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 $$\sum_{k=1}^p \lambda_k \circ a_k = e$$.

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

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

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

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

Hence the result.