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Let $S$ be a linearly independent set. Let $S'$ be a proper subset of $S$.

$Span$$(S') \neq Span(S)$

Let $v$ be an element of $S$ which is not contained in $S'$; such an element must exist because $S'$ is a proper subset of S. Since $v \in S$, we have $v \in$ $span$$(S)$. Now suppose that $v$ were also in $span$$(S')$. This would mean that there existed vectors $v_1, \ldots, v_n \in S'$ (which in particular were distinct from $v$) such that

$v = a_1v_1 + a_2v_2 + . . . + a_nv_n$,

or in other words

$(−1)v + a_1v_1 + a_2v_2 + . . . + a_nv_n = 0$.

But this is a non-trivial linear combination of vectors in S which sum to zero (it’s nontrivial because of the $−1$ coefficient of $v$). This contradicts the assumption that $S$ is linearly independent. Thus $v$ cannot possibly be in $span$$(S')$. But this means that $span$$(S')$ and $span$$(S)$ are different, and we are done.

So, we have $v = a_1v_1 + a_2v_2 + . . . + a_nv_n$. We can subtract $v$ from both sides of the equation and then argue that elements of $span$$(S')$ are linearly dependent.

Suppose a vector $v = a_1v_1 + a_2v_2 + . . . + a_nv_n$ is linearly independent. Can't we just subtract $v$ from both sides of the equation and say that the linear combination is linearly dependent every time?

Thanks.

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  • $\begingroup$ "we are done" ? With what ? And one single vector cannot be linearly independent, though if it is the zero vector it is lin. dependent. $\endgroup$ – DonAntonio Apr 21 '14 at 15:53
  • $\begingroup$ It's a part of bigger proof that I didn't want to post in full. Would take up too much space. $\endgroup$ – idontgetit Apr 21 '14 at 15:58
  • $\begingroup$ But then don't write that as people reading your post may think you forgot to add some info... $\endgroup$ – DonAntonio Apr 21 '14 at 15:58
  • $\begingroup$ I edited my OP. Now it should be more clear. $\endgroup$ – idontgetit Apr 21 '14 at 16:56
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Consider $a_{0}v - \sum_{i=1}^{n} a_{i}v_{i} = 0$. The vector $v$ is linearly independent with $v_{1}, ..., v_{n}$ if and only if $a_{0}, ..., a_{n}$ are all $0$ is the only linear combination satisfying the one I gave above.

So what we're really saying is this. If we have a set of linearly independent vectors, then no vector $v_{1}$ in the set $S$ can be formed as a linear combination of the vectors $v_{2}, ..., v_{n}$. If it could, then we would have two linear equations $-v_{1} + \sum_{i=2}^{n} a_{i}v_{i} = 0$ and $\sum_{i=1}^{n} 0v_{i} = 0$. Clearly, having two such equations violates the definition linear independence. Hence, the contradiction.

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