# 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.