I am trying to minimize the below function, can it be considered quasiconvex and/or convex?
$$F(\mathbf{x})= \frac{|\mathbf{a}^T\mathbf{\Sigma}\mathbf{x}|}{\mathbf{x}^T\mathbf{\Sigma}\mathbf{x}}$$
where $\mathbf{a},\, \mathbf{x}$ are real-valued vectors and $\mathbf{\Sigma}$ is a positive-definite matrix.
I know that the numerator is convex (absolute value of affine) and the denominator is convex and positive.
On
No, this function is not quasiconvex; as I pointed out last time you asked the question, the domain is not convex. This prevents sublevel sets being convex, and so the function cannot be quasiconvex.
As you are trying to minimise the function, the unfortunate issue is that the function has no minimum. Consider the sequence of vectors: $$\mathbf{x}_n = -\frac{1}{n}\mathbf{a}.$$ Then $$F(\mathbf{x}_n) = \frac{-\frac{1}{n}\mathbf{a}^\top \mathbf{\Sigma}\mathbf{a}}{\frac{1}{n^2}\mathbf{a}^\top \mathbf{\Sigma}\mathbf{a}} = -n \to -\infty.$$ Thus, the function is unbounded below, and no minimum, or even finite infimum, exists.
This ratio is in general not (quasi)convex. Consider dimension $n = 1$ and $\Sigma = a = 1$. You end up with $$ F(x) = \frac{|x|}{x^2} = \frac{1}{|x|}. $$
This cannot be (quasi)convex, since its sublevel sets are not convex. For example: $$ F_{1} := \left\{x \mid F(x) \leq 1 \right\} = \{x \mid |x| \geq 1 \} = (-\infty, -1] \cup [1, \infty), $$
which is a disjoint union of intervals and as such nonconvex.