# Is there a "formula" for the point in $\text{SO}(n)$ which is closest to a given matrix?

$$\newcommand{\Sig}{\Sigma}$$ $$\newcommand{\dist}{\operatorname{dist}}$$ $$\newcommand{\distSO}{\dist(#1,\SO)}$$ $$\newcommand{\distO}{\text{dist}(#1,\On)}$$ $$\newcommand{\tildistSO}{\operatorname{dist}_{\til d}(#1,\SO)}$$ $$\newcommand{\SOn}{\operatorname{SO}_n}$$ $$\newcommand{\On}{\operatorname{O}_n}$$ $$\newcommand{\GLp}{\operatorname{GL}_n^+}$$ $$\newcommand{\GLtwo}{\operatorname{GL}_2^+}$$ $$\newcommand{\GLm}{\operatorname{GL}_n^-}$$ $$\newcommand{\sig}{\sigma}$$ $$\newcommand{\diag}{\operatorname{diag}}$$

Let $$A$$ be a real $$n \times n$$ matrix, with negative determinant. Suppose that the singular values of $$A$$ are pairwise distinct. Then, it can be proved that there exist a unique special orthogonal matrix $$Q(A)$$ which is closest to $$A$$ (w.r.t the Frobenius distance).

I want to find out if there exist a "formula" for $$Q(A)$$, say in terms of positive roots, inverses, matrix multiplication etc. Is there any hope for such a thing?

By a formula, I do not really mean a "closed-form" formula. A closely related example of what I am looking for is the orthogonal polar factor of an invertible matrix:

If $$A=OP$$, where $$O \in \On$$, and $$P$$ is symmetric positive-definite, then $$P=\sqrt{A^TA}$$ (here $$\sqrt{}$$ is the unique symmetric positive-definite square root) and $$O=O(A)=AP^{-1}=A(\sqrt{A^TA})^{-1}$$. I consider this is an acceptable formula.

(Comment: The orthogonal factor $$O(A)$$ is the closest orthogonal matrix to $$A$$).

Edit 2:

After some more thinking, I think the idea of a "reasonable formula" might be a bit hopeless: If we had any such "reasonable" formula, we could probably extend it continuously to all of $$\GLm$$. However, such a continuous extension does not exist:

Set $$A_n=\begin{pmatrix} -1 & 0 \\\ 0 & 1+\frac{1}{n} \end{pmatrix},B_n=\begin{pmatrix} -(1+\frac{1}{n}) & 0 \\\ 0 & 1 \end{pmatrix}$$.

Then $$Q(A_n)=\text{Id},Q(B_n)=-\text{Id}$$, while $$A_n ,B_n$$ both converge to $$\begin{pmatrix} -1 & 0 \\\ 0 & 1 \end{pmatrix}$$.

This phenomena implies that perhaps the best we can do is to find "partial expressions", as in Dap's answer.

Here is what I know: Let $$A=U\Sig V^T$$ be the singular values decomposition of $$A$$; we can assume that $$\Sig = \diag\left( \sig_1,\dots\sig_n \right)$$ where $$\sigma_1$$ is the smallest singular value of $$A$$, and that $$U \in \SOn,V \in \On,\det V=-1$$.

Set $$\Sig':=\diag\left( -\sig_1,\dots\sig_n \right)$$, and rewrite $$A= U\Sig V^T = U (\Sig \diag\left( -1,1,1\dots ,1 \right)) (\diag\left( -1,1,1\dots ,1 \right) V ^T ) =U \Sig' \tilde V^T,$$ where $$\tilde V \in \SOn$$ is defined by requiring $$\diag\left( -1,1,1\dots ,1 \right) V ^T=\tilde V^T$$.

Then, it turns out that $$Q(A)=U\tilde V^T$$.

Specifically, we have $$\dist(A,\SOn)= \dist(U \Sig' \tilde V^T ,\SOn)= \dist( \Sig' ,\SOn)=d(\Sig' ,\text{Id})\\=(\sig_1+1)^2 + \sum_{i=2}^n \left( \sig_i-1 \right)^2,$$ and one can prove that $$\text{Id}$$ is the unique closest matrix in $$\SOn$$ to $$\Sig'$$. (It is important here that $$\sigma_1$$ is the smallest singular value of $$A$$).

