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If we have linearly independent vectors $v_1, v_2, ..., v_n$ and create a new collection of vectors $v_1', v_2',...,v_n'$ such that each $v_i'$ is a linear combination of $v_1, v_2, ..., v_n$.

Are vectors $v_1', v_2',...,v_n'$ linearly independent as well? I believe the answer is no, at least in general. How can one prove it easily?

Because $v_1, v_2, ..., v_n$ are independent, the equation $a_1v_1 + a_2 v_2 + ... + a_nv_n = 0$ is satisfied only if all scalars $a_i$ are zero.

If we wrote a similar equation for $v_1'$, namely $b_1v_1' + b_2 v_2' + ... + b_nv_n' = 0 $, after expressing each $v_i'$ as a linear combination of $v$ vectors, the new coefficients would have to be zero. Therefore the original coefficients $b_i$ don't have to be all zeros.

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  • $\begingroup$ A simple counter-example is $v'_1=\dots =v'_n$ $\endgroup$
    – Augustin
    Feb 12, 2016 at 15:13

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Take for $v_1,\ldots,v_n$ any basis of your vector space (for instance the standard basis of $\Bbb R^n$). Now any vector is a linear combination of those vectors, so if your result were true, any $n$-tuple of vectors would be linearly independent. Certainly you can think of some $n$-tuple of dependent vectors (unless $n=0$), so the staement cannot be true.

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Consider $v_i'=v_1+\cdots +v_n, i=1,\dots, n.$ Clearly, $v_i'$ are linear combination of $v_i's$ and not linearly independent.

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