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For $M \in Gl_n$ the polar decomposition is of the form $M=U\Sigma $ with $U$ an orthogonal matrix and $\Sigma$ symmetric positive definite.

-preliminary question: if $M$ is singular do we still have such a decomposition with $\Sigma$ just positive? -If $M$ is skew-symmetric, can we say something about $\Sigma$ and $U$?

Thanks

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We can define the polar decomposition of a rectangular matrix of maximal rank. For example, let $A\in M_{mn}$ with $m\geq n$ and $rank(A)=n$. Since $A^TA$ is symmetric $>0$, let $S=\sqrt{A^TA}$, $Q=AS^{-1}\in M_{mn}$. Then $Q^TQ=I_n$ and $Q$ is pseudo-orthogonal (the $n$ columns of $Q$ form an orthonormal system in $\mathbb{R}^m$).

If $K\in GL_{2p}$ is skew symmetric ($n=2p$ is even), then $K=Pdiag(a_1U,\cdots,a_pU)P^{-1}$ where $P$ is orthogonal, $U=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and $a_i>0$. Thus the polar decomposition is $K=QS$ where $Q=Pdiag(U,\cdots,U)P^{-1}$ and $S=Pdiag(a_1I_2,\cdots,a_pI_2)P^{-1}$.

EDIT. When $A\in M_n$ is singular, one can define "A" polar decomposition (it is not unique) as follows: let $(A_k)_k$ be a sequence of invertible matrices that converges to $A$ and $A_k=Q_kS_k$ be their (unique) polar decomposition. Since $O(n)$ is compact, there exists a subsequence of $(Q_k)$ that converges to $Q\in O(n)$. Then $A=Q\sqrt{A^TA}$ is such a decomposition.

If $A$ is skew symmtric, then $A$ is orthogonally similar to $D=diag(a_1U,\cdots,a_pU,0_{n-2p})$ and some decompositions of $D$ are $D=diag(U,\cdots,U,R)diag(a_1I_2,\cdots,a_pI_2,0_{n-2p})$, where $R\in O(n-2p)$.

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  • $\begingroup$ Thank you . Nothing can be said about the case where the dimension is odd (even if it's singular)? $\endgroup$ – Chevallier May 19 '16 at 18:19

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