Let $V = S^{n-1}(\mathbb{C}^2)$ be the $n-1$ th symmetric tensor power of $\mathbb{C}^2$. Thus $V = \mathbb{C}^n$ as a complex vector space. Suppose there given a matrix $A$ of the form
$A = A_1 \odot \cdots \odot A_{n-1}$,
where each $A_i$ is an Hermitian $2$ by $2$ matrix which is positive semidefinite, and the symmetric tensor product $\odot$ is over $\mathbb{C}$.
Question: is $A$ also hermitian positive semidefinite?
As a remark, it is clear that for any
$v = S^{n-1}(\psi)$
for some $\psi \in \mathbb{C}^2$, we have
$v^* A v \geq 0$. But this is as far as I got.