3
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

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 $p\geq 1$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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 $a_{ii}=b_{ii}=1$? $\endgroup$
    – D_809
    Commented Jul 12, 2019 at 22:41
  • $\begingroup$ 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. $\endgroup$
    – Paata Ivanishvili
    Commented Jul 13, 2019 at 5:48
  • $\begingroup$ @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. $\endgroup$
    – Paata Ivanishvili
    Commented Jul 14, 2019 at 6:04
  • $\begingroup$ 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? $\endgroup$
    – D_809
    Commented Jul 14, 2019 at 10:57
  • $\begingroup$ @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. $\endgroup$
    – Wolfgang
    Commented Jul 14, 2019 at 11:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .