We have three vectors T={v_1,v_2,v_3}, and the set of their sums S={v_1+v_2,v_1+v_3,v_2+v_3} for v_i in V, for V any vector space. I have to prove that for rational coefficients a,b,c in bbbQ, the following statement is true: T linearly independent <=> S linearly independent => was quite easy and I had no problem calculating it. However, <= is where I faced the problem and I would appreciate any help and besides that, does <= is true for real coefficients.

I came across a problem while proving a very simple statement.
We have three vectors $T=\left\{{v}_{1},{v}_{2},{v}_{3}\right\}$, and the set of their sums $S=\left\{{v}_{1}+{v}_{2},{v}_{1}+{v}_{3},{v}_{2}+{v}_{3}\right\}$ for ${v}_{i}\in V$, for V any vector space. I have to prove that for rational coefficients $a,b,c\in \mathbb{Q}$, the following statement is true:
T linearly independent $⇔$ S linearly independent
$⇒$ was quite easy and I had no problem calculating it.
However, $⇐$ is where I faced the problem and I would appreciate any help and besides that, does $⇐$ is true for real coefficients.
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

Mario Monroe
Note that the proof of $⇐$ is the same as that of $⇒$ only that
$T=\left\{\frac{1}{2}\left({w}_{1}+{w}_{2}-{w}_{3}\right),\frac{1}{2}\left({w}_{1}+{w}_{3}-{w}_{2}\right),\frac{1}{2}\left({w}_{2}+{w}_{3}-{w}_{1}\right)\right\}$
where $S=\left\{{w}_{1},{w}_{2},{w}_{3}\right\}$