# Sufficient and necessary conditions for $y_1 = |x_1|^2$, $y_2 = x_1 \overline{x_2}$, $y_3 = \overline{x_1} x_2$, $y_4 = |x_2|^2$ in terms of $y$'s

Let $$\mathbb C$$ and $$\mathbb R$$ denote the fields of complex and real numbers, respectively. Suppose $$x_1, x_2 \in \mathbb C$$, and \begin{align} y_1 & = |x_1|^2 \tag 1 \\ y_2 & = x_1 \overline{x_2} \tag 2 \\ y_3 & = \overline{x_1} x_2 \tag 3 \\ y_4 & = |x_2|^2 \tag 4 \end{align} where $$\overline{x}$$ denotes the complex conjugate of $$x$$. Are there any sufficient and necessary conditions, in terms of $$y_1$$, $$y_2$$, $$y_3$$ and $$y_4$$, for all the above conditions to hold?

The following are among the necessary conditions. \begin{align} y_1, y_4 & \in \mathbb R_+ \tag 5 \\ y_2 & = \overline{y_3} \tag 6 \\ y_1 y_4 & = y_2 y_3 \tag 7 \end{align} where $$\mathbb R_+$$ denotes the nonnegative subset of $$\mathbb R$$. I think something is still missing, although I could not find a counterexample.

## 1 Answer

Your conditions are sufficient with the condition (5) replaced by $$y_1,y_4\in\mathbb{R}_{\geq 0}$$ (I just noticed that you had fixed this condition). To show that your conditions are also sufficient, we first deal with the trivial cases $$y_1=0$$ or $$y_4=0$$. Without loss of generality, let $$y_1=0$$. Then $$y_2=\bar{y}_3\text{ and }|y_2|^2=|y_3|^2=y_2y_3=y_1y_4=0$$ imply that $$y_2=y_3=0$$. Hence, we may take $$x_1:=0$$ and $$x_2:=\sqrt{y_4}$$. All possible solutions $$(x_1,x_2)\in\mathbb{C}\times\mathbb{C}$$ in this case take the form $$(x_1,x_2)=\big(0,\sqrt{y_4}\,\exp(\text{i}\theta)\big)\,,$$ where $$\theta\in[0,2\pi)$$.

We now assume that $$y_1>0$$ and $$y_4>0$$. Take $$x_1:=\sqrt{y_1}$$ and $$x_2:=\dfrac{y_3}{\sqrt{y_1}}$$. Ergo, (1) and (3) follow immediately. To show (2) we have $$x_1\bar{x}_2=\sqrt{y_1}\left(\frac{\bar{y}_3}{\sqrt{y_1}}\right)=\bar{y}_3=y_2\,.$$ Additionally, $$y_1y_4=y_2y_3=|y_3|^2$$ implies that $$y_4=\frac{|y_3|^2}{y_1}=\left(\frac{\bar{y}_3}{\sqrt{y_1}}\right)\left(\frac{y_3}{\sqrt{y_1}}\right)=\bar{x}_2x_2=|x_2|^2\,.$$ In fact, all solutions $$(x_1,x_2)\in\mathbb{C}\times\mathbb{C}$$ are of the form $$\left(x_1,x_2\right)=\left(\sqrt{y_1}\,\exp(\text{i}\theta),\frac{y_3}{\sqrt{y_1}}\,\exp(\text{i}\theta)\right)$$ for some $$\theta\in[0,2\pi)$$.

• Thanks very much for your efforts! The answer is very insightful. – B. Groeger Jan 7 at 22:10