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$\newcommand{\psym}{\text{Psym}_n}$ $\newcommand{\sym}{\text{sym}}$ $\newcommand{\Sym}{\operatorname{Sym}}$ $\newcommand{\Skew}{\operatorname{Skew}}$ $\renewcommand{\skew}{\operatorname{skew}}$ $\newcommand{\GLp}{\operatorname{GL}_n^+}$

Denote by $\psym$ the space of symmetric positive-definite $n \times n$ matrices, and by $\GLp$ the group of real $n \times n$ invertible matrices with positive determinant.

Let $P:\GLp \to \psym$ map each matrix $A$ to its unique positive factor in the polar decomposition, i.e. $P(A)=\sqrt{A^TA}$.

I am trying to find a nice "closed-form algebraic expression" for the differential $dP_A$, where $A \in \psym$ is symmetric positive-definite. (So I am fine with using positive square roots, but not integral formulas or vectorization operations like here or here).

In other words: I want to find a formula for $dP_A(B)$, where $A$ is positive-definite and $B$ is an arbitrary matrix. Here is a partial result:

$dP_A(B)=\operatorname{sym}(B) \iff B \in V_P:=\{B\in M_n \, | \, BP + P B^T=B^T P+ P B\}$.

Proof: Differentiating $P^2=A^TA$ we get $ \dot PP + P \dot P = B^TA + A^TB.$ Since we assumed $A \in \psym$, we have $A=P$ at time $t=0$, so our equation becomes

$$ \dot PP + P \dot P = B^TP + PB,$$

and $\dot P$ is the unique solution of this equation. Now it is easy to verify that $\dot P=\operatorname{sym}(B)$ is a solution if and only if:

$$ \frac{B+B^T}{2}P + P \frac{B+B^T}{2} = P B + B^T P \iff BP + P B^T=B^T P+ P B \iff B \in V_p $$


Note that the presence of the "symmetrization operator" is natural here; $B \to dP_A(B)$ is something which eats arbitrary matrices and returns symmetric matrices. (We also have the special case where $B$ is symmetric, and then $dP_A(B)=B$; this is also immediate from the fact that $P_{\psym}=Id_{\psym}$ and $B \in T_A{\psym}$).

The result mentioned above does not cover all the cases, since in general $V_p \subsetneq M_n$.

Unfortunately, I couldn't come up with an expression for the general case. Since $dP_A$ is linear, and we always have $\sym \subseteq V_P$, it suffices to consider skew-symmetric matrices $B$ as inputs.

However, I don't see a clear pattern fur such matrices. In the simplest case of $n=2$, and

$$A=\begin{pmatrix} \sigma_1 & 0 \\\ 0 & \sigma_2 \end{pmatrix}, B=\begin{pmatrix} 0 & a \\\ -a & 0 \end{pmatrix}, \, \, \text{ where } \, \, \sigma_1 \ge \sigma_2 $$ a short calculation shows that

$$ \dot P=dP_A(B)=\begin{pmatrix} 0 & a\frac{\sigma_1-\sigma_2}{\sigma_1+\sigma_2} \\\ a\frac{\sigma_1-\sigma_2}{\sigma_1+\sigma_2} & 0 \end{pmatrix}=\frac{\sigma_1-\sigma_2}{\sigma_1+\sigma_2}\begin{pmatrix} 0 & a \\\ a & 0 \end{pmatrix}.$$

So, the map $B \to dP_A(B)$, considered as a map $\text{skew} \to \sym$ is of the form $\begin{pmatrix} 0 & a \\\ -a & 0 \end{pmatrix} \to c(A)\begin{pmatrix} 0 & a \\\ a & 0 \end{pmatrix}$, where $C(A)=\frac{\sigma_1-\sigma_2}{\sigma_1+\sigma_2}=\sqrt{1-4\frac{\det A}{(\text{tr}A)^2}}$ is a constant depending only on $A$.

I don't see an immediate way to write $\begin{pmatrix} 0 & a \\\ -a & 0 \end{pmatrix} \to \begin{pmatrix} 0 & a \\\ a & 0 \end{pmatrix}$ in "algebraic way", i.e. involving only matrix addition, multiplication, and square roots.

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