Closest conformal matrix to a given matrix

Question

$\newcommand{\SOn}{\operatorname{SO}_n}$ $\newcommand{\COn}{\operatorname{CO}_n}$ $\newcommand{\Sym}{\operatorname{Sym}_n}$ $\newcommand{\Skew}{\operatorname{Skew}_n}$ $\newcommand{\dist}{\operatorname{dist}}$ $\newcommand{\Sig}{\Sigma}$ $\newcommand{\sig}{\sigma}$ $\newcommand{\al}{\alpha}$ $\newcommand{\id}{\operatorname{Id}}$

Let $A \in \text{GL}_+$ be an $n \times n$ real matrix with positive determinant. Is there always a closest conformal matrix to $A$? (I measure the distance between matrices using the Euclidean distance). Is the closest matrix unique?

By using SVD we can restrict to the case where $A$ is positive diagonal .

Edit:

Let $\Sig=\text{diag}(\sigma_1,\dots,\sigma_n)$ be diagonal with positive entries. Suppose all the $\sig_i$ are different, and that $\sig_1 < \sig_2 < \dots \sig_n$. I prove below that if a minimizer exists, then it is $\bar \Sig=\frac{\sum_{i=1}^n\sigma_i}{n}\id.$

However, the question of existence of a minimizer still remains. Since the conformal group is not compact, it's non-trivial. We can restrict to given ball of course, but if it contains $0$ then we have a problem, since the intersection with the conformal group won't be closed.

Conjecture: A minimizer always exist and unique. Equivalently, for any positive diagonal matrix $\Sig$ its closest matrix is given by $\bar \Sig$.

I proved it for the case where the entries of $\Sig$ are all different, assuming the existence of a minimizer. (I can adapt the argument to the case of multiplicities but it's a bit cumbersome).

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Proof the closest matrix is diagonal: (If someone finds an easier proof that would be nice).

We denote the (special) conformal group by $\COn$, i.e $$ \COn = \{Q \in M_n \, | \, Q^TQ=(\det Q)^{\frac{2}{n}}\id,\det Q >0 \}.$$

Let $\Sig=\text{diag}(\sigma_1,\dots,\sigma_n)$ be diagonal with positive entries. If a closest matrix $Q \in \COn$ exists then I prove below it must be diagonal (hence a scalar times the identity, since it's conformal).

So, the problem is reduced to minimizing $\sum (\sigma_i - \sigma_{\rm new})^2$. Thus, the closest matrix is $$Q=Q(\Sig)=\bar \Sig=\frac{\sum_{i=1}^n\sigma_i}{n}.$$

In particular the closest matrix is unique.

Suppose $Q \in \COn$ satisfies $$ d( \Sig,Q) = \dist( \Sig ,\COn), \tag{1}$$

and let $\al(t)=\Sig+t\left( Q-\Sig \right)$ be the minimizing geodesic from $\Sig$ to $Q$. Since a minimzing geodesic from a point to a submanifold intersects that submanifold perpendicularly, we get

$$ \dot \al (1) = (Q- \Sig) \perp T_Q \COn $$

The map $X \to Q^{-1}X$ is homothety of $\text{GL}_+$ (endowed with the Euclidean Riemannian metric), so it preserves orthogonality. Thus,

$$ \id-Q^{-1}\Sig=Q^{-1}(Q-\Sig) \in (T_{\id}\COn)^{\perp} \subseteq (\Skew)^{\perp}=\Sym, $$

which implies $Q^{-1}\Sig \in \Sym$, or $$ Q^{-1}\Sig = \Sig (Q^{-1})^T. \tag{2}$$

Since $Q \in \COn$, $Q^TQ=(\det Q)^{\frac{2}{n}}\id$. So, $Q^{-T}=(\det Q)^{-\frac{2}{n}}Q$. Plugging this into $(2)$ we obtain

$$ Q^{-1}\Sig = \Sig (\det Q)^{-\frac{2}{n}}Q, $$ or equivalently $$ \big( (\det Q)^{-\frac{1}{n}}Q \big)^{-1} \Sig = \Sig \cdot (\det Q)^{-\frac{1}{n}}Q \tag{3}.$$

