Every $A \in \text{GL}\_n(\mathbb{R})$ has a unique [Polar decomposition](https://en.wikipedia.org/wiki/Polar_decomposition): 


$A=O\_AP\_A$, $O \in \operatorname{O}\_n, P \in \operatorname{Psym}\_n$.
In particular the orthogonal factor is given by $$O\_A=A(\sqrt{A^TA})^{-1}.$$ **Question:**


Let $A,B \in \operatorname{GL}\_n^+$. Does there exist a positive constant $c<1$ such that $$ |AB-O\_{AB}| \ge c|AB-O\_AO\_B|$$


(Where the norm $| \cdot |$ is the standard Euclidean/Frobenius norm)


**Motivation:**


$O\_A$ is the [closest matrix in $\text{SO}\_n$ to $A$](https://math.stackexchange.com/questions/1477855/distance-of-a-matrix-from-the-orthogonal-group); We can think of it as the isometric projection of $A$. This projection is not multiplicative in general\*\* (i.e $O\_{AB} \neq O\_AO\_B $ for some $A,B$, for a concrete example see [here](https://math.stackexchange.com/a/1910835/104576)). 


My question concerns the boundedness of the "error" (in computing the distacne from $\text{SO}\_n$) when using $O\_A O\_B$ instead of $O\_{AB}$.




---


\*\* Indeed, see [this previous question](https://mathoverflow.net/questions/248842/when-does-isometric-projection-respect-multiplication?rq=1) of mine, which concerns the analysis of the pairs of matrices satisfying the multiplicativity property.



Bounding the non-multiplicativity of isometric projection
If the constant is allowed to depend upon the dimension, the estimate is simple. Let $A=O\_AP\_A,B=O\_BP\_B$, then $AB=O\_AO\_B(O\_B^\*P\_AO\_B)P\_B$ and we are left with showing that if the product of two positive definite self-adjoint operators $X=O\_B^\*P\_AO\_B$ and $Y=P\_B$ is $\delta$-close to a unitary operator $V$, then it is $C\delta$-close to $I$. Now, we do not really know much about $XY$ except that it is conjugate to $X^{1/2}YX^{1/2}$, so all eigenvalues are real positive. Thus, it will suffice to show that if a $\delta$-perturbation of a unitary operator $U$ has real positive eigenvalues, then $U$ is $C\delta$-close to the identity. However, if it were not the case, there would be an eigenvalue of $U$ that is $\frac 12 C\delta$ far from the positive semi-axis. If you now take the Gershgorin disks of radius $\delta$ and if $C>5n$, say (where $n$ is the matrix size), then there would be a connected component of the union of the Gershgorin disks that would not cross the positive semi-axis, so some eigenvalues would be confined there. 


I'm still curious if we can get a dimension-independent bound, so don't hurry to accept this answer ;-)


***Edit***. Since there are some difficulties in understanding, let me go into some details (all of which are totally classical).


**Gershgorin-Rouche theorem** Let $A\_t\quad (t\in[0,1])$ be a continuous family of matrices and $K$ a compact set on the complex plane (with decent boundary, if you want, though it is irrelevant). if the boundary of $K$ is free from the eigenvalues of $A\_t$ for all $t\in(0,1)$, then all $A\_t$ have the same number of eigenvalues in $K$. In particular, if $A\_0$ is normal and $A\_t=A\_0+tQ$, then each *connected component* $K$ of the union of closed disks of radius $\|Q\|$ centered at the eigenvalues of $A\_0$ has exactly as many eigenvalues of $A\_1$ in it as of $A\_0$. 


**Proof** The first part is just the classical Rouche theorem about zeroes of analytic functions applied to the characteristic polynomials of $A\_t$.
To show that second part, let us just show that no $A\_t$ can have an eigenvalue $\lambda$ on the boundary of $K$. Indeed, for any non-zero vector $x$, we have $|(A\_0-\lambda I)x|\ge \|Q\||x|$ (here we use the normality of $A\_0$) and $|tQx|<\|q\||x|$, so, by the triangle inequality $|(A\_t-\lambda I)x|>0$.


Now, take $A\_0=U$ here and let $K$ be the connected component of the union of disks of radius $\delta$ containing the "faraway" (from the positive real semi-axis) eigenvalue of $U$. Then any $\delta$-perturbation of $U$ must have eigenvalues in $K$. Note that those eigenvalues *do not* need to be close to any particular eigenvalue of $U$, but they are certainly *not* real positive, and that's all I need to get a contradiction. 


