Suppose v1,…,vmv_1,\dots,v_m is linearly independent in VV and w∈Vw \in V. Prove that

dim⁡span(v1+w,…,vm+w)≥m−1\dim \text{span}(v_1 + w,\dots,v_m + w) \ge m - 1

It's asking us to prove that adding ww can at most decrease the dimension by one.
Intuitively this makes sense since we still have access to the differences vj−vkv_j - v_k but not individual vjv_j

If v1+w,…,vm+wv_1+w,\dots,v_m+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)w \in \text{span}(v_1,\dots,v_m).

Write w=a1v1+⋯+amvmw = a_1v_1+\dots+a_mv_m, then

b1(v1+w)+⋯+bm(vm+w)=0b_1(v_1+w) + \dots + b_m(v_m+w) = 0

Becomes (let S=b1+⋯+bmS = b_1+\dots+b_m)

S(a1v1+⋯+amvm)=−(b1v1+⋯+bmvm)S(a_1v_1+\dots+a_mv_m) = -( b_1v_1+\dots+b_mv_m)

Which only happens if

(Sa1−b1)v1+⋯+(Sam−bm)vm=0(Sa_1 - b_1)v_1 + \dots + (Sa_m-b_m)v_m = 0

Since v1,…,vmv_1,\dots,v_m are independent we must have Saj−bj=0Sa_j-b_j = 0 for all 1≤j≤m1 \le j \le m. The only way for this to happen is bj=aj/Sb_j = a_j/S meaning there is exactly one nonzero combination that gives zero. Let b1,…,bmb_1,\dots,b_m be this combination, with bjb_j being the first nonzero coefficient. This implies we can remove vjv_j without changing the span since vjv_j is a combination of other vv's

vj=−(b1v1+⋯+bj−1vj−1+bj+1vj+1+⋯+bmvm)/bjv_j = -(b_1v_1+\dots + b_{j-1}v_{j-1} + b_{j+1}v_{j+1} + \dots +b_mv_m)/b_j

Since there's only one combination that gives zero, after removing vjv_j the list (v1+w,…,vm+w)(v_1+w,\dots,v_m+w) is independent. and so the dimension of the span is m−1m-1.