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I am reading the paper Semidefinite descriptions of the convex hull of rotation matrices by James Saunderson, Pablo A. Parrilo and Alan S. Willsky. On page 2, it says:

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  1. I "guess" the set of rotation matrices is closed. The intuition tells me that it is

    $$\{X \mid X^TX=I_n\} \cap\{X\mid \text{det}(X) = 1\}$$
    since both sets are closed, the intersection of them is closed.

  2. However, to prove conv $SO(n)$ is closed, from the following:
    Is the convex hull of closed set in $R^{n}$ is closed?
    there is no guarantee that the convex hull of a closed set is closed.

My question is:

  1. How to show the set of rotation matrices is compact? (convex hull of compact set is compact.)
  2. Why it contains the origin? ($0_{n\times n}$?) I believe it is not.
Rodrigo de Azevedo
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sleeve chen
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  • One might guess they lie on a sphere and if $M$ is a rotation matrix then so is $-M$. Not sure about the first guess, but surely the second is correct--that would give you the origin in the hull. – MPW Sep 02 '16 at 02:22
  • @MPW But by the definition of convex hull, the coefficients sum up to one, you cannot let them be all zero to obtain $0_{n\times n}$. That is my reason – sleeve chen Sep 02 '16 at 03:59
  • Not all zero -- half to $M$, half to $-M$, zero everywhere else, sum is zero. – MPW Sep 02 '16 at 04:38
  • @MPW Maybe not since $-M$ may not be a rotation matrix. This is because $\text{det } (M) $ will be $-1$, which is required by being a rotation matrix, if it is a $3\times 3$ matrix or odd dimension. – sleeve chen Sep 02 '16 at 04:53

1 Answers1

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You're right, $SO(n)$ is closed for the reason you give. To show that it is compact, just note that, as a subset of $\mathbb{R}^{n^2}$, it is bounded (every entry in an orthogonal matrix is less than or equal to $1$ in absolute value.)

As for the convex hull, the result is true for all $n > 1$. Since $SO(1) = \{1\}$ is a single point, it coincides with its convex hull; clearly it does not contain $0$. If $n$ is even, then $-I \in SO(n)$. Since $0 = (1/2)I + (1/2)(-I$, we can express the zero matrix as a convex combination of two elements of $SO(n)$. It follows that $0$ lies in the convex hull of $SO(n)$.

Finally, suppose that $n$ is an odd integer greater than $1$. As noted in the comments, $SO(n)$ contains the matrices $M_1, \ldots, M_n$, where $M_i$ is the diagonal matrix whose diagonal entries are given by: $1$ in the $(i,i)$ entry and $-1$ otherwise. Since ${\rm det}(M_i) = 1(-1)^{n - 1} = 1$, and $M_i^T M_i = I$, each $M_i \in SO(n)$. Using these matrices, we can express the zero matrix as a convex combination: $0 = \Big(\sum_{i = 1}^n \dfrac{1}{2n-2}M_i \Big) + \dfrac{n-2}{2n-2}I$. The result follows.

Note: Thanks to Rahul for correcting my erroneous argument for $n$ odd.

PeterJL
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    For odd $n$, consider the diagonal matrices $M_i$ with $i$th diagonal entry equal to $1$ and all other diagonal entries equal to $-1$. Then $\sum_{i=1}^n M_i + (n-2)I = 0$. –  Sep 02 '16 at 23:38
  • I find the other way to prove boundedness is by its eigenvalue. $SO(n)$'eigenvalue is bounded since: http://mathworld.wolfram.com/RotationMatrix.html . Your answer gives another nice way to think about it. – sleeve chen Sep 03 '16 at 06:48