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Suppose $a,b$ and $c$ are linearly independent vectors in a vector space $V$. How can I prove that $a+b$ or $b+c$ are also linearly independent?

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    $\begingroup$ Linearly independent w.r.t which vectors? Obviously not $a,b$ and $b,c$. $\endgroup$
    – Jacob
    Sep 30, 2012 at 16:31
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    $\begingroup$ I think what is intended is that you show that the set $\{a+b,b+c\}$ is a linearly independent set. $\endgroup$ Sep 30, 2012 at 16:32
  • $\begingroup$ Well that's the problem. the question is just giving a,b and c. no actual values. $\endgroup$
    – Nima
    Sep 30, 2012 at 16:33
  • $\begingroup$ You don't need actual values-the problem is that $a+b$ and $b+c$ are just single vectors, and linear independence of one vector on its own is trivial. So as @AndréNicolas said a more reasonable question is to prove that $a+b$ and $b+c$ are linearly independent, as was proved below. $\endgroup$ Sep 30, 2012 at 16:36
  • $\begingroup$ Linear independence of one vector on its own is trivial. To be more precise, linear independence of one vector is equivalent to this vector being non-zero, i.e., $v$ is linearly independent if and only if $v\ne0$. $\endgroup$ Sep 30, 2012 at 16:39

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If I understand your question correctly, you want to show that if $a$, $b$, $c$ are linearly independent, then $a+b$ and $b+c$ are linearly independent.

Just look at the definitions.

You know that $x_1a+x_2b+x_3c=0$ implies $x_1=x_2=x_3=0$. (This is the definition of linear independence for three vectors.)

You ask whether $y_1(a+b)+y_2(b+c)=0$ implies $y_1=y_2=0$.

Just simplify this to get: $y_1 a + (y_1+y_2)b +y_2c=0$. This implies that $y_1=y_1+y_2=y_2=0$. The condition $y_1+y_2=0$ is redundant there, but we have shown that $y_1=y_2=0$.

This means that the vectors $a+b$, $b+c$ are linearly independent.

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  • $\begingroup$ Can we generalize this? If a finite collection of vectors $\{u,v,w\}$ is LI, then any collection of linear combinations generated by $\{u,v,w\}$ is also LI. $\endgroup$
    – johnny09
    Sep 20, 2019 at 1:54
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    $\begingroup$ @johnny09 You can generalize it, but it is not true that any linear combinations would be independent. For example, $u+v+w$, $v+w$, $w$ are linearly independent, since $(1,1,1)$, $(0,1,1)$, $(0,0,1)$ are linearly independent. More generally $a_{11}u+a_{12}v+a_{13}w$, $a_{21}u+a_{22}v+a_{23}w$, $a_{31}u+a_{32}v+a_{33}w$ are linearly independent if $(a_{11},a_{12},a_{13})$, $(a_{21},a_{22},a_{23})$, $(a_{31},a_{32},a_{33})$ are linearly independent. Basically, we are just looking at coordinates of the vectors w.r.t. the basis $u$, $v$, $w$ of $\operatorname{span}(u,v,w)$. $\endgroup$ Sep 20, 2019 at 12:19
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    $\begingroup$ It works in a similar way if we have more than three vectors. (Also the number of linear combinations can be arbitrary.) $\endgroup$ Sep 20, 2019 at 12:21

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