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The empty set is linearly independent. Also, $S_1$ is linearly independent if $S_1 = S_2.$ Suppose $v_1, \ldots, v_n \in S_2.$ Then $a_i = 0$ for all $i$ in $a_1v_1 + \ldots + a_nv_n = 0.$ Now consider $a_1v_1 + \ldots a_{n -1}v_{n-1}= -a_nv_n.$ Since $a_n = 0, -a_nv_n = 0.$ Thus $S_1$ is linearly independent if $S_1 \subseteq S_2.$ Does this make sense?

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  • $\begingroup$ The later part doesn't make much sense to me. I can't understand your proof when $S_1\ne \emptyset,S_2$. $\endgroup$ – BigbearZzz May 26 '16 at 12:47
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No that's incorrect. The point is that if you take any collection of elements $v_1,\dots,v_n\in S_1$, having $a_1v_1+\dots+a_nv_n=0$ implies $a_i=0$ for each $i$ because the $v_i$ are elements of $S_2$ by definition

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