What is the connection between $\rho$ and $\sigma$ if $\rho\rho^T=\sigma\sigma^T$? I want to prove that there exists a Borel function $R(\rho,\sigma)$ with values in $M^{d\times d}$ defined on $D=\lbrace(\rho,\sigma)\in M^{d\times d}\times M^{d\times d}\,: \rho\rho^T=\sigma\sigma^T\rbrace$ such that $\sigma=\rho R(\rho,\sigma)$ and $RR^T=I$. 
My idea is: Diagonalize $\rho\rho^T=\sigma\sigma^T=QDQ^T$ where Q is an orthogonal matrix.
It's obvious that $\sigma=UQ\sqrt{D}$ with U orthogonal matrix satisfies the request but is this the only possibility? I would appreciate any possible help. Thank you in advance.
 A: Edit: I rewrote this post. I left the original post bellow, because there are comments related to this original post.
Let $\Im(A)$ denote the image of $A$ and $M_k$ the set of real matrices of order $k$. 
Proposition: Let $\rho,\sigma\in M_k$. 
These matrices satisfy $\rho\rho^T=\sigma\sigma^T$
if and only if $\rho R(\rho,\sigma)=\sigma$, such that $R(\rho,\sigma)R(\rho,\sigma)^T$ is an orthogonal projection onto the $\Im(\rho^T)$. Moreover, $R(\rho,\sigma)$ can be chosen to be a borel function of $\rho$ and $\sigma$.
Remark: Notice that if $\rho$ is invertible then $\Im(\rho^T)=\mathbb{R}^k$ and the only orthogonal projection onto $\Im(\rho^T)=\mathbb{R}^k$ is $Id$. Thus, $R(\rho,\sigma)R(\rho,\sigma)^T=Id$.
Proof: Suppose $\rho R(\rho,\sigma)=\sigma$. Then $\rho R(\rho,\sigma)R(\rho,\sigma)^T\rho^T=\sigma\sigma^T$.
Now, since $R(\rho,\sigma)R(\rho,\sigma)^T$ is an orthogonal projection onto the $\Im(\rho^T)$ then $R(\rho,\sigma)R(\rho,\sigma)^T\rho^T=\rho^T$. Thus,  $\rho R(\rho,\sigma)R(\rho,\sigma)^T\rho^T=\rho\rho^T=\sigma\sigma^T.$
Next, suppose $\rho\rho^T=\sigma\sigma^T$. Then, $\Im(\rho)=\Im(\sigma)$.
Let $\rho^+$ be the pseudo inverse of $\rho$. Thus, $\rho\rho^+$ is an orthogonal projection onto the $\Im(\rho)=\Im(\sigma)$.
Thus, $\rho\rho^+\sigma=\sigma$. Define $R(\rho,\sigma)=\rho^{+}\sigma$.
Notice that $R(\rho,\sigma)R(\rho,\sigma)^T=\rho^{+}\sigma\sigma^T(\rho^{+})^T=\rho^{+}\rho\rho^T(\rho^{+})^T=\rho^{+}\rho(\rho^{+}\rho)^T=(\rho^{+}\rho)^2=\rho^{+}\rho$.
Remind that $\rho^{+}\rho$ is an orthogonal projection onto $\Im(\rho^T)$. 
Finally, $R(\rho,\sigma)=\rho^{+}\sigma$ is borel because $f(\rho)=\rho^+$ is borel, since it is a pointwise limit of borel functions: 
$f(\rho)=\displaystyle\lim_{h\rightarrow 0+} f_h(\rho)$, where $f_h(\rho)=(\rho^T\rho+hId)^{-1}\rho^T$.   
(See http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse )
$\square$

Original Post:
For the real case, it is possible to obtain a borel function $R(\rho,\sigma)$ such that 
$\rho R(\rho,\sigma)=\sigma$.
Let $\rho,\sigma$ be real matrices of order k. Let $\Im(A)$ denote the image of $A$. 
If $\rho\rho^T=\sigma\sigma^T$ then $\Im(\rho)=\Im(\sigma)$.
Let $\rho^+$ be the pseudo inverse of $\rho$. Thus, $\rho\rho^+$ is an orthogonal projection onto the $\Im(\rho)=\Im(\sigma)$.
Thus, $\rho\rho^+\sigma=\sigma$.
Let us define $R(\rho,\sigma)=\rho^{+}\sigma$. Notice that
the function $R(\rho,\sigma)$ is borel because $f(\rho)=\rho^+$ is borel, since it is a pointwise limit of borel functions: 
$f(\rho)=\displaystyle\lim_{h\rightarrow 0+} f_h(\rho)$, where $f_h(\rho)=(\rho^T\rho+hId)^{-1}\rho^T$.   
(See http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse ) 
