$AA^T$ can be non singular only if the columns in $A$ are linear independent and they span the column space of $A$. Because the columns are linear independent then $A$ can be reduced to echelon form and $A$ will have $m$ pivots only if $m < n$. Because we have $m$ pivots, $\text{rank}(A) = m$.
Is this a valid prove for If $A$ is $m\times n$ matrix and $AA^T$ is non singular show that $\text{rank}(A) = m$?
Thanks ^_^
You are right but you don't need to consider RREF, simply observe that since $AA^T$ is m-by-m matrix thus if it is not singular it is full rank with rank=m and thus $rank(A)=rank(A^TA)=m$.
This is simplified by viewing both as linear maps.
Note that $A:\mathbb R^n \to \mathbb R^m$ while $A^t:\mathbb R^m \to \mathbb R^n$.
In particular, $AA^t: \mathbb R^m \to \mathbb R^m$. If this is nonsingular, it is an isomorphism, so $A$ must be a surjection, and hence $\mathrm{Rank}(A) \leq m$, but $A^t$ has the same rank, and is an injection, so $\mathrm{Rank}(A) \geq m$.
Show that
$$\text{rank}(AB) \leq \min\{\text{rank}(A),\text{rank}(B)\}$$
Then, since $\text{rank}(A)=\text{rank}(A^T)$ we have
$$m=\text{rank}(AA^T) \leq \text{rank}(A)$$
On the other hand, if $A$ is $m\times n$ then
$$\text{rank}(A)\leq \min\{m,n\}\leq m$$
