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

Question 1

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

$rank(A^TA) \le min(rank(A^T),rank(A))$

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

$A^TA = \begin{bmatrix}a_1^T \\ a_2^T \\ \ldots \\ a_n^T \end{bmatrix} \begin{bmatrix}a_1 a_2 \ldots a_n \end{bmatrix} = \begin{bmatrix}A^Ta_1 A^Ta^2 \ldots A^Ta_n\end{bmatrix}$

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?

Question 2

Consider the following:
$A^TAx=\mathbf 0$
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.

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

Answer

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