Multiplicativity of "isometric projections" and commutation of matrices

Question

Every $A \in \text{GL}_n^+(\mathbb{R})$ (an invertible matrix with positive determinant) has a unique Polar decomposition: $A=OP$ where $O \in \text{SO}_n$, $P$ symmetric positive definite. The orthogonal polar factor is $O(A)=A(\sqrt{A^TA})^{-1}$ and the positive factor is $P(A)=\sqrt{A^TA}$.

Question:

Assume $O(A)O(B)=O(AB)$. Does it imply the $P(A),P(B)$ commute?

What about the converse? Does $[P(A),P(B)]=0 \Rightarrow O(A)O(B)=O(AB)$?

(It is easy to see a sufficient condition for $O(A)O(B)=O(AB)$ is that $P(A)$ commutes with $O(B),P(B)$. This question is a step towards seeing whether this condition is necessary) .

Motivation:

This question has a geometric nature, since $O(A)$ is the closest matrix in $\text{SO}_n$ to $A$ (w.r.t to the Frobenius norm). Thus, it reflects the "rotational part" of $A$; We can think of it as the best "isometric approximation" of $A$. If we know these two approximations for $A,B$ it is natural to compose them and to ask if we obtained in this way the best approximation to $AB$.

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Partial result:

If $B > 0$ is symmetric positive definite, then $O(A)O(B)=O(AB) \iff [P(A),B]=[P(A),P(B)] =0$

Proof:

Let $A=O_AP_A$ be the polar decomposition of $A$. Then:

$$ O_AO_B=O_{AB} \iff O_A=O_{AB}=O_{O_AP_AB} \iff O_AP_AB= O_A\hat P$$ for some $\hat P >0$.

Thus, $ O_AO_B=O_{AB} \iff $ $P_AB > 0 \iff [P_A,B]=0$.

Answer 1

A fairly random computation with Mathematica suggests that the answer is no. A numerical counterexample is $$A=\begin{pmatrix}1&0.7\\0.2&4.9296967794602\end{pmatrix},\,\,\,\,B=\begin{pmatrix}1&0.3\\0.2&5\end{pmatrix},$$ which you can check with the commands

m1={{1, 0.7}, {0.2, 4.9296967794602}}; m2={{1, 0.3}, {0.2, 5}}; m3=m1.m2; {u1, w1, v1} = SingularValueDecomposition[m1]; {u2, w2, v2} = SingularValueDecomposition[m2]; {u3, w3, v3} = SingularValueDecomposition[m3]; a1=u1.Transpose[v1]; a2=u2.Transpose[v2]; a3=u3.Transpose[v3]; a1.a2-a3
p1=v1.w1.Transpose[v1]; p2=v2.w2.Transpose[v2]; p3=v3.w3.Transpose[v3]; p1.p2-p2.p1

(Here, m1 is A, m2 is B, m3 is AB. The polar decompositions are m1=a1.p1, m2=a2.p2, m3=a3.p3, and we check that a1.a2-a3 is zero up to precision while p1.p2-p2.p1 is non-zero.)

Answer 2

heptagon showed that probably $O(A)O(B)=O(AB)$ does not imply $[P(A),P(B)]=0$.

I will show the converse is also false, i.e it is possible that $[P(A),P(B)]=0$, but $O(A)O(B) \neq O(AB)$.

We take $P(A)=P(B)=P$, i.e $A=O_AP,B=O_BP$, so $AB=O_APO_BP$.

Notice that $O(A)O(B)=O(AB)$ implies $O_APO_BP=AB=O_AO_B\tilde P$, so

$$ PO_BP = O_B\tilde P,$$ and in particular $(O_B)^{-1}PO_BP$ should be a symmetric matrix. However, there are matrices in $\operatorname{SO}$ which do not satisfy this property for some positive definite $P$.

For example, we can take

$O_B=\begin{pmatrix} \frac{1}{\sqrt 2}& \frac{1}{\sqrt 2} \\ -\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{pmatrix},\,\,\,\,P=\begin{pmatrix}2&0\\0&3\end{pmatrix},$ ($O_B$ is a rotation by $\frac{\pi}{4}$)

Then, a calculation shows:

$$ (O_B)^{-1}PO_BP= \begin{pmatrix}5&-\frac{3}{2}\\-1&\frac{15}{2}\end{pmatrix}$$

which is not symmetric.