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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...

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  • $\begingroup$ Maybe you would be interested in the generalization of this fact: the Russo-Dye theorem. This is proved in most books that talk about C*-algebras. Davidson's book, for example. You can also look at Popa's quantitative version of it, in free access. $\endgroup$ – Julien Jul 3 '13 at 0:06
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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.

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