Question: A theorem in my LA book is characterized as follows:

An indexed set $S=\{\vec{v\_1},...,\vec{v\_p}\}$ of two or more vectors is linearly dependent if and only if at least one vector is a linear combination of the others. **In fact, if S is linearly dependent and $\vec{v\_1} \neq \vec{0}$, then some $\vec{v\_j}$ (with $j > 1$) is a linear combination of the preceding vectors $\vec{v\_1},...,\vec{v\_{j-1}}$.**

I don't get the boldface part at all. It seems oddly specific. Why does it have to be a linear combination of the **preceding** vectors? What if the vectors that $\vec{v\_j}$ is a linear combination of are scattered around in the set, both preceding and succeeding $\vec{v\_j}$?

Answer: 
There is this proposition:

> 
> If a linear independent set of vectors $S$ is given, a vector $\vec v$ not in $S$ is given, such that $S\cup\{\vec v\}$ is linearly dependent, then $\vec v=\sum\_{i=1}^nc\_i\vec u\_i$ for a finite number of vectors $u\_i\in S$.
> 
> 
> 

You may prove this yourself. If you cannot, comment and I will include a proof.

To prove your boldface part, look at the sequence of sets:

$\{\vec{v\_1}\},\{\vec{v\_1},\vec{v\_2}\},\{\vec{v\_1},\vec{v\_2},\vec{v\_3}\},\{\vec{v\_1},\vec{v\_2},\vec{v\_3},\vec{v\_4}\},\dots ,\{\vec{v\_1},\vec{v\_2},\dots,\vec{v\_p}\}$

The first one is linearly independent. The next one may or may not be. If it is linearly dependent, you know that $\vec{v\_2}$ is a linear combination of $\vec{v\_1}$; otherwise, look at the next set to see if it is linearly dependent after adding one more vector. If $\{\vec{v\_1},\vec{v\_2}\}$ is linearly independent while $\{\vec{v\_1},\vec{v\_2},\vec{v\_3}\}$ is not, you know that $\vec{v\_3}$ is a linear combination of $\vec{v\_1},\vec{v\_2}$, by applying the proposition in quote. If $\{\vec{v\_1},\vec{v\_2},\vec{v\_3}\}$ is linearly independent, look at the next set $\{\vec{v\_1},\vec{v\_2},\vec{v\_3},\vec{v\_4}\}$.

Repeat until you find the vector $\vec{v\_j}$ which upon adding to the set $\{\vec{v\_1},\dots,\vec{v\_{j-1}}\}$ turns the linearly independent set into a linearly dependent set. You can always find one such $\vec{v\_j}$, because your set $\{\vec{v\_1},\vec{v\_2},\dots,\vec{v\_p}\}$ is assumed to be linearly dependent. Then you know $\vec {v\_j}$ is a linear combination of $\vec{v\_1},\dots,\vec{v\_{j-1}}$, by applying the proposition in quote.
You are wanting to know why the statement specifies $$ v\_j = \sum\_{k=1}^{j-1} c\_k v\_k $$

because actually it is possible that $$v\_j = c\_{j+1}v\_{j+1} + \sum\_{k=1}^{j-2} c\_k v\_k$$ 

The answer is because without loss of generality, we can solve for the biggest indexed vector, $v\_{j+1}$: $$ v\_{j+1} = \frac{1}{c\_{j+1}}v\_j - \sum\_{k=1}^{j-2} \frac{1}{c\_{j+1}}c\_k v\_k $$

And now we made index j+1 satisfy the theorem conditions