Let $V$ be a vector space over a scalar field $K$.
Let $S=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\}$ be the set of vectors in $V$, where $n \geq 2$.

Then prove that the set $S$ is linearly dependent if and only if at least one of the vectors in $S$ can be written as a linear combination of remaining vectors in $S$.

$(\implies)$: If $S$ is linearly dependent, then a vector is a linear combination

Suppose that the set $S=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\}$ is linearly dependent. Then there exists scalars $c_1, c_2, \dots, c_n$ in $K$ such that
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots+c_n\mathbf{v}_n=\mathbf{0}\]
and not all of $c_1, \dots, c_n$ are zero.
(This is just the definition of linearly dependency.)

Since at least one of $c_1, \dots, c_n$ is nonzero, without loss of generality we may assume that $c_1\neq 0$.
Then we have
\begin{align*}
c_1\mathbf{v}_1=-c_2\mathbf{v}_2-c_3\mathbf{v}_3-\cdots -c_n\mathbf{v}_n
\end{align*}
and we divide this by $c_1\neq 0$ and obtain
\begin{align*}
\mathbf{v}_1=-\frac{c_2}{c_1}\mathbf{v}_2-\frac{c_3}{c_1}\mathbf{v}_3-\cdots -\frac{c_n}{c_1}\mathbf{v}_n.
\end{align*}
Therefore, we have written the vector $\mathbf{v}_1$ as a linear combination of $\mathbf{v}_2, \mathbf{v}_3, \dots, \mathbf{v}_n$.

$(\impliedby)$: If a vector is a linear combination, then $S$ is linearly dependent

Suppose now that some vector of $S$ is a linear combination of the remaining vectors.
Without loss of generality, we may assume that $\mathbf{v}_1$ is a linear combination of the remaning vectors. So we have
\[\mathbf{v}_1=c_2\mathbf{v}_2+c_3\mathbf{v}_3+\cdots+c_n\mathbf{v}_n,\]
for some scalars $c_2, c_3, \dots, c_n$ in $K$.

This implies that we have
\[\mathbf{v}_1-c_2\mathbf{v}_2-c_3\mathbf{v}_3-\cdots-c_n\mathbf{v}_n=\mathbf{0},\]
and this is a nontrivial linear combination of vectors in $S$ because the coefficient of $\mathbf{v}_1$ is $1 \neq 0$.
Thus, the set $S$ is linearly dependent.

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