# Is there a smooth polar decomposition for non-invertible matrices?

Every $n \times n$ real matrix $A$ has a polar decomposition $A=OP$, where $O \in O_n, P$ is symmetric positive semi-definite.

$P$ is uniquely determined by $A$, by $P(A)=\sqrt{A^TA}$, and when $A$ is invertible $O$ is also unique, given by $O=A(\sqrt{A^TA})^{-1}$.

When $A$ is not invertible, then $O$ is non-unique.

Question:

Can we choose a smooth polar factor $O(A)$ for all non-zero matrices $A$ with non-negative determinant?

More precisely, denote by $M_n^+$ the set of matrices with non-negative determinant.

Is there a smooth function $O:M_n^+ \setminus \{0\} \to O_n$ such that for every $A \in M_n^+ \setminus \{0\}$, $A=O(A)P(A)=O(A)\sqrt{A^TA}$?

Remarks:

1) We know that such a function, if exists, must give to each $A \in GL_n^+$ its unique polar factor.

2) The reason we exluded the zero matrix is that continuity implies two contradictory evaluations:

$O(0)=\lim_{t \to 0^+} O(\left(\begin{matrix}t & 0 \\ 0 & t\end{matrix}\right))=I$,

$O(0)=\lim_{t \to 0^+}=O(\left(\begin{matrix}0 & -t \\ t & 0\end{matrix}\right))=\left(\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right)$

3) The reason we had to restrict to $M_n^+ \setminus \{0\}$ (instead of working with all $M_n \setminus \{0\}$) is connectedness issues.

$M_n\setminus \{0\}$ is connected (for $n >1$). However for $O|_{O_n}=Id_{O_n}$, so if we insisted to take the domain to be all $M_n \setminus \{0\}$, the image would be $O_n$ which is disconnected.

The restriction in fact implies that $O$ is a function into $SO_n$.

• Can you make a similar remark like point (2) about matrices which split into $2\times 2$ blocks down the diagonal? – Ben McKay May 4 '16 at 13:44
• Hmm.. I think you are right. Then the answer is negative. – Asaf Shachar May 4 '16 at 13:49
• maybe you can define a smooth polar factor for all matrices with $rank(A)\geq n-1$. Consider its SVD and change the singular values by $1$. – user35593 May 4 '16 at 14:09
• @user35593 Is this set of matrices a submanifold with boundary? I am actually unsure about whether or not the substet of matrices with non-negative determinant is a submanifold with boundary. – Asaf Shachar May 4 '16 at 15:27

To take user35593's comment to fruition:

There is a unique continuous extension of the polar decomposition from $GL^+_n$ to the portion of its boundary consisting of matrices with nullity 1.

### Why is the polar decomposition non-unique when $A$ is non-invertible?

Suppose $A = OP$ and $P$ has non-trivial kernel. Then if $\Omega$ is any orthogonal matrix that acts as the identity on $\mathrm{ker}(P)^\perp$, then $O\Omega P$ is another polar decomposition. And it is easy to see that this freedom is the extent of non-uniqueness: if $A = OP = \tilde{O} P$, then $O^{-1} \tilde{O}$ is an orthogonal matrix that acts as the identity on $\mathrm{ker}(P)^\perp$.

When the kernel of $A$ (equivalently kernel of $P$) is one dimensional, then the only possibility for $\Omega$ is the matrix that is the identity on $\mathrm{ker}(P)^\perp$ and $-1$ on $\mathrm{ker}(P)$.

### Proof of claim

Now, since $\det A = \det O \det P$ and we know that $\det P \geq 0$, when we restrict to $GL_n^+$ we must have that $\det O > 0$. That is, on $M_n^+$ we can require the matrix $O$ in the polar decomposition to be not only orthogonal, but in fact in $SO_n$. And for rank $n-1$ matrices only one of $O$ and $O\Omega$ can fulfill that criterion. And it is easy to check that this is the continuous extension from $GL_n^+$.

Remark: if you extend from $GL_n^-$ you will pick up the other one. An illustrative example is the rank $1-1 = 0$ matrix $(0)\in M_1$. Approaching from $M_1^+$ the continuous limit of the polar decomposition gives $$(0) = (1)(0)$$ while approaching from $M_1^-$ the continuous limit of the polar decomposition gives $$(0) = (-1)(0).$$

### Differential geometry

To answer this question in the comments: no, when $n > 1$ the set $M_n^+$ is should not be thought of as a (smooth) submanifold with boundary; it is better described as a manifold with corners. This is precisely what you outlined in your post with the different limiting directions approaching the zero matrix.

But in a neighborhood of (topological) boundary points which have rank $n-1$, the set $M_n^+$ does look like a manifold with boundary: in a neighborhood of a rank $n-1$ matrix, the function $\det: M_n\to \mathbb{R}$ has non-vanishing derivative, and so can be used as a defining function for a hypersurface.

• Thanks! Your answer was very helpful. Do you think this continuous extension is in fact smooth (as a map from the manifold with boundary $\{A \in M_n^+ | \, \, \operatorname{rank}(A) \ge n-1\}$ to the manifold $SO_n$? – Asaf Shachar May 5 '16 at 17:52
• I don't see any reason why not. In the normal direction to the boundary the derivative of this map is 0. – Willie Wong May 5 '16 at 18:27