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.

Question
Vectors and spaces
asked 2021-01-17
Let \(\displaystyle{v}_{{1}},{v}_{{2}},\ldots.,{v}_{{k}}\) be vectors of Rn such that
\(\displaystyle{v}={c}_{{1}}{v}_{{1}}+{c}_{{2}}{v}_{{2}}+\ldots+{c}_{{k}}{v}_{{k}}={d}_{{1}}{v}_{{1}}+{d}_{{2}}{v}_{{2}}+\ldots+{d}_{{k}}{v}_{{k}}\).
for some scalars \(\displaystyle{c}_{{1}},{c}_{{2}},\ldots.,{c}_{{k}},{d}_{{1}},{d}_{{2}},\ldots.,{d}_{{k}}\).Prove that if \(\displaystyle{c}{i}\ne{d}{i}{f}{\quad\text{or}\quad}{s}{o}{m}{e}{i}={1},{2},\ldots.,{k}\),
then \(\displaystyle{v}_{{1}},{v}_{{2}},\ldots.,{v}_{{k}}\) are linearly dependent.

Answers (1)

2021-01-18
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
\(\displaystyle{\left({c}_{{j}}-{d}_{{j}}\right)}{v}_{{j}}=-{\left({c}_{{1}}{d}_{{1}}\right)}{v}_{{1}}-\ldots-{\left({c}_{{{j}-{1}}}-{d}_{{{j}-{1}}}\right)}{v}_{{{j}-{1}}}\)
\(\displaystyle-{\left({c}_{{{j}+{1}}}-{d}_{{{j}+{1}}}\right)}{v}_{{{j}+{1}}}-\ldots-{\left({c}_{{k}}-{d}_{{k}}\right)}{v}_{{k}}\)
\(\displaystyle{v}_{{j}}=\frac{{{d}_{{1}}-{c}_{{1}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{1}}+\ldots+\frac{{{d}_{{{j}-{1}}}-{c}_{{{j}-{1}}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{{j}-{1}}}\)
\(\displaystyle+\frac{{{d}_{{{j}+{1}}}-{c}_{{{j}+{1}}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{{j}+{1}}}+\ldots+\frac{{{d}_{{k}}-{c}_{{k}}}}{{{c}_{{j}}-{d}_{{j}}}}{v}_{{k}}\)
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