TITLE: $A^TA$ is non-singular if and only if $A$ has full row rank
QUESTION [3 upvotes]: Suppose that $A$ is an $m \times n$ matrix. I want to show that $A^TA$ is non-singular if and only if $A$ has full rank.
Now `full rank' can mean two things. I know that saying that $A$ has full rank means that $\text{rank}(A)=\min\{m,n\}$. If $m \geq n$ then $\text{rank}(A)=n$, i.e., $A$ has full column rank, so $A$ is injective. I have proven this case with no issue.
What I am struggling to prove is that $A^TA$ is non-singular if and only if $A$ has full row rank, i.e., in the case where $m <n$, $\text{rank}(A)=m$, meaning that $A$ is surjective. 
Thank you, in advance.

REPLY [2 votes]: Note that:
if $x \neq0 \in Null(A)$
$$A^TAx=0$$
thus $A^TA$ is singular
if $x \neq0$ and $A^TA$ is singular
$$A^TAx=0 \implies x^TA^TAx=0 \implies Ax=0$$
thus $x\in Null(A)$
thus
$A^TA$ is singular $\iff$ A is not full column rank
that's equivalent to

$A^TA$ is not-singular $\iff$ A is full column rank

NOTE
if A is full row rank and $n\neq m \implies Null(A)\neq {0}$ thus $A^TA$ is singular
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