$A$ is in general a rectangular matrix of size $(n\times d)$. The product $AA^T$ will be invertible when it is a full rank matrix, and I have read this happens when $A^T$ has at least $d$ linearly independent rows, "so that any point in $\mathbb{R^d}$ can be reached by a linear combination of such $d$ rows".
I wanted to please ask why this assures full rank of the stated product and what is the connection between its rank and this requirement.
On
My own attempt:
The product $AA^T$ can be thought of as a sum involving the column vectors of $A$, $a_i$: $$AA^T=\sum_{i=1}^{d}a_i({a_i}^T)$$
Each of these terms is a matrix of rank one, and - as long as the vectors being added are linearly independent - the sum of rank one terms has a rank equal to the number of terms. We therefore need (exactly) $d$ of such terms for the product to be of rank $d$.
When $A^t$ is injective, $AA^t$ is injective as well, because $x^tAA^tx=\|A^tx\|^2\not=0$