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Show that any set of vectors containing a linearly dependent subset is again linearly dependent.

I think you're supposed to show this by contradiction, but not sure how.

I tried:

Let V be a linearly independent set of vectors {$v_1,...v_n$} such that $a_1v_1+...+a_nv_n$=0 and $a_1=...=a_n=0$. Then, V is a subset of {$v_1,...,v_{n+1}$} and $a_1v_1+...+a_{n+1}v_{n+1}=0$ where $a_1=...=a_{n+1}=0$. Therefore, the set {$v_1,...,v_{n+1}$} is again linearly independent.

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  • $\begingroup$ If the coefficients are all zero, you will get zero regardless of whether or not your set is independent. Hence the calculation given does not make any useful progress. $\endgroup$ – vadim123 Jan 31 '16 at 19:31
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Suppose $\{x_1,\ldots, x_m\}$ is dependent. Then there are coefficients $a_1,\ldots, a_m$, not all zero, such that $$a_1x_1+\cdots+a_mx_m=0$$ Now, let $n\ge m$, and consider the set $\{x_1,\ldots, x_m,x_{m+1},\ldots, x_n\}$. Now, take $b_1=a_1, b_2=a_2,\ldots, b_m=a_m$, and $b_{m+1}=b_{m+2}=\cdots=b_n=0$. Since the $a_i$'s were not all zero, the $b_i$'s are not all zero either. We have $$b_1x_1+\cdots+b_nx_n=a_1x_1+\cdots a_mx_m=0$$ Hence $\{x_1,\ldots,x_n\}$ is dependent as well.

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  • $\begingroup$ Which is to say, "Prove the contrapositive" $\endgroup$ – pjs36 Jan 31 '16 at 19:34
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    $\begingroup$ I proved the statement given in the title, directly, using the definition of a set of vectors being dependent. $\endgroup$ – vadim123 Jan 31 '16 at 19:35
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    $\begingroup$ @pjs36: As far as I can see, this is a direct proof. $\endgroup$ – Henning Makholm Jan 31 '16 at 19:35
  • $\begingroup$ Yep, I completely misread everything - wow! $\endgroup$ – pjs36 Jan 31 '16 at 19:35

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