# Closest conformal matrix to a given matrix

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

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).

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.

• For diagonal matrices you are just minimizing $\sum (\sigma_i - \sigma_{\rm new})^2$, which does yield the arithmetic mean.
– user856
Jul 29, 2017 at 16:51
• You might be in the right direction, but this is not so simple. You first need to prove the closest conformal matrix to a (positive) diagonal matrix is itself diagonal. (This seems intuitive, since there is no "reason" to rotate, as the original matrix is already "aligned", but that is no proof). Jul 29, 2017 at 21:54

## 1 Answer

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

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 $\lambda=\frac{\sum_{i=1}^n\sig_i}{n}$.