Prove that if two vectors are linearly dependent, one of them is a scalar multiple of the other.
Solution
1. Let \(u, v\) be linearly dependent.
setup.
2. Then there exist scalars \(a, b\), not both zero, with \(au + bv = 0\).
definition of linear dependence.
3. If \(a \neq 0\), then \(u = -\frac{b}{a} v\), so \(u\) is a scalar multiple of \(v\).
solve \(au + bv = 0\) for \(u\).
4. If \(b \neq 0\), then \(v = -\frac{a}{b} u\), so \(v\) is a scalar multiple of \(u\).
solve \(au + bv = 0\) for \(v\).
5. Hence one of \(u, v\) is a scalar multiple of the other.
3 and 4 cover the two cases (at least one of \(a, b\) is nonzero).
6. Q.E.D.
5.