[Math] $A^TA$ is non-singular if and only if $A$ has full row rank

Question

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.

Answer 1

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