Linearly Dependent Sequence of Vector Space

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Theorem

Let $\struct {G, +}$ be a group whose identity is $\mathbf 0$.

Let $\struct {G, +, \circ}_K$ be a $K$-vector space.

Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct non-zero vectors of $G$.


Then $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent if and only if:

$\exists p \in \closedint 2 n: a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$


Proof

Necessary Condition

Suppose $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent.

We have by hypothesis that the set of all integers $r \in \closedint 1 n$ such that $\sequence {a_k}_{1 \mathop \le k \mathop \le r}$ is linearly dependent is not empty.

Let $p$ be its smallest element.

Then from Singleton is Linearly Independent‎, $p \ge 2$, as $a_1 \ne \mathbf 0$ and hence $\set {a_1}$ is linearly independent.

Also, there exist scalars $\lambda_1, \ldots, \lambda_p$, not all of which are zero, such that $\ds \sum_{k \mathop = 1}^p \lambda_k \circ a_k = \mathbf 0$.


Aiming for a contradiction, suppose $\lambda_p = 0$.

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

Then $\sequence {a_k}_{1 \mathop \le k \mathop \le p-1}$ is linearly dependent.

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

So, because:

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

we must have:

$\ds a_p = \sum_{k \mathop = 1}^{p - 1} \paren {-{\lambda_p}^{-1} \lambda_k} \circ a_k$

and thus $a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$.

$\Box$


Sufficient Condition

Now suppose that $a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$.

Then:

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

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

$\forall k \in \closedint 1 n: \lambda_k = \begin{cases}

\mu_k & : k < p \\ -1 & : k = p \\ 0 & : k > p \\ \end{cases}$

Then:

$\ds \sum_{k \mathop = 1}^n \lambda_k \circ a_k = \mathbf 0$

Hence the result.

$\blacksquare$


Sources

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