TITLE: Unitary factor in polar decompositions
QUESTION [8 upvotes]: Let $A, B$ be $n$-square (Hermitian) positive definite matrices. Let $AB=U|AB|$ be the polar decomposition of $AB$. So $U$ is unitary (called the unitary factor of $AB$). What is the optimal constant $c$ such that $\|I-U\|\le c$, where the norm is the usual spectral norm?
I want to have some understanding on the behaviour of the unitary factor for certain   classes of matrices (e.g. matrices with real eigenvalues). Perhaps this is well known, any pointer to the existing papers is welcome.

REPLY [2 votes]: I think that the  part $(a)$ of  proposition $2.4$ of this paper   shows that for $n$ sufficiently  large one  can construct  a $n \times n$ unitary  matrix $U=-I_{2}\oplus U'$,  which can  be  decomposed as the  product of three positive  matrices. For  such  $U$ we have $\parallel U-I\parallel=2$. In fact one  can take a unitary  matrix $U'$  with  $Det \;U'=1$ such that $0$ lies in the interior of  the  convex hull of the eigenvalues of $U$. This convex hull is equal to $W(U)$, the numerical range of $U$.
In fact  if $U=ABC$  for  positive  $A,B,C$ then $UC^{-1}=AB$  so $C^{-1}=|AB|$ is the positive  factor of $U $ in its   polar  decomposition. Of  course such  $U$ has  $-1$ as an eigenvalue.
An explicit  example is the  following 
$$\begin{pmatrix}-1& 0&0&0\\0&-1&0&0\\0&0&\sqrt{3}/2&-1/2\\0&0& 1/2&\sqrt{3}/2 \end{pmatrix} $$
According to the  aboved  linked paper this  matrix  is  a  product of  three positive  matrices. This matrix, with  distance $2$ from  the identity matrix $I_{4}$, is  the  unitary  factor  of  polar  decomposition of $AB$ for two  positive  matrices $A,B$. So  according to the notation $C_{n}$ in the  answer  by Loup  Blanc, we  have $C_{n}=2, \; \forall n\geq 4$.