Let $\lVert A\rVert = \left(\sum_{i,j=1}^n \left\| a_{ij} \right\|^2\right)^{\frac{1}{2}} = \sqrt{\operatorname{tr}(A^\top A)}$ be the Frobenius norm on $n \times n$ matrices.

Fix $A \in GL_n(\mathbb{R})$.

1) Is there a formula for the $dist(A,O(n))$?

where $dist^2(A,O(n)) =\underset{X \in O(n)}{\text{min}} \|A - X\|^2$

(O(n) is the orthogonal group , i.e matrices satisfying $X^TX=I_d$, The minimum exists since $O(n)$ is compact)

Note: In general we don't always have a unique minimizer (i.e there can be more than one orthogonal matrice which is closest to $A$ in $o(n)$), at least if we consider non-invertible matrices $A$.

For example, $A=0$ is in the same distance from each element of $o(n)$ since the Frobenius norm of any isometry is $\sqrt n$.

Question: Can we prove the minimizer is unique (if $A \in GL_n(\mathbb{R})$)? If so, is the function: $GL_n(\mathbb{R}) \to \mathbb{R}^{n^2} \, , \, A \to X(A)$ where $X(A)$ is the minimizer, smooth? Can we provide an explicit formula for it?

I have tried using Lagrange's multipliers, but so far with no success. (I couldn't determine if the gradients of all the $n^2$ constraints are always linearly independent.

Remark: \begin{align} \|A - X\|^2 &= \mathrm{tr} \left( (A-X)^t (A-X) \right) \\\ &= \mathrm{tr} \left( (A^t -X^t) (A-X) \right) \\\ &= \mathrm{tr} (A^tA -A^tX - X^tA + X^tX) \\\ &= \mathrm{tr} (A^tA) - \mathrm{tr}(A^tX) - \mathrm{tr}((A^tX)^t) + \mathrm{tr} (I_d)\\\ &= n + \mathrm{tr} (A^tA) - 2\mathrm{tr}(A^tX) \end{align}

so minimizing $\|A - X\|$ is equivalent to maximizing $\mathrm{tr}(A^tX)$. (In particular, the objective function to optimize is linear and not quadratic)

  • $\begingroup$ Is there any special properties of- $\mathcal{GL}_n(\mathbb{R})$? $\endgroup$ – Rajat Oct 13 '15 at 10:16
  • $\begingroup$ @Rajada : You mean why I want the given input matrix to be invertible? Actually I am not sure this is essential. One could think of the distance from any matrix. However, I am indeed more interested in finding information about the distance of linear automorphisms of $\mathbb{R}^n$, and care less about non-invertible transformations. $\endgroup$ – Asaf Shachar Oct 13 '15 at 12:17
  • 1
    $\begingroup$ @Rajada: I found one possible reason for focusing upon invertible matrices only - this is because I hope for a unique minimizer. I am interested in properties of the minimizer function (like is it smooth?). I have edited the question to stress these points. $\endgroup$ – Asaf Shachar Oct 13 '15 at 12:33
  • 1
    $\begingroup$ Note: the answer coincides with the orthogonal matrix in (either) polar decomposition of $A$. $\endgroup$ – Ben Grossmann Dec 11 '16 at 17:09

Let us prove Omnomnomnom's suspicion. Fix $A \in M_n(\mathbb{R})$. Following the OP's observation, we want to maximize the linear functional $\varphi \colon M_n(\mathbb{R}) \rightarrow \mathbb{R}$ given by $\varphi(X) = \mathrm{tr}(A^tX)$ subject to the constraint $X^t X = I$. First, let us assume that $A$ is a diagonal matrix with non-negative entries on the diagonal. Then, given an orthogonal $X$ we have

$$ \varphi(X) = \mathrm{tr}(A^tX) = \sum_{i = 1}^n a_{ii} x_{ii} \leq \sum_{i=1}^n a_{ii} = \mathrm{tr}(A^t) = \varphi(I) $$

(because $-1 \leq x_{ij} \leq 1$ for all $1 \leq i, j \leq n$). Thus, the minimizer in this case is just $X = I$. Note that if $a_{ii} > 0$ for all $1 \leq i \leq n$ then the proof shows that $X = I$ is the unique maximizer and if some $a_{ii} = 0$ then clearly there can be infinitely many maximizers.

