Final answer:
If subset s1 of set s2 is linearly dependent, then s2 is also linearly dependent because the nontrivial linear combination that exists in s1 is also part of s2. The addition of more elements to s2 cannot create independence, thus maintaining or increasing the linear dependence.
Step-by-step explanation:
Proving Linear Dependence in Sets
To prove that a nonempty subset s1 of a finite set s2 being linearly dependent implies that s2 is also linearly dependent, we must understand the definitions involved. Linear dependence in a set implies that there is a nontrivial linear combination of its elements that equals zero. In other words, there exist scalars, not all of which are zero, such that when multiplied by elements of the set and summed together, the result is zero.
If subset s1 of set s2 is linearly dependent, meaning there exists a nontrivial combination of its elements that sums to zero, then this combination also exists when considering set s2 since s1 is a part of s2. Any linear combination that works for s1 will naturally be a part of the linear combinations available for s2, maintaining the linear dependence. Therefore, we cannot escape the fact that s2 is at a minimum as dependent as s1 is, which implies s2's linear dependence.
This can be better understood by acknowledging that linear dependence is a property that does not diminish by adding more elements to a set. Adding elements can introduce more dependencies or maintain the existing ones, but cannot create independence where there wasn’t any. Consequently, if a subset s1 is dependent, so is the larger set s2 that contains it.