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Moreover if $A$ is regular, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$
Suppose $A$ is not regular. It holds that $$x^TAA^Tx=(A^Tx)^T(A^Tx)= \|A^Tx\|^2_2 \ge 0$$ Therefore $AA^T$ is positive semidefinite. Is this argument enough, or am I missing something?