When will $\operatorname{det}\left(A\cdot A^{\top}\right)=0$? I am writing a small computer program to solve certain linear algebra equations as part of a larger program.  For two of my functions I need to evaluate $\left(A\cdot A^{\top}\right)^{-1}$.  This got me wondering.  Is there any significance to $\operatorname{det}\left(A\cdot A^{\top}\right)=0$ ?  When will this happen, and is it common in the real world?  In my particular case it is unlikely that $A$ itself would ever be a square matrix.  Just throwing that out there.
If $A$ is $m\times n$ real matrix and $\operatorname{rank}A = k\le\min(m,n)$, then $AA^T$ is $m\times m$ matrix of rank $k$. Thus $AA^T$ is invertible if and only if $\operatorname{rank}A = m$ (in particular: if $m &gt; n$, then $AA^T$ must be singular.)



For complex matrices it does not hold in general, as baharampuri suggests. However for complex matrix $A$, $\operatorname{rank} A = \operatorname{rank} AA^*$ holds where $A^*$ is the conjugate transpose of $A$.
For A being a n x n matrix :
det(A * A^T)=det(A)*det(A^T)=det(A)^2
and det(A)=0 implies that A is not invertible.