Suppose v1,…,vmv_1, \dots, v_m is linearly independent in VV and w∈Vw \in V. Prove that if v1+w,…,vm+wv_1 + w, \dots, v_m + w is linearly dependent, then w∈span(v1,…,vm)w \in \text{span}(v_1,\dots,v_m).


Suppose v1+w…,vm+wv_1+w\dots,v_m+w is linearly dependent, then

a1(v1+w)+⋯+am(vm+w)=0a_1(v_1+w) + \dots + a_m(v_m+w) = 0

With some aj≠0a_j \ne 0. Distribute and move the ww's to one side

a1v1+⋯+amvm=−(a1+⋯+am)wa_1v_1 + \dots + a_mv_m = -(a_1+\dots+a_m)w

Since the vv's are linearly independent, and some aj≠0a_j \ne 0 we must have a1v1+⋯+amvm≠0a_1v_1 + \dots + a_mv_m \ne 0. Therefor −(a1+⋯+am)w≠0-(a_1+\dots+a_m)w \ne 0 which allows us to divide by the constant term to get

w=b1v1+⋯+bmvmw = b_1v_1 + \dots + b_m v_m

For bk=−ak/(a1+⋯+am)b_k = -a_k/(a_1 + \dots + a_m). Thus w∈span(v1,…,vm)w \in \text{span}(v_1,\dots,v_m).