 # Let v_1,...,v_n be linealy independent vectors in RR^m and let w=a_1v_1+...+a_nv_n with real numbers a_1,...a_n, be linear combination of these vectors. Prove if a_1 != 0, then the vector w,v_2,...,v_n are linearly independent So far, what I have thought is forming a formula like this: x_1w+x_2v_2+...+x_nv_n=0, because v_1,...,v_n is linearly independent vectors, so x_2=...=x_n=0, then now we need to prove x_1=0 so these vector will be linearly independent. But from here, I don't know how to process further more in this problem, can anyone help me out with a way to solve this or a better alternative solution is appreciate. pleitatsj1 2022-08-07 Answered
Let ${v}_{1},...,{v}_{n}$ be linealy independent vectors in ${R}^{m}$ and let $w={a}_{1}{v}_{1}+...+{a}_{n}{v}_{n}$ with real numbers ${a}_{1},...{a}_{n}$, be linear combination of these vectors.
Prove if ${a}_{1}\ne 0$, then the vector $w,{v}_{2},...,{v}_{n}$ are linearly independent
So far, what I have thought is forming a formula like this: ${x}_{1}w+{x}_{2}{v}_{2}+...+{x}_{n}{v}_{n}=0$, because ${v}_{1},...,{v}_{n}$ is linearly independent vectors, so ${x}_{2}=...={x}_{n}=0$, then now we need to prove ${x}_{1}=0$ so these vector will be linearly independent. But from here, I don't know how to process further more in this problem, can anyone help me out with a way to solve this or a better alternative solution is appreciate.
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If w has a contribution from ${v}_{1}$, then any linear combination of $w,{v}_{2},\dots ,{v}_{n}$ can be written as a linear combination of ${v}_{1},\dots ,{v}_{n}$. That's suppose ${x}_{1}w+{x}_{2}{v}_{2}+\cdots +{x}_{n}{v}_{n}=0$. Then ${x}_{1}{a}_{1}{v}_{1}+\left({x}_{1}{a}_{2}+{x}_{2}\right){v}_{2}+\cdots +\left({x}_{1}{a}_{n}+{x}_{n}\right){v}_{n}=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{1}=\left({x}_{1}{a}_{2}+{x}_{2}\right)=\cdots =\left({x}_{1}{a}_{n}+{x}_{n}\right)=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{1}={x}_{2}=\cdots ={x}_{n}=0$