Assume $A$ is an arbitrary $m\times n$ real matrix.
Is $\det(AA^T)$ always positive? Is it non-negative or it can have any value?
Edit:
It seems I have to emphasis that $m \ne n$ i.e. matrix is non-squared.
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$\begingroup$If $A$ has a singular value decomposition $$ A = VDU^{-1} $$ (note since $A$ is rectangular, $V \neq U$ but $V^T = V^{-1}$ and $U^T = U^{-1}$ and finally, $D^T=D$ since $D$ is diagonal) then $$ A^T = (U^T)^{-1}D V^T = UDV^T $$ and now $$ AA^T = VD^2 V^T $$ so $\det(AA^T) = \det D^2$ which must be real but cuold be negative is one of the singular values of $A$ is complex.
$\endgroup$ 2 $\begingroup$In the case n=m (square matrix)
$$det(AB)=det(A)det(B)$$
and
$$det(A^T)=det(A)$$
yes, it is positive
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