Polar decomposition for matrix-valued functions Given a continuous matrix-valued function, say $A_\theta$, of a variable $\theta\in I\subseteq \mathbb{R}$. Is it always possible to find a polar decomposition which is a continuous function of $\theta$, i.e., a decomposition of $A$ of the form 
$$A_\theta = U_\theta R_\theta$$ 
where $U_\theta$ has orthonormal columns, $R_\theta$ is positive definite and $U_\theta$,$R_\theta$ are both are continuous function of $\theta$? 
References which address this problem (or similar ones) are welcome!
 A: *

*Assume that for every $\theta$, $A_{\theta}$ is invertible; then the answer is yes. Indeed, in this case $R,U$ are unique: $R_{\theta}=\sqrt{A_{\theta}^TA_{\theta}},U_{\theta}=AR_{\theta}^{-1}$ and are continuous.

*Otherwise, the decomposition is not unique; that follows is a particular polar decomposition of $A$. A SVD of $A_{\theta}$ is $W_{\theta}\Sigma_{\theta} V_{\theta}^T$; the columns of $W_{\theta}$ are eigenvectors of $A_{\theta}A_{\theta}^T$, the columns of $V_{\theta}$ are eigenvectors of $A_{\theta}^TA_{\theta}$ and the diagonal of $\Sigma_{\theta}$ is composed with the square-roots of the eigenvalues of $AA^T$. Then $A_{\theta}=U_{\theta}R_{\theta}$ with $U_{\theta}=W_{\theta}V_{\theta}^T$ and $R_{\theta}=V_{\theta}\Sigma_{\theta} V_{\theta}^T=\sqrt{A_{\theta}^TA_{\theta}}$ that is continuous. When $A_{\theta}$ is invertible, $W_{\theta},V_{\theta}^T$ are not unique but their product $U_{\theta}$ is unique and continuous; when $A_{\theta}$ is not invertible, I don't know if the uniqueness remains.
EDIT. (Answer to Jacquard). Point 1 remains true if we assume that, for every $\theta$, $A_{\theta}$ is a $m\times n$ matrix with $m\geq n$ and $rank(A_{\theta})=n$. About point 2, assume that there is an isolated $\theta_0$ with rank $<n$; then $R_{\theta_0}$ is only $\geq0$ and I don't think that we can choose (in worst cases) $U_{\theta_0}$ s.t. the function $U_{\theta}$ is continuous in $\theta_0$; yet, I did not write any counter-examples.
Note that if $m\leq n$ and $rank(A)=m$, then you can write $A=RU$ where the rows of $U$ are orthonormal.
