Where can I find the proof of the fact that the convex hull of the set of orthogonal matrices is the set of matrices with norm not greater than one?

It is easy to show that a convex combination of orthogonal matrices has norm (I mean the norm as operators) not larger than $1$. The reverse seems quite tricky...


Solution 1:

Presumably the norm in question is the operator $2$-norm, i.e. the largest singular value. Now suppose $\|A\|\le1$. Consider the singular value decomposition $A=U\Sigma V^T$, where $U,V$ are real orthogonal and $\Sigma=\operatorname{diag}(\mathbf{s})$ for some $\mathbf{s}\in[0,1]^n$ is a diagonal matrix with entries between $0$ and $1$. Since products of orthogonal matrices are orthogonal, it suffices to show that $\Sigma$ is a convex combination of orthogonal matrices, but clearly, $\mathbf{s}$ is a convex combination of the extreme points $\mathbf{s}_1,\mathbf{s}_2,\ldots,\mathbf{s}_{2^n}$ of $\{-1,1\}^n$. Therefore $\Sigma$ is a convex combination of $\operatorname{diag}(\mathbf{s}_1),\,\operatorname{diag}(\mathbf{s}_2),\,\ldots,\,\operatorname{diag}(\mathbf{s}_{2^n})$, which are real orthogonal.