Convex hull of orthogonal matrices

Question

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...

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. — Julien, Jul 03 '13 at 00:06
Kyle A. Gallivan, P.-A. Absil, Note on the convex hull of the Stiefel manifold, July 19, 2010. — Rodrigo de Azevedo, Mar 31 '23 at 17:24

score 16 · Accepted Answer · answered Jul 03 '13 at 00:03

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.

score 1 · Answer 2 · answered Mar 31 '23 at 13:29

I want to slightly extend the answer of @user1551 to the unitary case because --- although the proof idea stays the same --- in the complex case every contraction is the arithmetic mean of just two unitaries (cf. also page 42 of Bhatia's book "Positive Definite Matrices", 2007).

Proposition. Let $n\in\mathbb N$ and $A\in\mathbb C^{n\times n}$ be given. If $\|A\|_{2\to 2}\leq 1$, then there exist $U_1,U_2\in\mathbb C^{n\times n}$ unitary such that $A=\frac12(U_1+U_2)$.

Here $\|\cdot\|_{2\to 2}$ denotes, as before, the operator norm with respect to the $2$-norm on $\mathbb C^n$, i.e. the largest singular value of the input.

Proof. As before the easiest way to show this is to use the singular value decomposition $A=U{\rm diag}(s_1,\ldots,s_n)V$ where $U,V$ are unitary and $s_1,\ldots,s_n\in[0,1]$. With this we for each $j=1,\ldots,n$ define $$z_j^+:=s_j+i\sqrt{1-s_j^2}\qquad \quad z_j^-:=s_j-i\sqrt{1-s_j^2}\,, $$ as well as \begin{align*} U_1&:=U{\rm diag}(z_1^+,\ldots,z_n^+)V\\ U_2&:=U{\rm diag}(z_1^-,\ldots,z_n^-)V\,. \end{align*} Obviously $|z_j^\pm|=1$ so $U_1,U_2$ are unitary as product of three unitaries, and $U_1+U_2=U{\rm diag}(2s_1,\ldots,2s_n)V=2A$.$\quad\square$

Convex hull of orthogonal matrices

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