You haven't mentioned what v is so I'm going to ignore it.

We know that

\(\displaystyle{\sum_{{{i}={1}}}^{{k}}}{c}_{{i}}{v}_{{i}}={\sum_{{{i}={1}}}^{{k}}}{d}_{{i}}{v}_{{i}}\)

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

\(\displaystyle{\sum_{{{i}={1}}}^{{k}}}{\left({c}_{{i}}-{d}_{{i}}\right)}{v}_{{i}}={0}\)

Now suppose that for some \(\displaystyle{j}\in{\left\lbrace{1},{2},\ldots,{k}\right\rbrace}\), we have \(\displaystyle{c}_{{j}}\ne{d}_{{j}}\). Then \(\displaystyle{c}_{{j}}-{d}_{{j}}\ne{0}\).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

\(\displaystyle{\sum_{{{i}={1}}}^{{k}}}{a}_{{i}}{v}_{{i}}={0}\)

such that not all of the \(\displaystyle{a}_{{i}}'{s}{a}{r}{e}{0}.{N}{a}{m}{e}{l}{y}{a}_{{1}}={c}_{{1}}-{d}_{{1}},\ldots,{a}_{{k}}={c}_{{k}}-{d}_{{k}}\).

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 \(\displaystyle{c}_{{j}}-{d}_{{j}}\ne{0}:{w}{e}{c}{a}{n}\div{b}{y}{c}_{{j}}-{d}_{{j}}\).

Doing so we get

\(\displaystyle{\left({c}_{{j}}-{d}_{{j}}\right)}{v}_{{j}}=-{\left({c}_{{1}}{d}_{{1}}\right)}{v}_{{1}}-\ldots-{\left({c}_{{{j}-{1}}}-{d}_{{{j}-{1}}}\right)}{v}_{{{j}-{1}}}\)

\(\displaystyle-{\left({c}_{{{j}+{1}}}-{d}_{{{j}+{1}}}\right)}{v}_{{{j}+{1}}}-\ldots-{\left({c}_{{k}}-{d}_{{k}}\right)}{v}_{{k}}\)

\(\displaystyle{v}_{{j}}=\frac{{{d}_{{1}}-{c}_{{1}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{1}}+\ldots+\frac{{{d}_{{{j}-{1}}}-{c}_{{{j}-{1}}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{{j}-{1}}}\)

\(\displaystyle+\frac{{{d}_{{{j}+{1}}}-{c}_{{{j}+{1}}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{{j}+{1}}}+\ldots+\frac{{{d}_{{k}}-{c}_{{k}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{k}}\)

We know that

\(\displaystyle{\sum_{{{i}={1}}}^{{k}}}{c}_{{i}}{v}_{{i}}={\sum_{{{i}={1}}}^{{k}}}{d}_{{i}}{v}_{{i}}\)

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

\(\displaystyle{\sum_{{{i}={1}}}^{{k}}}{\left({c}_{{i}}-{d}_{{i}}\right)}{v}_{{i}}={0}\)

Now suppose that for some \(\displaystyle{j}\in{\left\lbrace{1},{2},\ldots,{k}\right\rbrace}\), we have \(\displaystyle{c}_{{j}}\ne{d}_{{j}}\). Then \(\displaystyle{c}_{{j}}-{d}_{{j}}\ne{0}\).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

\(\displaystyle{\sum_{{{i}={1}}}^{{k}}}{a}_{{i}}{v}_{{i}}={0}\)

such that not all of the \(\displaystyle{a}_{{i}}'{s}{a}{r}{e}{0}.{N}{a}{m}{e}{l}{y}{a}_{{1}}={c}_{{1}}-{d}_{{1}},\ldots,{a}_{{k}}={c}_{{k}}-{d}_{{k}}\).

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 \(\displaystyle{c}_{{j}}-{d}_{{j}}\ne{0}:{w}{e}{c}{a}{n}\div{b}{y}{c}_{{j}}-{d}_{{j}}\).

Doing so we get

\(\displaystyle{\left({c}_{{j}}-{d}_{{j}}\right)}{v}_{{j}}=-{\left({c}_{{1}}{d}_{{1}}\right)}{v}_{{1}}-\ldots-{\left({c}_{{{j}-{1}}}-{d}_{{{j}-{1}}}\right)}{v}_{{{j}-{1}}}\)

\(\displaystyle-{\left({c}_{{{j}+{1}}}-{d}_{{{j}+{1}}}\right)}{v}_{{{j}+{1}}}-\ldots-{\left({c}_{{k}}-{d}_{{k}}\right)}{v}_{{k}}\)

\(\displaystyle{v}_{{j}}=\frac{{{d}_{{1}}-{c}_{{1}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{1}}+\ldots+\frac{{{d}_{{{j}-{1}}}-{c}_{{{j}-{1}}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{{j}-{1}}}\)

\(\displaystyle+\frac{{{d}_{{{j}+{1}}}-{c}_{{{j}+{1}}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{{j}+{1}}}+\ldots+\frac{{{d}_{{k}}-{c}_{{k}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{k}}\)