Theorem – Some element of linearly dependent set of vectors (in a finite dimensional vector space) can be written as linear combination of other vectors in that set.

Theorem

Let $V$ be a finite dimensional vector space and let $B\subset{V}$ be a linearly dependent subset of $V$ . Then, the following are equivalent,

Some vector in $B$ can be written as a linear combination of other vectors in $B$ ,
The set $B$ is linearly dependent.

Proof

We will prove the implication in one direction only, as the reverse is simply the same working but in the opposite direction, as all steps will be equivalences.

We will first prove that $(1) \implies (2)$ . We assume that a set of vectors $B$ is linearly dependent, so that we may write

$a_1 \mathbf{b_1} + a_2 \mathbf{b_2} + ... + a_n \mathbf{b_n} = \mathbf{0}$

Where $b_i$ are vectors in $B$ . Since, by definition of linear dependence, we know that this implies that there exists a $b_i \ne 0$ in $B$ . Let us consider this particular $b_i$ . If we subtract this vector and coefficient product from both sides, and then divide by $a_i$ , we obtain the following,

$\cfrac{a_1}{a_i} \mathbf{b_1} + \cfrac{a_2}{a_i} \mathbf{b_2} + ... + \cfrac{a_n}{a_i} \mathbf{b_n} - \frac{a_i}{a_i} \mathbf{b_n}= - \mathbf{b_n}$

Now, since we have $- \mathbf{b_n} + \mathbf{b_n} = 0$ on the RHS, we know that the RHS does not contain the $\mathbf{b_n}$ . We have thus shown that $(1) \implies (2)$ . The reverse implication is simply the same logic in reverse, which completes the proof. $\square$