How does the phase of the determinant of a complex matrix relate to the phase of the original matrix entries?

Question

Suppose you have a complex-valued rank-2 tensor of the form:

$$\mathbf{A} = \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$

where $a_{ij} \in \mathbb{C}$ are complex numbers. If the phase ($\phi_{ij}$) of $a_{ij}$ is bounded (for example, bounded on $0 < \phi_{ij} < \pi/2 $), then will the phase of $\mathrm{det}(\mathbf{A})$ also be bounded?

I define the determinant as

$$\mathrm{det}(\mathbf{A}) = a_{11}a_{22}-a_{21}a_{12}$$

For complex values, I think this can be simplified to:

$$\mathrm{det}(\mathbf{A}) = |a_{11}||a_{22}|e^{i(\phi_{11}+\phi_{22})}-|a_{21}||a_{12}|e^{i(\phi_{21}+\phi_{12})}$$

Can this be simplified into some simpler form (e.g. $re^{i\alpha})$?

In a more general sense, if we know something about the phases of the matrix entries (e.g. bounds, quadrant, etc.), can we automatically deduce similar facts about the phase of the determinant?

Answer 1

In general, there is very little you can expect here. Your simplification tells you that you are interested in complex numbers of the form $r e^{i\alpha} + r' e^{i(\alpha'+\pi)}$ for some given angles $\alpha,\alpha'$ and arbitrary $r,r'\in\mathbb{R}_{>0}$. That means you are looking at a sector in the complex plane (=convex cone spanned by $e^{i\alpha}$ and $e^{i(\alpha'+\pi)}$). The angles $\alpha,\alpha'$ are arbitrary in $A+A$ where $A$ is the set of phases your matrix entries are allowed to have.

The $+\pi$ part is of course the problem here. Imagine for example that $A=(0,\epsilon)$. Then $A+A=(0,2\epsilon)$. The sectors spanned by all the corresponding rays already cover the complex plane so that no non-trivial restriction for the determinant can be proven.