Given an arbitrary matrix $A=\left[a_{i,j}\right]\in\mathbb{C}^{2\times2} $

Now define the set $$ \mathcal K = \left\lbrace z\in\mathbb C : \left|z-a_{1,1}\right|\cdot\left|z-a_{2,2}\right|=\left|a_{1,2}\right|\cdot\left|a_{2,1}\right| \right\rbrace$$

And the set of so-called equiradial matrices

$$\omega(A) = \left\lbrace B\in\mathbb C^{n\times n} : b_{1,1}=a_{1,1}, b_{2,2}=a_{2,2}, \left|b_{2,1}\right|=\left|a_{2,1}\right|, \left|b_{1,2}\right|=\left|a_{1,2}\right| \right\rbrace$$

Now I want to show, that the set of all spektra of those matrices $B$ is equal to $\mathcal K$

Necessarily $B$ is of the form $$B = \begin{pmatrix} a_{1,1} & \left|a_{1,2}\right|\exp{i\psi_1} \\ \left|a_{2,1}\right|\exp{i\psi_2} & a_{2,2} \end{pmatrix}$$ where $\psi_1$, $\psi_2$ are some real numbers.

Now fix one of those matrices $B$ and calculate the eigenvalues gives, putting them into the defining equality in $\mathcal K$ demonstrates, that they really belong to $\mathcal K$

But I am stuck on the other direction. Now I want to fix any $z \in \mathcal K$ and show that there is a corresponding matrix $B$ in $\omega(A)$ that has this $z$ as an eigenvalue. It's quite obvious that there must exist such an $B$ but how to chose $\psi_1$ and $\psi_2$ depending on $z$?


Suppose that $z\in\mathcal{K}$ is fixed so that $$|a_{1,1}-z||a_{2,2}-z|=|a_{1,2}||a_{2,1}|.$$ We may write $a_{1,1}-z$ and $a_{2,2}-z$ in polar form as $r_1e^{i\psi_1}$ and $r_2 e^{i\psi_2}$, respectively. From this it follows that $|a_{1,1}-z||a_{2,2}-z|=r_1r_2$, and hence \begin{align*} (a_{1,1}-z)(a_{2,2}-z)=(r_1e^{i\psi_1})(r_2 e^{i\psi_2})&=r_1r_2 e^{i(\psi_1+\psi_2)}\\ &=|a_{1,1}-z||a_{2,2}-z|e^{i(\psi_1+\psi_2)}=|a_{1,2}||a_{2,1}|e^{i(\psi_1+\psi_2)}. \end{align*} This proves that $z$ is a root of the characteristic polynomial of $$B=\begin{pmatrix} a_{1,1} & |a_{1,2}|e^{i\psi_1}\\ |a_{2,1}|e^{i\psi_2} & a_{2,2} \end{pmatrix}$$ as desired.


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