# Prove minimal spanning set implies linear independence

Let V = span{$v_1$,...,$v_k$}. Prove that {$v_1$,...,$v_k$} is a minimal spanning set of V if and only if $v_1$,...,$v_k$ are linearly independent.

I am trying to do this by proving the contrapositive, i.e. $v_1$,...,$v_k$ are linearly dependent if and only if {$v_1$,...,$v_k$} is not a minimal spanning set of V.

Linear dependence implies there exists $c_1$,...,$c_k$, not all zero, such that $c_1$$v_1+...+c_k$$v_k$ = $0$

How can I use this to prove that {$v_1$,...,$v_k$} is not a minimal spanning set of V?

If $B=\{v_1,\dots,v_k\}$ is not a minimal spanning set, then we can remove WLOG $v_k$ and $\{v_1,\dots,v_{k-1}\}$ is still a spanning set. We may assume $v_k\neq 0$, otherwise the original set was clearly not linearly independent. So $v_k=c_1v_1+\cdots+c_{k-1}v_{k-1}$ and the original set was not linearly independent.
Conversely, if $v_i$ are linearly dependent, then WLOG assume $v_k$ is in the span of $\{v_1,\dots,v_{k-1}\}$. If $B$ was not a spanning set we are done. If $B$ was a spanning set, then it cannot be minimal because $v_k\in\text{Span}\{v_1,\dots,v_{k-1}\} \implies \{v_1,\dots,v_{k-1}\}\;\;\text{is a spanning set}$.