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? – user1551 Jun 22 '16 at 19:15
• @gerw I modify it. – sleeve chen Jun 22 '16 at 19:21