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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!

Ludwig
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1 Answers1

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  1. 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.

  2. 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.

  • Thank you! Just a couple of clarifications: (i) if $A_\theta$ is a full-rank (for every $\theta$) rectangular matrix-valued function then your argument still applies, right? (ii) I didn't get your point about uniqueness: in the second case (non-invertible $A_\theta$ for some $\theta$) is the decomposition unique or not? – Ludwig Dec 08 '15 at 10:02
  • Is the second point really necessary? The OP wants $U_\theta$ to have orthonormal columns and $R_\theta$ to be positive definite. So, presumably $A_\theta$ must have full column rank and hence $A_\theta^T A_\theta$ is always positive definite. – user1551 Dec 08 '15 at 11:55
  • @ user1551, you are right. I skip from one thing to another: can you have a look on http://math.stackexchange.com/questions/1558591/if-a-matrix-is-triangular-is-there-a-quicker-way-to-tell-if-it-is-can-be-diagon/1559880?noredirect=1#comment3185285_1559880 –  Dec 08 '15 at 12:12