Suppose v1,…,vm is linearly independent in V and w∈V. Prove that
dimspan(v1+w,…,vm+w)≥m−1
It's asking us to prove that adding w can at most decrease the dimension by one.
Intuitively this makes sense since we still have access to the differences vj−vk but not individual vj
If v1+w,…,vm+w is linearly independent the result clearly holds. So let's suppose they're dependent, In 2a/10 we showed this implies w∈span(v1,…,vm).
Write w=a1v1+⋯+amvm, then
b1(v1+w)+⋯+bm(vm+w)=0
Becomes (let S=b1+⋯+bm)
S(a1v1+⋯+amvm)=−(b1v1+⋯+bmvm)
Which only happens if
(Sa1−b1)v1+⋯+(Sam−bm)vm=0
Since v1,…,vm are independent we must have Saj−bj=0 for all 1≤j≤m. The only way for this to happen is bj=aj/S meaning there is exactly one nonzero combination that gives zero. Let b1,…,bm be this combination, with bj being the first nonzero coefficient. This implies we can remove vj without changing the span since vj is a combination of other v's
vj=−(b1v1+⋯+bj−1vj−1+bj+1vj+1+⋯+bmvm)/bj
Since there's only one combination that gives zero, after removing vj the list (v1+w,…,vm+w) is independent. and so the dimension of the span is m−1.