Is
$$ det(AA^T) = det (A^TA) $$
for any rectangular matrix true?
Consider $A=[1\cdots 1]\in M_{1,n}(\mathbb{R})$ and $n\ge2$.
We have $A^TA=n$ and so $\det(A^TA)=n$
But $AA^T$ has rank $1<n$ hence $\det(AA^T)=0$
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Consider $A=[1\cdots 1]\in M_{1,n}(\mathbb{R})$ and $n\ge2$.
We have $A^TA=n$ and so $\det(A^TA)=n$
But $AA^T$ has rank $1<n$ hence $\det(AA^T)=0$