A vector in a linearly dependent indexed set is a linear combination of the PRECEDING vectors?

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 1

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.

Answer 2

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