Prove that V=span{x1,x2,…,xn} if and only if {x1,x2,…,xn} is linearly independent.

Question

Let $\{ {\bf x}_1, {\bf x}_2, \dots, {\bf x}_n \}$ be any set of $n$ vectors in an $n$-dimensional vector space $V$. Prove that $V=\operatorname{span}\{ {\bf x}_1, {\bf x}_2, \dots, {\bf x}_n \}$ if and only if $\{ {\bf x}_1, {\bf x}_2, \dots, {\bf x}_n \}$ is linearly independent.

Answer 1

Let $S=\{x_1,...,x_n\} $ be a set of generators for the n-dimensional vector space $V $. If S were not linearly independent, one of its element is a linear combination of the others, say $x_1=\displaystyle \sum_{2\leq i \leq n} a_i x_i $ . So you can take $x_1$ off from the set. The new set $S^{\prime}=\{x_2, ...,x_n\} $ still is a set of generators for V because $V=Span (S)=Span (S^{\prime} )$, thus $dim (V)=dim(S^{\prime})\leq n-1$, or V is not n-dimensional.