Comment:

I prefer a formula which does not involve directly the singular vectors of $$A$$, since I want to understand how smoothly does $$Q(A)$$ varies with $$A$$. (The formula for the orthogonal factor mentioned above immediately implies that it is a smooth function of the matrix, once one knows that the positive square root is smooth). Finally, note that while $$Q(A)=U\tilde V^T$$, the orthogonal factor satisfies $$O(A)=UV^T$$).

Edit: I found out that the minimizer $$Q(A)$$ indeed changes smoothly; This follows from the fact that locally, we can choose the matrices $$U,V$$ smoothly. However, I think that an elegant formula would still be a nice thing to have. (Even though we do not need one to establish smoothness).

A more abstract discussion about smoothness of minimizers can be found here.

• the $Q(A)=UV^T$ formula is proven here: math.stackexchange.com/questions/2215359/… Mar 26 '19 at 9:09
• alternatively, the polar decomposition $A=OP$ with $O$ orthogonal and $P$ non-negative provides the minimiser $Q(A)=O=A(A^TA)^{-1/2}$; is that the type of "formula" you would want? Mar 26 '19 at 9:14
• Yes, exactly; but I am not really sure such a thing is possible in this "modified" problem. (when you pass from $O(A)=UV^T$ to $Q(A)=U\tilde V^T$). Mar 26 '19 at 9:22
• For $n=2$ the solution is $\begin{pmatrix} cos(\alpha) & -sin(\alpha)\\ sin(\alpha) & cos(\alpha) \end{pmatrix}$ with $\alpha=atan2(a_{21}-a_{12},a_{11}+a_{22})$ Mar 26 '19 at 12:58
• Mar 27 '19 at 6:07

Not quite an answer, but we can compute $$Q$$ using a single suitably chosen real parameter $$\lambda.$$ For any $$\lambda>0,$$ for all matrices $$A$$ with singular values $$\sigma_1\leq\sigma_2\leq\cdots\leq\sigma_n$$ satisfying $$0<\sigma_1<\lambda<\sigma_2,$$ $$Q(A)=O(A)O(A^TA-\lambda^2 I)$$
where $$I$$ is the identity matrix, and $$O(M)=M(\sqrt{M^TM})^{-1}$$ as defined in the question. To prove this, let $$A=U\Sigma V^T$$ as in the question. Then $$\Sigma^2-\lambda^2I=|\Sigma^2-\lambda^2I|\operatorname{diag}(-1,1,1,\dots,1)$$ where $$|\Sigma^2-\lambda^2I|$$ is a certain positive define diagonal matrix. This gives
$$O(A)O(A^TA-\lambda^2 I)=O(U\Sigma V^T)O(V|\Sigma^2-\lambda^2 I| \tilde{V}^T)=UV^TV\tilde{V}^T=U\tilde{V}^T.$$
If you were interested in integral formulas (still depending on $$\lambda$$): $$Q$$ of the form $$Q(A)=A \cdot f(A^TA)$$ where $$f$$ acts on the spectrum of $$A^TA$$ by taking $$z>0$$ to $$z^{-1/2}$$ for $$z>\lambda$$ and to $$-z^{-1/2}$$ for $$z<\lambda.$$ The holomorphic functional calculus then gives an integral formula for $$f,$$ locally in $$A.$$
Here are some relationships between the expressive power of various parameters. If you happened to know the smallest two singular values you could plug in $$\lambda=\sqrt{\sigma_1(A)\sigma_2(A)}.$$ This is the inverse square root of the singular norm of $$\bigwedge^2 A^{-1},$$ which can be defined via spectral radius using $$M^TM,$$ and spectral radius has a sort of formula, Gelfand's formula. On the other hand, for symmetric $$A$$ with $$\sigma_1(A)<\sigma_2(A),$$ the matrix $$\tfrac12(O(A)-Q(A))$$ is projection onto the right singular vector corresponding to $$\sigma_1(A).$$