Let $A(\theta):\mathbb{R}\rightarrow \mathbb{R}^{n\times n}$ be a analytic matrix-valued function such that
$A(\theta)=A^T(\theta)\geq 0,\quad \forall \theta\in\mathbb{R}$
For each fixed $x\in\mathbb{R}^n$, $x\neq 0$, $x^TA(\theta)x\not\equiv 0$
What can I say about the set where $A(\theta)>0$? Can the set where $A(\theta)$ is singular have an accumulation point?
Since $A(\theta)$ is symmetric, it has orthonormal eigenvectors. Since for each fixed $x \neq 0$, $x^T A(\theta) x \not\equiv 0$ we can assume that eigenvectors are also non-constant analytic. One example for $A(\theta)$ can be $$ A(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} f(\theta) & 0 \\ 0 & g(\theta) \end{bmatrix} \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} $$ where $f,g \geq 0$ are analytic.
Ultimately, it boils down to the zeros of $f$ and $g$. The zeros of an analytic function is either the entire real line (when the function is identically zero) or it consist of isolated points. So, the set where $A(\theta)$ is singular is either the entire real line or consist of isolated points. Furthermore, these isolated points can be selected arbitrarily by proper selection of $f$ and $g$.