# A question about linear independence

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.

• "we are done" ? With what ? And one single vector cannot be linearly independent, though if it is the zero vector it is lin. dependent. – DonAntonio Apr 21 '14 at 15:53
• It's a part of bigger proof that I didn't want to post in full. Would take up too much space. – idontgetit Apr 21 '14 at 15:58
• But then don't write that as people reading your post may think you forgot to add some info... – DonAntonio Apr 21 '14 at 15:58
• I edited my OP. Now it should be more clear. – idontgetit Apr 21 '14 at 16:56

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.