If $n$ vectors are linearly independent, is their span $\mathbb{R}^n$? Have $n$ vectors in $\mathbb{R}^n$.
If the $n$ vectors are linearly independent, can we conclude that their span is $\mathbb{R}^n$?
 A: You can use the theorem that every linearly independent set in a vector space $V$ can be extended to a basis of $V$. Since the dimension of $\mathbb{R}^n$ is simply $n$, the extension of the $n$ vectors to a basis is trivial (i.e. the vectors are unchanged); these vectors therefore already span $\mathbb{R}^n$.
A: A useful result in linear algebra is:
Proposition. Let $W$ be a subspace of a finite-dimensional vector space $V$. If $\dim W=\dim V$, then $W=V$.
If we are given linearly independent $v_1,\dotsc,v_n\in\Bbb R^n$, then $\DeclareMathOperator{Span}{Span}\Span\{v_1,\dotsc,v_n\}$ is an $n$-dimensional subspace of $\Bbb R^n$. Since $\dim \Bbb R^n=n$, the proposition implies that $\Span\{v_1,\dotsc,v_n\}=\Bbb R^n$.
Can you prove this proposition?
A: Hint: if the span of $m$ linearly independent vectors $\{v_1,\dots,v_m\}$ is a proper subspace $U$ of $\mathbb{R}^n$, then, for every vector $v\in\mathbb{R}$, $v\notin U$, the set $\{v_1,\dots,v_m,v\}$ is linearly independent.
Can you see there's a contradiction if $m=n$?
