Let v1,v2,….,vk be vectors of Rn such that v=c1v1+c2v2+…+ckvk=d1v1+d2v2+…+dkvk. for some scalars c1,c2,….,ck,d1,d2,….,dk.Prove that if ci≠diforsomei=1,2,….,k, then v1,v2,….,vk are linearly dependent.

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Let v1,v2,….,vk be vectors of Rn such that  v=c1v1+c2v2+…+ckvk=d1v1+d2v2+…+dkvk.  for some scalars c1,c2,….,ck,d1,d2,….,dk.Prove that if ci≠diforsomei=1,2,….,k,  then v1,v2,….,vk are linearly dependent.

Talisha · Accepted Answer

You haven&#039;t mentioned what v is so I&#039;m going to ignore it.We know that∑i=1kcivi=∑i=1kdivifor some ci,di(I&#039;m just writing what you wrote, but in a way that saves me some space). With a little rearranging, we thus see that∑i=1k(ci−di)vi=0Now suppose that for some j∈{1,2,…,k}, we have cj≠dj. Then cj−dj≠0.Depending on how your class defines a linearly dependent set of vectors, you might be done at this point.By supposition, there is a solution to the equation∑i=1kaivi=0such that not all of the ai′sare0.Namelya1=c1−d1,…,ak=ck−dk.But let&#039;s say your class defines linearly dependent as meaning that at least one of the vectors is expressible as a linear combination of the others. Then we just move all of the terms except the jth one to the RHS.And here&#039;s what we get by saying cj−dj≠0:wecan÷bycj−dj.Doing so we get(cj−dj)vj=−(c1d1)v1−…−(cj−1−dj−1)vj−1−(cj+1−dj+1)vj+1−…−(ck−dk)vkvj=d1−c1cj−djv1+…+dj−1−cj−1cj−djvj−1+dj+1−cj+1cj−djvj+1+…+dk−ckcj−djvk

Let v_1,v_2,....,v_k be vectors of Rn such that v=c_1v_1+c_2v_2+...+c_kv_k=d_1v_1+d_2v_2+...+d_kv_k. for some scalars c_1,c_2,....,c_k,d_1,d_2,....,d_k.Prove that if ci != di for some i = 1, 2,....,k, then v_1,v_2,....,v_k are linearly dependent.

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