What is the number of degrees of freedom (or dimension?) of $AA^H$, where $A$ is a $4\times4$ complex matrix and $A^H$ the Hermitian conjugate? I'm counting a complex number as 2 degrees of freedom.
Since $AA^H$ is Hermitian, the degrees of freedom should be at most 16 (2*6 off-diagonal + 4 diagonal), but I do not know how to show that it is not fewer.
Is there any anti-involution which would make the degrees of freedom of $AA^*$ equal to 12?
Yes, it is $12$. The point is that every positive definite (Hermitian) matrix has a square root, and the set of positive definite matrices is open in the space of Hermitian matrices (since positive definiteness can be characterized by a finite number of determinants being positive).
[EDIT] Oops, I mean it is $16$. For $n \times n$ Hemitian matrices the dimension over the reals is $n + n(n-1) = n^2$, $n$ for the diagonal elements and $n (n-1)$ for the real and imaginary off-diagonal elements.
