Does there exist $p>1$ such that for all $n\geq 2$, if $(a\_{ij})$ and $(b\_{ij})$ are symmetric positive semidefinite $n\times n$ matrices and $a\_{ij}, b\_{ij}\geq 0$ then $\bigl(\|(a\_{ij},b\_{ij})\|\_p\bigr)=\bigl((a\_{ij}^p+b\_{ij}^p)^{1/p}\bigr)$ is also positive semidefinite? 


Maybe, a simpler question: is it true for $p=2$?


**Edited:** Original question did not have the condition $a\_{ij}, b\_{ij}\geq 0$. If we take $b\_{ij}=0$, it is possible that $(|a\_{ij}|)$ is not positive semidefinite when $(a\_{ij})$ is.



Hadamard $\ell_p$ sum of two symmetric positive semidefinite matrices
Looks like there any many counterexamples already when $n=3$. 


Take $u=(0,1,1)$, and $v = (1,2,0)$. Consider rank-1 matrices $A=u^{T}u$ and $B=v^{T}v$. Then the $\ell^{p}$ Hadamard matrix is 


$$
\begin{pmatrix}
1 & 2 & 0\\
2 & (1+2^{2p})^{1/p} & 1\\
0 & 1 & 1
\end{pmatrix}
$$
 whose determinant is $(1+2^{2p})^{1/p} - (1+2^{2}) <0$ for all $p \in (1,\infty)$. The last inequality follows from the superadditivity of the map $x \mapsto x^{p}$ on $[0, \infty)$ for $p\geq 1$. 


