# 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.

Let ${v}_{1},{v}_{2},\dots .,{v}_{k}$ be vectors of Rn such that
$v={c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\dots +{c}_{k}{v}_{k}={d}_{1}{v}_{1}+{d}_{2}{v}_{2}+\dots +{d}_{k}{v}_{k}$.
for some scalars ${c}_{1},{c}_{2},\dots .,{c}_{k},{d}_{1},{d}_{2},\dots .,{d}_{k}$.Prove that if $ci\ne dif\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}somei=1,2,\dots .,k$,
then ${v}_{1},{v}_{2},\dots .,{v}_{k}$ are linearly dependent.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Talisha

You haven't mentioned what v is so I'm going to ignore it.
We know that
$\sum _{i=1}^{k}{c}_{i}{v}_{i}=\sum _{i=1}^{k}{d}_{i}{v}_{i}$
for some ${c}_{i},{d}_{i}$(I'm just writing what you wrote, but in a way that saves me some space). With a little rearranging, we thus see that
$\sum _{i=1}^{k}\left({c}_{i}-{d}_{i}\right){v}_{i}=0$
Now suppose that for some $j\in \left\{1,2,\dots ,k\right\}$, we have ${c}_{j}\ne {d}_{j}$. Then ${c}_{j}-{d}_{j}\ne 0$.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
$\sum _{i=1}^{k}{a}_{i}{v}_{i}=0$
such that not all of the ${a}_{i}^{\prime }sare0.Namely{a}_{1}={c}_{1}-{d}_{1},\dots ,{a}_{k}={c}_{k}-{d}_{k}$.
But let'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's what we get by saying ${c}_{j}-{d}_{j}\ne 0:wecan÷by{c}_{j}-{d}_{j}$.
Doing so we get
$\left({c}_{j}-{d}_{j}\right){v}_{j}=-\left({c}_{1}{d}_{1}\right){v}_{1}-\dots -\left({c}_{j-1}-{d}_{j-1}\right){v}_{j-1}$
$-\left({c}_{j+1}-{d}_{j+1}\right){v}_{j+1}-\dots -\left({c}_{k}-{d}_{k}\right){v}_{k}$
${v}_{j}=\frac{{d}_{1}-{c}_{1}}{{c}_{j}-{d}_{j}}{v}_{1}+\dots +\frac{{d}_{j-1}-{c}_{j-1}}{{c}_{j}-{d}_{j}}{v}_{j-1}$
$+\frac{{d}_{j+1}-{c}_{j+1}}{{c}_{j}-{d}_{j}}{v}_{j+1}+\dots +\frac{{d}_{k}-{c}_{k}}{{c}_{j}-{d}_{j}}{v}_{k}$