91Ó°ÊÓ

Assume \(S = \{0\}\) or there exist distinct vectors \(v, u_1, u_2, \ldots, u_n\) in \(S\) such that \(v\) is a linear combination of \(u_1, u_2, \ldots, u_n\). We need to show that \(S\) is linearly dependent. If \(S = \{0\}\), then there is a trivial linear combination of the zero vector itself, so the set is linearly dependent. Now, assume there exist distinct vectors \(v, u_1, u_2, \ldots, u_n\) in \(S\) such that \(v\) is a linear combination of \(u_1, u_2, \ldots, u_n\). This means that we can write: \(v = c_1u_1 + c_2u_2 + \cdots + c_nu_n\) for some coefficients \(c_i\), not all equal to zero. Since \(v\) can be written as a linear combination of other vectors in the set, the set \(S\) is linearly dependent. Hence, the first direction is proved.

Assume \(S\) is linearly dependent. We need to show that either \(S = \{0\}\) or there exist distinct vectors \(v, u_1, u_2, \ldots, u_n\) in \(S\) such that \(v\) is a linear combination of \(u_1, u_2, \ldots, u_n\). Since \(S\) is linearly dependent, there exists at least one non-trivial linear combination of vectors in \(S\) that equals zero, i.e., \(c_1v_1 + c_2v_2 + \cdots + c_kv_k = 0\) for some coefficients \(c_i\), not all equal to zero, and vectors \(v_1, v_2, \ldots, v_k \in S\). If \(k=1\), we have only one non-zero coefficient \(c_1\) and \(v_1 \in S\) with \(c_1v_1 = 0\). Since \(c_1 \neq 0\), it implies \(v_1 = 0\), and \(S=\{0\}\) as required. If \(k > 1\), let \(c_p\) be a non-zero coefficient; without loss of generality, assume \(c_1 \neq 0\). Then we have: \(c_1v_1 = -c_2v_2 - \cdots - c_kv_k\) Now, divide by \(c_1\): \(v_1 = -\frac{c_2}{c_1}v_2 - \cdots - \frac{c_k}{c_1}v_k\) In this case, we have \(v_1\) as a linear combination of other vectors in the set \(S\), which are \(v_2, \ldots, v_k\). Hence, \(v_1\) and \(v_2, \ldots, v_k\) are the required distinct vectors as stated in the problem statement. Hence, both directions of "if and only if" are proved, completing the proof.