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.

  • $\begingroup$ 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

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.

  • $\begingroup$ Which is to say, "Prove the contrapositive" $\endgroup$ – pjs36 Jan 31 '16 at 19:34
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.