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?


1 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.


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