# 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 $\mathbb{R}^n$?

• Yes${}{}{}{}{}{}$. – user63181 May 26 '14 at 23:00
• How do you define span? Is this a conclusion that might follow from a definition you're familiar with? – Muphrid May 26 '14 at 23:01
• @Muphrid: My concept of Span is pretty basic. To me, it's just the set of all vectors resulting from all linear combinations of the $n$ vectors. – Zol Tun Kul May 26 '14 at 23:02

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?

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 $\mathbb{R}^n$.

• That assumes some theorems about dimension, of course - namely, that dimension is well-defined... – Thomas Andrews May 26 '14 at 23:22
• True, although that follows easily from the theorem that linearly independent sets are no longer than spanning sets. If $B_1$ and $B_2$ are bases of $V$, then $B_1$ can be no longer than $B_2$ and the converse is true, so their lengths must be equal. – Michael M May 27 '14 at 1:06

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 $m=n$?