Suppose $\mathcal{D}_N$ denote an $N\times N$ doubly stochastic matrix, given any element $M\in \mathcal{D}_N$ , the singular value decomposition for $M$ is $ M=USV'$
where $U$ and $V$ are two $N\times N$ orthogonal matrix and $S$ is a $N \times N$ diagonal matrix
Let $P$ be the 'closest' orthogonal matrix to $M$, i.e. $P=\arg\min_{X\in\mathcal{O}}||X-M||_F^2$,$\mathcal{O}$ represents the $N\times N$ orthogonal matrix set. Note such $P$ may be not unique. In this case, we choose any of it. On conclusion about $P$ is $P=UV'$, where $U$ and $V$ are defined before(although can be not unique, we just choose any of them)
$M_1 \in \mathcal{D}_N$, which is 'closest' to $P$. More specifically
$ M_1 = \arg\min_{X\in\mathcal{D}} ||X - P||_F ^2 $ Similarly, If $M_1$ is not unique, we choose any of it(This should not happen actually. Since we may image it as a 'ball' approaching a 'polygon', should have only one minimum)
My question is :
The statement: $M_1=M$ if and only if $M$ is a permutation matrix
Does this statement always hold true?
Actually, if $M$ is a permutation matrix, $M_1=M$, this is obvious, since $S=I$, and $P=M$. However, does another direction always hold true? If so, how to prove this, otherwise, how to give a counter-example?
Thanks for any suggestions!