Is a convex set of permutation matrices $n\times n$ ($\mathbb{P}_{n}\cap \mathbb{C}$) a singleton?

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$\mathbb{C}$ is a closed convex set.

In detail,

$\min_{X\in\mathbb{P}_{n}\cap\mathbb{C}} f(x):= x^{T}Wx + c^{T}x$

$x:=vec(X)\in \mathbb{R}^{n^2}$

What is wrong if I remove the $\cap\mathbb{C}$ as the below?

$\min_{X\in\mathbb{P}_{n}} f(x):= x^{T}Wx + c^{T}x$

Another statement is as the title,

If a subset of permutation matrices is convex set, is this subset Singleton, which has only one element in the set?

@Robert Israel Yeah, I also think it ends up a singleton, but it should not be like that..

@John Hughes, thanks for quick response! Have you looked at the cases n=1,2,3 and tried writing out all the permutation matrices?

Yes, n=1, 0, 1 leads 1/2 1/3 many arbitrary convex combination element if the set have 0 and 1. So, possible answer is {0}, {1}. They are singleton.

n=2, $P_{2}=\begin{pmatrix} 0 & 1\\ 1 & 0\\ \end{pmatrix} $ or $P_{2}=\begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix} $ The same analogy, {${P_{2}}$} with one element is only possible.

n=3, n=4... I think if there is any pairs which is not matched as (0,0) or (1,1), it causes the element which is out of Permutation matrices. I cannot get any valid subset of Permutation matrices which has over 1 element.

Have you looked at convex combinations of pairs of permutation matrices in these low-dimensional cases? Did it lead to any insight?

So the answer is the above.