Polar set of orthogonal matrices set is nuclear norm ball

The set of orthogonal matrices is defined as:

$$\mathcal{O}(n) = \{X\in \mathbb{R}^{n\times n}:X^TX=I\}$$

The polar of $\mathcal{O}(n)$:

$$\mathcal{O}(n)^o = \{Y\in \mathbb{R}^{n\times n}:\langle Y,X\rangle\leq 1, \forall X\in \mathcal{O}(n)\}$$

i.e., the set of linear functionals that take value at most one on $\mathcal{O}(n)$. The definition of polar cone is the general definition.

How to prove the polar of $\mathcal{O}(n)$ is the nuclear norm ball?
i.e. $$\mathcal{O}(n)^o = \{Y\in \mathbb{R}^{n\times n}:\sum_{i=1}^n\sigma_i(Y)\leq 1\}$$

I try to consider how to go to $\sum\sigma_i\leq1$ from $\langle Y,X\rangle=\text{tr}(YX)\leq 1$; however, I cannot find a way to break through.

By user1551's suggestion:

Let $YX=U\Sigma V^T$, where $Y\in \mathcal{O}(n)^o, X\in \mathcal{O}(n)$.

By Convex hull of orthogonal matrices, can I say:

$\|YX\|_2^2=\langle Y,X \rangle \leq 1$ so I get $\Sigma$'s diagonal elements $\in [0,1]^n$. If this is true, how to say the SVD of $Y$, particularly $\Sigma(Y)$?

Note: $\|\cdot\|_2$ is the spectral norm (the largest singular value).

• You should note that this is note the polar cone, but the polar set. The polar cone is defined with $\le 0$ (instead of $\le 1$) and, as the name suggests, it is a cone.
– gerw
Jun 22 '16 at 19:11
• Does the boxed statement follow directly from singular value decomposition? Jun 22 '16 at 19:15
• @gerw I modify it. Jun 22 '16 at 19:21
• Do you agree with my edits? Mar 15 '21 at 21:10

I will call $$\mathcal{A} = \{Y\in \mathbb{R}^{n\times n}:\langle Y,X\rangle\leq 1, \forall X\in \mathcal{O}(n)\}$$ and $$\mathcal{B} = \{Y\in \mathbb{R}^{n\times n}:\sum_{i=1}^n\sigma_i(Y)\leq 1\}$$, denote the nuclear norm by $$\|\cdot\|_1$$ and the operator norm by $$\|X\|_{\infty}$$ (the indexing comes from viewing them as special cases of the Schatten norms).

Part 1: $$\mathcal{B} \subseteq \mathcal{A}$$

This direction follows from the Hölder's inequality. Let $$Y \in \mathcal{B}$$ and $$X \in \mathcal{O}(n)$$ then by Hölder's inequality we have \begin{aligned} \langle Y, X \rangle &\leq \|Y\|_1 \|X\|_{\infty}\\ &= \|Y\|_1\\ &\leq 1. \end{aligned} Thus $$Y \in \mathcal{A}$$ and $$\mathcal{B} \subseteq \mathcal{A}$$.

Part 2: $$\mathcal{A} \subseteq \mathcal{B}$$

This direction is a little more tricky. To solve it we need the following lemma which gives a variational characterization of the nuclear norm.

Lemma: Let $$Y \in \mathbb{R}^{n\times n}$$ then $$\|Y\|_1 = \sum_{i=1}^n\sigma_i(Y) = \max_{X \in \mathcal{O}(n)} \langle Y,X\rangle.U$$ Proof$$\quad$$ Consider the singular value decomposition $$Y = UDV$$ where $$U,V \in \mathcal{O}(n)$$ and $$D$$ is the diagonal matrix containing the singular values of $$Y$$. Then let $$X = UV$$ and we see \begin{aligned} \langle Y, X\rangle &= \mathrm{Tr}[Y^T X] \\ &= \mathrm{Tr}[V^T D U^T UV] \\ &= \mathrm{Tr}[VV^T D U^T U] \\ &= \mathrm{Tr}[D] \\ &= \sum_{i=1}^n \sigma_i(Y). \end{aligned} Thus $$\sum_i \sigma_i(Y) \leq \max_{X \in \mathcal{O}(n)} \langle Y,X\rangle$$. The other inequality follows directly from the Hölder's inequality used in Part 1, i.e. $$\langle Y, X\rangle \leq \|Y\|_1$$ for any $$X \in \mathcal{O}(n)$$. $$\tag*{\blacksquare}$$

Now the result follows fairly quickly. Given $$Y \in \mathcal{A}$$ we know $$\langle Y , X \rangle \leq 1$$ for all $$X \in \mathcal{O}(n)$$. However, from the above lemma we also know that \begin{aligned} \sum_{i=1}^n \sigma_i(Y) &= \max_{X \in \mathcal{O}(n)} \langle Y,X\rangle \\ &\leq 1 \end{aligned} Thus $$Y \in \mathcal{B}$$ and $$\mathcal{A} \subseteq \mathcal{B}$$ and we are done!