}\) still spans \(V\) (any linear combination using \(v_j\) can be rewritten using the other vectors).
3.
5. This contradicts the hypothesis that no proper subset spans \(V\).
4.
6. Hence \(v_1, \ldots, v_n\) are linearly independent.
5.
7. Q.E.D.
1 (a) and 6 (b).