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:
-
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. -
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:
- How to show the set of rotation matrices is compact? (convex hull of compact set is compact.)
- Why it contains the origin? ($0_{n\times n}$?) I believe it is not.
Best Answer
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.