TITLE: Is $\text{rank} (AA^*)=\text{rank}(A)$ for all nonsquare matrices?
QUESTION [5 upvotes]: If $A$ is a $m\times n$ type matrix with $m\geq n$ then 
$$
rank (A^*A)=rank (A).
$$
Is maybe also true in general that 
$$
rank (A^*A)=rank (A) ?
$$
Thanks
Edit.
My question is different from the question about $rank(A^TA)=rank A$, because concerns complex conjugate transposed matrice instead of transposed.

REPLY [7 votes]: if $*$ is the transpose operation, then yes. here is why. we can show that $$Ax = 0 \, \text{iff}  \,  A^T Ax = 0. \tag 1$$ this is true because $$0 = |Ax|^2 = (Ax)^TAx = x^T(A^T A) x. $$
then $(1)$ implies $$N(A) = N(A^T A)$$ together will the nullity theorem $$\dim N(A) + rank(A) = n$$ should give you $$rank(A) = rank(A^T A). $$