Equivalence of Frobenius norm and trace norm According to [1], [2] and other related publications, the following holds for any matrix $X$:
$$\| X\|_\Sigma=\min_{X=UV'}\|U\|_\mathrm{Fro}\|V\|_\mathrm{Fro}=\min_{X=UV'}\frac{1}{2}(\|U\|_\mathrm{Fro}^{2}+\|V\|_\mathrm{Fro}^2)$$
where $\|\cdot\|_\Sigma$ is the trace (nuclear/Ky-Fan) norm and $\|\cdot\|_\mathrm{Fro}$ is the Frobenius norm.
Can anyone show why this equality is true?
In the publications, it is filed under "Preliminaries" and one of the few Lemmas without proof. I find this relationship very fundamental and interesting, but could not find it anywhere else, let alone proof.
Thank you for any help!
What I already have:
If I understand the 'hint' in ref. 1 (above) correctly, $\min_{X=UV'}\|U\|_\mathrm{Fro}\|V\|_\mathrm{Fro}$ is minimized by $U=\hat{U}\sqrt{\Lambda}$ and $V=\hat{V}\sqrt{\Lambda}$, where $X=\hat{U}\Lambda \hat{V'}$ is the singlular value decomposition of $X$ (Page 75, Lemma 8 in ref. 1 (above)).
 A: 2nd equality: the $\le$ is just AM-GM inequality.  To get $=$, replace $U$ by $\lambda U$ and $V$ by $\lambda^{-1}V$ so that the norms of $U$ and $V$ become the same.
1st equality: the hint tells you how to get $\ge$.  So suppose that $X = UV$.  Let the singular decomposition of $X = Q \Lambda P^*$, where $Q$ and $P$ are unitary, and $\Lambda$ is diagonal with non-negative entries.  Then $\Lambda = Q^* U V P$, and so $$\|X\|_\Sigma = \text{trace}(\Lambda) \le \|Q^* U\|_F \|V P\|_F = \|U\|_F \|V\|_F .$$
A: Let $U=\hat{U}\sqrt{\Lambda}$ and $V=\hat{V}\sqrt{\Lambda}$, where $X=\hat{U}\Lambda \hat{V'}$ is the singlular value decomposition of $X$.
1. Equation: $\| X\|_\Sigma=\min_{X=UV'}\|U\|_\mathrm{Fro}\|V\|_\mathrm{Fro}$
$\|U\|_\mathrm{Fro}\|V\|_\mathrm{Fro}=\sqrt{\mathrm{trace}((\hat{U}\sqrt{\Lambda})(\hat{U}\sqrt{\Lambda})^\intercal)\mathrm{trace}((\hat{V}\sqrt{\Lambda})(\hat{V}\sqrt{\Lambda})^\intercal)}=\sqrt{\mathrm{trace}(\hat{U}\sqrt{\Lambda}\sqrt{\Lambda}^\intercal\hat{U}^\intercal)\mathrm{trace}(\hat{V}\sqrt{\Lambda}\sqrt{\Lambda}^\intercal\hat{V}^\intercal)}=$
Using the facts that (i) $\Lambda$ is a diagonal matrix (ii) the trace of multiplied matrices does not change under cyclic rotations (iii) $\hat{U}$ and $\hat{V}$ are unitary: 
$=\sqrt{\mathrm{trace}(\hat{U}\Lambda\hat{U}^\intercal)\mathrm{trace}(\hat{V}\Lambda\hat{V}^\intercal)}=\sqrt{\mathrm{trace}(\hat{U}^\intercal\hat{U}\Lambda)\mathrm{trace}(\hat{V}^\intercal\hat{V}\Lambda)}=\sqrt{\mathrm{trace}(\Lambda)\mathrm{trace}(\Lambda)}=\mathrm{trace}(\Lambda)=\| X\|_\Sigma$
2. Equation: $\min_{X=UV'}\|U\|_\mathrm{Fro}\|V\|_\mathrm{Fro}=\min_{X=UV'}\frac{1}{2}(\|U\|_\mathrm{Fro}^{2}+\|V\|_\mathrm{Fro}^2)$
The AM GM inequality states that the geometric mean and the arithmetic mean are equal if the numbers being averaged are all the same. It is therefore enough to show that $\|U\|_\mathrm{Fro}^{2}=\|V\|_\mathrm{Fro}^2$.
As in 1.:
$\|U\|_\mathrm{Fro}^{2}=\|\hat{U}\sqrt{\Lambda}\|_\mathrm{Fro}^2=\mathrm{trace}((\hat{U}\sqrt{\Lambda})(\hat{U}\sqrt{\Lambda})^\intercal)=\mathrm{trace}(\Lambda)=\|V\|_\mathrm{Fro}^{2}$
The only thing missing: why does $U=\hat{U}\sqrt{\Lambda}$ and $V=\hat{V}\sqrt{\Lambda}$ minimize $\min_{X=UV'}\|U\|_\mathrm{Fro}\|V\|_\mathrm{Fro}$?
