Convex hull of rotation matrices is closed and contains the origin 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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*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:   

  
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*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.
  

 A: 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.
