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A problem in a textbook goes as follows: Show that if S is a linearly independent set of vectors, then so is every nonempty subset of S.

Is it acceptable to say that since every vector in S is linearly independent, then any vectors that make up a subset of S will also be linearly independent? Or is that logically incorrect?

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    $\begingroup$ Be careful with what you mean by 'every vector is linearly independent'. It's always the case that $\{\mathbf v\}$ is a linearly independent set for all $0 \neq \mathbf v \in V$... I would rather argue that if $S' \subseteq S$ was a linearly dependent set, then viewing the vectors in $S'$ as elements of $S$, we'd have that $S$ was linearly dependent - contradiction. $\endgroup$ – ah11950 Apr 20 '14 at 13:10
  • $\begingroup$ And the empty set is linearly independent too. $\endgroup$ – Git Gud Apr 20 '14 at 13:12
  • $\begingroup$ Oh, I didn't think of that. Thank you. $\endgroup$ – Oscar Flores Apr 20 '14 at 13:13
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    $\begingroup$ "every vector in $S$ is linearly independent" makes no sense. $\endgroup$ – Valentin Waeselynck Apr 20 '14 at 13:25
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    $\begingroup$ It is important that you realise that linear independence is a property only of sets (or families) of vectors, not of individual vectors. It is for this reason that you argument is invalid. The danger is in reasoning in the opposite direction: the fact that every vector belongs to a linearly independent subset does not imply that the whole set of vectors is linearly independent $\endgroup$ – Marc van Leeuwen Apr 20 '14 at 13:37
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Linearly independent according to the formal definition means, if $S=\{s_1,\ldots,s_n\}$ then $S$ is linearly independent when: supposing \begin{equation}\mathbf{0}=a_1s_1+\cdots+a_ns_n, \quad (1)\end{equation} we must have \begin{equation}a_1=\cdots=a_n=0. \quad (2)\end{equation}

Now suppose to the contrary, that there is some subset $\{s_1,\ldots,s_k\}$ which is linearly dependent, that is, we can find $b_1,\ldots,b_k$, not all zero, so that $\mathbf{0}=b_1s_1+\cdots+b_ks_k$. But then $\mathbf{0}=b_1s_1+\cdots+b_ks_k+a_ls_l+\cdots+a_ns_n$ contradicting $(2)$.

So we must have $\{s_1,\ldots,s_k\}$ is linearly independent.

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You don't need to use logic here, you can also prove it by contradiction.

Let $S=\{v_i\}_{1=1}^n$, and let $0<m<n$. Then every non empty subset of $S$ is $S_m=\{v_i\}_{1=1}^m$, up to relabeling of the elements of $S$. Then, if $S_m$ was linearly independent, there would exist $\{a_1, a_2, ..., a_m\}$, not all zero scalars such that $\Sigma_{i=1}^m a_iv_i=\vec 0$. Extending the set of $a_i$ with $\{a_{m+1},... ,a_n\}$ all zero yields that $\Sigma_{i=1}^n a_iv_i=\vec 0$, which contradicts the linear independence of $S$.

So $S_m$ are linearly independent for any $m$.

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  • $\begingroup$ By using en.wikipedia.org/wiki/Proof_by_contradiction you do in fact make use of logic (just as a btw) $\endgroup$ – Christiaan Hattingh Apr 20 '14 at 13:41
  • $\begingroup$ @christiaan-hattingh Yes, every proof is logic. I think that in the context of the question, it is clear I was talking about the heuristic kind of proof that was suggested, and not the whole discipline of logical thinking. $\endgroup$ – Robin Apr 20 '14 at 13:44

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