Let $\mathbb{S}\_n$ denote the set of $n \times n$ symmetric positive semidefinite matrices. I am trying to figure out whether $k: \mathbb{S}\_n \times \mathbb{S}\_n \to \mathbb{R}\_+$ defined as:


$$k(A, B) = \text{Tr}[(A^{1/2} B A^{1/2})^{1/2}]$$


is a positive definite kernel. Can anyone find a counterexample showing it is not? Or prove it is indeed a positive definite kernel?


**Some analysis so far:** Having failed to find a counterexample numerically, I am trying to show:
$$\sum\_{ij} k(A\_i, A\_j) x\_i x\_j \geq 0$$
holds for any $A\_1, \dots, A\_m \in \mathbb{S}\_n$ and $x\_1, \dots, x\_m \in \mathbb{R}$.
Without loss of generality we can assume that $x\_1, \dots, x\_m \in \{-1, +1\}$ since the change of $x\_i \mapsto \text{sign}(x\_i)$ and $A\_i \mapsto x\_i^2 A\_i$ preserves the value on the left hand side above and all matrices remain positive semidefinite after this change of variables.


Let $\mathcal{P}$ denote the set of indices where $x\_i x\_j = +1$ and let $\mathcal{N}$ denote the set of indices where $x\_i x\_j = -1$.
Then we just need to show the following inequality holds:
$$\sum\_{i=1}^m \text{Tr}[A\_i] + 2 \sum\_{(i,j)\in\mathcal{P}} k(A\_i, A\_j) \geq 2 \sum\_{(i,j)\in\mathcal{N}} k(A\_i, A\_j)$$
This is where I'm stuck, or maybe a different approach is needed.



Is $k(A, B) = \text{Tr}[(A^{1/2} B A^{1/2})^{1/2}]$ a positive definite kernel?
Counterexample for $n = 2$ :


Let $A\_k$ be the orthonormal projection on the span of $$(\cos(2 \pi (k-1) / 5), \sin(2 \pi (k-1) / 5))^\mathsf{T} , \quad k = 1...5.$$


Then $k(A\_k,A\_l) = \vert \cos(2 \pi (k-l) / 5) \vert$ .


The corresponding matrix has the eigenvalue $-0.11803398874989484820458683436563811772$ with multiplicity $2$ .


