The problem itself does not say that [math]A[/math] is a square matrix (in fact it may be NOT a square one, one says that a matrix is invertible or not only when the matrix is square), so let's treat it as an [math]m[/math] by [math]n[/math] matrix, and the vector [math]x[/math] should be a [math]n[/math]-D column vector, [math] A^T A [/math] is a [math]n[/math]-square matrix.

Through
[math] x^T A^T A x = (A x)^T Ax = \| Ax \|^2 \geq 0 [/math]
we know that [math] A^T A [/math] is positive semidefinite.

1. When [math]m=n[/math], [math]A[/math] is a square matrix, and [math] A^T A [/math] is positive definite if and only if [math]A[/math] is invertible (full-rank, [math]r(A^T A)=m=n[/math]).
2. When [math]{m}<{n}[/math], [math] A^T A [/math] is not a full-rank matrix ([math]r(A^T A) \leq m<{n}[/math]), therefore it is ALWAYS positive semidefinite.
3. When [math]m>n[/math], [math]r(A^T A) \leq n[/math], hence it may be positive definite or only positive semidefinite, depending on specific examples.