Linearly Dependent Sequence of Vector Space

Theorem
Let $\left({G, +}\right)$ be a group whose identity is $e$.

Let $\left({G, +, \circ}\right)_K$ be a $K$-vector space.

Let $\left \langle {a_n} \right \rangle$ be a sequence of distinct non-zero vectors of $G$.

Then $\left \langle {a_n} \right \rangle$ is linearly dependent iff $\exists p \in \left[{2 \,. \, . \, n}\right]: a_p$ is a linear combination of $\left \langle {a_{p-1}} \right \rangle$.

Necessary Condition
Suppose $\left \langle {a_n} \right \rangle$ is linearly dependent.

By hypothesis, the set of all integers $r \in \left[{1 \,. \, . \, n}\right]$ such that $\left \langle {a_r} \right \rangle$ is linearly dependent is not empty.

Let $p$ be its smallest element.

Then from Singleton is Linearly Independent‎, $p \ge 2$, as $a_1 \ne e$ and hence $\left\{{a_1}\right\}$is linearly independent.

Also, there exist scalars $\lambda_1, \ldots, \lambda_p$, not all of which are zero, such that $\displaystyle \sum_{k=1}^p \lambda_k \circ a_k = e$.

Suppose $\lambda_p = 0$.

Then not all of $\lambda_1, \ldots, \lambda_{p-1}$ can be zero.

Then $\left \langle {a_{p-1}} \right \rangle$ is linearly dependent.

That contradicts the definition of $p$, so $\lambda_p \ne 0$.

So, because:
 * $\displaystyle \lambda_p \circ a_p = - \sum_{k=1}^{p-1} \lambda_k \circ a_k$

we must have:
 * $\displaystyle a_p = \sum_{k=1}^{p-1} \left({- \lambda_p^{-1} \lambda_k}\right) \circ a_k$

and thus $a_p$ is a linear combination of $\left \langle {a_{p-1}} \right \rangle$.

Sufficient Condition
Now suppose that $a_p$ is a linear combination of $\left \langle {a_{p-1}} \right \rangle$.

Then $\displaystyle a_p = \sum_{k=1}^{p-1} \mu_k \circ a_k$.

So we can assign values to $\lambda_k$ as follows:



\forall k \in \left[{1 \,. \, . \, n}\right]: \lambda_k = \begin{cases} \mu_k & : k < p \\ -1 & : k = p \\ 0 & : k > p \\ \end{cases} $

Then $\displaystyle \sum_{k=1}^n \lambda_k \circ a_k = e$.

Hence the result.