Is $AA^T$ a positive-definite symmetric matrix?

Hint: let $v$ be a non zero vector; then, setting $B=A^T$ for simplicity, $$ v^TB^TBv=(Bv)^T(Bv) $$ is positive if and only if $Bv\ne 0$. How can you ensure that $Bv\ne0$ if and only if $v\ne0$?

Conversely, if $AA^T$ is positive definite, what can you say about the rank of $A$?

So, what's a necessary and sufficient condition so that $AA^T$ is positive definite?

$AA^T$ is not necessarily positive definite, but it is positive semi-definite, meaning that $\langle x, AA^Tx \rangle \ge 0$ for all vectors $x$. To see this, note that $\langle x, AA^Tx \rangle = \langle A^Tx, A^Tx \rangle = \Vert A^Tx \Vert^2 \ge 0$. A counter example to positive definiteness is provided, when $n = 2$, by taking

$A= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}; \tag{1}$

then

$AA^T = A, \tag{2}$

so if $x = (0, 1)^T$,

$\langle x, AA^Tx \rangle = 0. \tag{3}$

It is easy to generalize this example by taking $A$ to be a diagonal matrix in $M_{n \times n}(\Bbb R)$ with at least one zero on the diagonal; many other generalizations are also possible.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

For a matrix to be positive definite we need $x^TMx \geq0$. Consider the case where the matrix A is not full rank so therefore has more rows than columns. It should be clear there exist a vector $x$ where $ x \neq 0 $, such that $x^TA = 0$. Thus we have $x^TAA^Tx = 0$ Therefore $AA^T$ cannot be strictly positive definite. (But you can show it is always semi-positive definite.)