Denote $P:=(\det Q)^{-\frac{1}{n}}Q$, and note that $P \in \SOn$. Then $(3)$ is equivalent to

$$ P^T\Sig=\Sig P. \tag{4}$$

It suffices to show $P$ is diagonal.

$$ P^T\Sig = \begin{pmatrix} \sig_1P_{11} & \sig_2 P_{21} & \cdots & \sig_nP_{n1} \\ \sig_1P_{12} & \sig_2 P_{22} & \cdots & \sig_nP_{n2} \\ \vdots & \vdots& \vdots & \vdots & \\ \sig_1P_{1n} & \sig_2 P_{2n} & \cdots & \sig_nP_{nn} \\ \end{pmatrix} = \begin{pmatrix} \sig_1 P_{11} & \sig_1 P_{12} & \cdots & \sig_1 P_{1n} \\ \sig_2P_{21} & \sig_2 P_{22} & \cdots & \sig_2P_{2n} \\ \vdots & \vdots& \vdots & \vdots & \\ \sig_nP_{n1} & \sig_n P_{n2} & \cdots & \sig_nP_{nn} \\ \end{pmatrix} = \Sig P \tag{5} $$

Looking at the first column of both matrices in $(5)$ and comparing (squared) norms we get $$ \sig_1^2 \sum_{i=1}^n P_{1i}^2=\sig_1^2=\sig_1^2 \sum_{i=1}^n P_{i1}^2=\sum_{i=1}^n \sig_i^2 P_{i1}^2. \tag{6} $$

Since we assumed $\sig_1$ is strictly smaller than all the other singular values, we deduce that $P_{i1} = 0$ for $i > 1$. Inserting this back to $(6)$, we get

$$ \sig_1^2 \sum_{i=1}^n P_{1i}^2=\sum_{i=1}^n \sig_i^2 P_{i1}^2= \sig_1^2 P_{11}^2, $$

So, we deduce $P_{1i}=0$ for $i >1$.

Now we continue in this way.

Answer 1

$\newcommand{\SOn}{\operatorname{SO}_n}$ $\newcommand{\COn}{\operatorname{CO}_n}$ $\newcommand{\Sym}{\operatorname{Sym}_n}$ $\newcommand{\Skew}{\operatorname{Skew}_n}$ $\newcommand{\dist}{\operatorname{dist}}$ $\newcommand{\Sig}{\Sigma}$ $\newcommand{\sig}{\sigma}$ $\newcommand{\al}{\alpha}$ $\newcommand{\id}{\operatorname{Id}}$

Here is complete answer. There always exist a unique minimizer.

Let $\Sigma=\text{diag}(\sigma_1,\dots,\sigma_n)$ be diagonal and positive. We want to solve $$ \min_{Q \in \COn} \| \Sig-Q\|^2=\min_{Q \in \COn} \| \Sig\|^2+ \|Q\|^2-2\langle \Sig,Q \rangle.$$

Since $\Sig$ is constant, this is equivalent to solving

$$\min_{Q \in \COn} F(Q), \, \text{ where } \, F(Q):=\|Q\|^2-2 \sum_{i=1}^n \sig_iQ_{ii}.$$

Any $Q \in COn$ can be written uniquely as $\lambda O$ for some $\lambda \in \mathbb{R}^+$, $O \in \SOn$. Since

$$ F(\lambda O)=\lambda^2n-2\lambda\sum_{i=1}^n \sig_iO_{ii},$$

it is obvious that $$ F(\lambda O) \ge F(\lambda \id).$$ So, we are reduced to optimizing $\lambda$, i.e solve

$$ \min_{\lambda \in \mathbb{R}^+} F(\lambda \id)=\min_{\lambda \in \mathbb{R}^+} \lambda^2n-2\lambda\sum_{i=1}^n \sig_i.$$

By differentiating we get that $\lambda=\frac{\sum_{i=1}^n\sig_i}{n}$.