# Hadamard $\ell_p$ sum of two symmetric positive semidefinite matrices

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.

## 1 Answer

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 $$p\geq 1$$.

• Thank you! I feel bad asking you this, since I should probably first play around with your example myself, but do you have an example when all $a_{ii}=b_{ii}=1$? – Jaime F. Jul 12 at 22:41
• In this case your question for $n=3, p=2$ reduces to the following one: let $0\leq a,b,c,x,y,z\leq 1$ be such that $1+2xyz\geq x^{2}+y^{2}+z^{2}$ and $1+2abc\geq a^{2}+b^{2}+c^{2}$. Then does it follow that $1+2\sqrt{\frac{a^{2}+x^{2}}{2}}\sqrt{\frac{b^{2}+y^{2}}{2}}\sqrt{\frac{c^{2}+z^{2}}{2}}\geq \frac{a^{2}+b^{2}+c^{2}+x^{2}+y^{2}+z^{2}}{2}$. I checked some boundary cases and looks like it is true. Also I can show that it is enough to consider the case $a\geq x, b\geq y, c\geq z$. I may think about it later, it is too late right now. – Paata Ivanishvili Jul 13 at 5:48
• @JaimeF. I am afraid there is no counterexample to your question (asked in the comment) when $p=2$ and $n=3$. If this is interesting to you I can write more details about this. The reason lies in the fact that the set $\{(a,b,c) \in [0,1]^{3} : 1+2\sqrt{abc}-a-b-c\geq 0\}$ is convex set. – Paata Ivanishvili Jul 14 at 6:04
• I'm very interested to know if this might be true for $p=2$ and all $n$. Do you have any intuition about this? Do you think it is worth asking a follow-up question with this extra constraint? – Jaime F. Jul 14 at 10:57
• @JaimeF. Yes, it is a good idea to ask that in a new question, as the current one has been answered and you have accepted the answer. – Wolfgang Jul 14 at 11:01