# Why is ATAA^TA invertible if AA has independent columns?

How can I understand that $A^TA$ is invertible if $A$ has independent columns? I found a similar question, phrased the other way around, so I tried to use the theorem

Given $rank(A) = rank(A^T) = n$ and $A^TA$ produces an $n\times n$ matrix, I can’t seem to prove that $rank(A^TA)$ is actually $n$.

I also tried to look at the question another way with the matrices

But I still can’t seem to show that $A^TA$ is invertible. So, how should I get a better understanding of why $A^TA$ is invertible if $A$ has independent columns?

Here, $Ax$, an element in the range of $A$, is in the null space of $A^T$. However, the null space of $A^T$ and the range of $A$ are orthogonal complements, so $Ax=\mathbf 0$.
If $A$ has linearly independent columns, then $Ax=\mathbf 0 \implies x=\mathbf 0$, so the null space of $A^TA=\{\mathbf 0\}$. Since $A^TA$ is a square matrix, this means $A^TA$ is invertible.