Redundant Vector Implies Linear Dependence

Theorem

If a set $S$ contains a Redundant Vector, then that set $S$ is Linearly Dependent

Contrapositive

If $S$ is Linear Independent then there are no redundant vectors

Proof

1. Suppose $span(\{w,\overrightarrow{v_1},\dots ,\overrightarrow{v_k}\})=span(\{\overrightarrow{v_1},\dots ,\overrightarrow{v_k}\})$
2. Then, since the spans are equal, $w$ must exist in the second span , perhaps as a Linear Combination
   1. For some $a_i\in F$, $1\overrightarrow{w}=a_1\overrightarrow{v_1}+\dots a_k{v}_k$
   2. $1\overrightarrow{w}-a_1\overrightarrow{v_1}-\dots a_k{v}_k=0$
   3. Note that above is a linear combination of the first set with a non-zero coefficient,
   4. Thus, the first set is Linearly Dependent