Hi


I am meeting a problem concerning semi-definite positive matrices, and I have no clue concerning them, the classical approaches I know have not given any result, maybe people used to manipulating them could help me...


Call $\mathscr{P}$ the convex set of symmetric SDP matrices $S$ of order $N\geq 1$ such that $|S\_{ij}|<1$ . Consider the mapping $$\Phi:S=(S\_{ij})\mapsto (\frac{2}{\pi}\arcsin(S\_{ij})).$$ 


I think it is a (well) known fact that $\Phi(\mathscr{P})\subset \mathscr{P}$. My question is the following: Is $\Phi(\mathscr{P})$ convex?


At first I thought that the answer was "no", but I checked that for $N=2$ the answer is "yes". I have no clue as how to test higher dimensions. Does anyone have a suggestion on a method?


Another possible formulation: Given two SDP matrices $A$ and $B$, and $a\in [0,1]$, is the matrix $C$ defined by $$C\_{ij}=\sin\left(a\arcsin(A\_{ij})+(1-a)\arcsin(B\_{ij})\right)$$ also SDP?



Mapping a subset of semi-definite matrices through arcsinus
Consider the SDP matrices


$$A = \pmatrix{1&0&-1&0\cr 0&1&0&-1\cr -1&0&1&0\cr0&-1&0&1\cr},\
B=\pmatrix{ 1&\sqrt{3}/2 & 1/2&0\cr \sqrt{3}/2&1& \sqrt{3}/2& 1/2\cr 1/2& \sqrt{3}/2&1& \sqrt{3}/2
\cr 0& 1/2& \sqrt{3}/2&1\cr}$$
where with $a=1/2$
$$ C = \Phi^{-1}(\Phi(A)/2 + \Phi(B)/2) = \pmatrix{ 1&1/2&-1/2&0\cr1/2&1&1/2&-1/2\cr-1/2&1/2&1&1/2\cr0&-1/2&1
/2&1\cr}$$
is not positive semidefinite.


