# Mapping a subset of semi-definite matrices through arcsinus

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?

• That should be $$C_{ij} = \sin(a \arcsin(A_{ij}) + (1-a) \arcsin(B_{ij})$$ – Robert Israel Apr 18 '12 at 18:40

$$A = \pmatrix{1&0&-1&0\cr 0&1&0&-1\cr -1&0&1&0\cr0&-1&0&1\cr},\ B=\pmatrix{ 1&\sqrt{3}/2 & 1/2&0\cr \sqrt{3}/2&1& \sqrt{3}/2& 1/2\cr 1/2& \sqrt{3}/2&1& \sqrt{3}/2 \cr 0& 1/2& \sqrt{3}/2&1\cr}$$ 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.
• Thank you! Did you try with $3\times 3$ matrices? How did you find this example? – kaleidoscop Apr 19 '12 at 7:01
• I looked at Toeplitz matrices of the form $$M(s,t) = \pmatrix{1 & s & t & 0\cr s & 1 & s & t\cr t & s & 1 & s\cr 0 & t & s & 1\cr}$$ and plotted the region in the $(s,t)$ plane where $\Phi^{-1}(M)(s,t)$ is SDP. It turned out not to be convex. The $3 \times 3$ case was convex. – Robert Israel Apr 19 '12 at 16:23