$A \in M_{n}(\mathbb{R})$ has inverse if and only if it's rank is n. I am having trouble trying to prove that a matrix $A \in M_{n}(\mathbb{R})$ has inverse if and only if it's rank($A$) = n. What I was trying to do is to prove it using the fact that 
$$rank(A) = \dim(\text{column-space of} \ A)$$
but I don´t really know how to continue. 
Thank you for your help.
 A: Let $a_1,\dots,a_n$ be the columns of $A$. Recall that
$$A \left(\begin{aligned}x_1\\x_2\\ \cdots\\x_n\end{aligned}\right) = x_1a_1+x_2a_2+\cdots+x_na_n.$$
Since $A$ has full rank, $\mathbb{R}^n=\mathrm{Col}(A)$. So for every column $e_i$ of the identity matrix, there is a column vector $b_i$ so that
$$Ab_i = e_i.$$
Putting all $b_i$'s together we get
$$A(b_1\,b_2\,\cdots\,b_n) = (e_1\,e_2\,\cdots\,e_n) = I_n.$$
Thus $A$ is invertible.
A: We can take a more general approach I guess with a linear transformation $T$. First show that $T$ is one to one iff $\text{dim}(ker \:T)=0$. Indeed Let $x\in \text{ker}(T)=\{\bar{0}\}$. Then $Tx=0$. But then $x=0$ by definition so that $T$ is one to one. Conversely if $T$ is one to one then $Tx=0\implies x=0$. But $Tx=0$ means $x\in \text{ker}(T)$. SO $x\in \text{ker}(T)\implies x=0$. This means $\text{ker}(T)=\{\bar{0}\}$. Now we know that $T$ is invertible iff it is one to one. This is easy to prove. Now since a linear transformation $T$ can be equivalently represented by its matrix $A$ with respect to a basis so with $V=M_n(\mathbb{R})$, we have our result.