Now, given an arbitrary $A$, use SVD decomposition to write $A = U \Sigma V^t$. Then, $A^t = V \Sigma U^t$ and given an orthogonal matrix $X$ we have

$$ \varphi(X) = \mathrm{tr}(V \Sigma U^t X) = \mathrm{tr}(V^t (V \Sigma U^t X) V) = \mathrm{tr}(\Sigma U^t X V) = \mathrm{tr}(\Sigma W(X)) $$

where $W(X) = U^t X V$ is also an orthogonal matrix. When $X$ runs over all orthogonal matrices, so does $W(X)$ and so we have reduced the problem to the previous case and $X$ such that $W(X) = U^t X V = I$ is a maximizer. Solving for $X$, we see that $X = UV^t$. If $A$ is invertible, the matrix $\Sigma$ has strictly positive entries on the diagonal and so by the previous case the maximizer is in fact unique. This does look surprising and this implies that while the SVD decomposition is highly non-unique in general, the product $UV^t$ should be unique, at least when $A$ is invertible. It would be interesting to check this directly.

Regarding your question about a formula, you can also decompose $A$ using polar decomposition as $A = OP$ where $O$ is orthogonal and $P$ is non-negative. By orthogonally diagonalizing $P$ with $P = UDU^t$, you obtain an $SVD$ decomposition $A = OUDU^t$ for $A$ showing that $O$ is a minimizer for the distance. If $A$ is invertible, then $O$ is determined uniquely and is given by $O = A(\sqrt{A^t A})^{-1}$ where $\sqrt{A^tA}$ is the unique positive root of $A^t A$. However, as $\sqrt{A^t A}$ doesn't really have an "explicit" formula, I'm not sure how much this helps.

  • 1
    $\begingroup$ Excellent job here $\endgroup$ – Ben Grossmann Oct 13 '15 at 17:10
  • $\begingroup$ I accepted levap's answer, although clearly there is a joint contribution to the solution from Omnomnomnom. However, I can accept only one answer, and this one actually contains a complete proof. $\endgroup$ – Asaf Shachar Oct 15 '15 at 17:19

I have a guess as to the minimizer. Perhaps there's a neat proof that this happens to be the right answer.

Note that $A$ has a singular value decomposition $$ A = U\Sigma V^T $$ where $\Sigma$ is in this case a square, diagonal matrix whose diagonal elements are the (strictly positive) singular values of $A$, and $U,V$ are orthogonal matrices.

My suspicion is that $$ dist(\Sigma,O(n)) = \|\Sigma - I\| $$ where $I$ is the identity matrix. By the orthogonal invariance of the Frobenius norm, we can equivalently state that $$ dist(A,O(n)) = \|A - UV^T\| = \sqrt{\sum_{j=1}^n [s_j(A)-1]^2} $$ If $UV^T$ does turn out to be a minimizer in general, there is no reason (at first glance) to believe that this minimizer is unique.

There is a condition that would guarantee a unique minimizer that I suspect is true. In particular, I think that $SO(n)$ is the boundary of its own convex hull. If that is the case, then the minimizer of the distance from any $X$ outside of this convex hull is unique.

Note that uniqueness does not hold for elements on the interior of the convex hull. In particular, every element of $O(n)$ is at the minimum distance from $X = 0$. I suspect it is this condition (rather than invertibility) that "causes" uniqueness to fail.

Edit: see levap's answer, which looks correct to me. It seems that a unique minimizer indeed exists if and only if $A$ is invertible.

  • $\begingroup$ Could you please tell me what is $s_j(A)$ and how have you got the formula after getting the minimizer, i.e., $ \|A - UV^T\| = \sqrt{\sum_{j=1}^n [s_j(A)-1]^2}$? $\endgroup$ – XYZABC Sep 17 '19 at 5:43
  • $\begingroup$ @XYZABC $s_j(A)$ is the $j$th singular value of $A$. Do you understand what I mean when I say in my answer "the orthogonal invariance of the Frobenius norm"? $\endgroup$ – Ben Grossmann Sep 17 '19 at 7:15
  • $\begingroup$ The orthogonal invariance of the Frobenius norm is it $\|AB\|_F=\|A\|_F$, where $\|A\|_F$ is the Frobenius norm of $A$? $\endgroup$ – XYZABC Sep 17 '19 at 8:29
  • $\begingroup$ @XYZABC More specifically, when $B$ is an orthogonal matrix we have $$ \|AB\|_F = \|BA\|_F = \|A\|_F $$ $\endgroup$ – Ben Grossmann Sep 17 '19 at 8:56
  • $\begingroup$ Sorry, but still I am not getting why does this square root coming into the picture and what to do with the invariance? $\endgroup$ – XYZABC Sep 17 '19 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.