1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.