If the columns of an $n \times n$ matrix are linearly independent, then the columns span $\mathbb{R}^{n}$ My textbook says that its true but I can't find a proof of this on the internet.

"If $A$ is an $n \times n$ matrix with linearly independent columns, then the columns of $A$ span $\mathbb{R}^{n}$."

 A: The dimension of a finite dimensional vector space is the maximal number of linearly independent vectors, and it is also the minimal number of a system of generators. A maximal system of linearly independent vectors is a basis. A minimal system of generators is a basis.
A: Notice that the columns of the matrix (lets call it $A$) are $n$ linearly independent vectors. On the other hand, any basis of $\mathbb{R}^n$ requires $n$ linearly independent vectors. You already have them: they are the vectors of $A$. 
Formally, let $x \in \mathbb{R}^n$ then
\begin{equation}
x = \sum\limits_{i=1}^{n} \alpha_i v_i
\end{equation}
where $\alpha_i$ denote scalars and $v_i$ denote linearly independent vectors. In particular there is a change of coordinates that transforms $\alpha_i v_i \to \beta_i A_i$ where $A_i$ are the columns of your matrix. And thus you can write any $x$ as: 
\begin{equation}
x = \sum\limits_{i=1}^{n} \beta_i A_i
\end{equation}
A: The idea is that, if this matrix is called $A$, and the columns are linearly independent, then $A$ is invertible. This means given any system $$Ax=b$$ we can solve for $x$ simply by multiplying by $A^{-1}$, so $$x=A^{-1}b$$
Since $b$ could have been any vector in $\mathbb{R}^n$, the columns span $\mathbb{R}^n$, because $Ax$ is just a linear combination of the columns of $A$.
