Let $A,B$ be two invertible commuting matrices (over the reals).

Given $A \in GL_n$, we denote by $O(A)$ its orthogonal polar factor, i.e the unique orthogonal matrix such that there exists a decomposition of $A$:

$$ A=O(A) \cdot P,$$ where $O(A) \in O_n, P $ is symmetric positive-definite.

Thus $O(A)=A(\sqrt{A^TA})^{-1}$.

Is it true that $O(A),O(B)$ commute?

I know that $O(A^{-1})=(O(A))^{-1}$.


It is false even for $n=2$. To see that, it suffices to randomly choose $A\in M_2(\mathbb{R})$ and after, $B\in M_2(\mathbb{R})$ s.t. $AB=BA$ and $\det(A)\det(B)<0$.

An instance is: $A=\begin{pmatrix}27&-76\\-93&-72\end{pmatrix},B=A^2-A+2I$.

  • $\begingroup$ Thanks. Can you please elaborate on why $\det(A)\det(B) <0$ implies the orthogonal factors do not commute? $\endgroup$ – Asaf Shachar Aug 3 '16 at 22:11
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    $\begingroup$ If $u\in O^+(2)\setminus\{\pm I\}$ and $v\in O^-(2)$ then $uv\not= vu$. $\endgroup$ – loup blanc Aug 3 '16 at 23:16
  • $\begingroup$ Thanks! By the way, is this fact you mentioned about non-commutativity of matrices in $O^+,O^-$ holds in higher dimensions than $2$? $\endgroup$ – Asaf Shachar Aug 4 '16 at 7:01
  • $\begingroup$ $O^+(2)$ is commutative; yet, if $n\geq 3$, then $O^+(n)$ is not and the condition $\det(A)\det(B)<0$ is useless. $\endgroup$ – loup blanc Aug 5 '16 at 13:56

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