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I'm searching for rather specific counter-example.

Some notation: $A(\alpha|\beta)$ is the sub matrix of $A$ with with rows $\alpha$ and columns $\beta$. $\textrm{det } A(\alpha|\alpha) =: \textrm{det } A(\alpha)$ are the principal minors of $A$. We define $\textrm{det } A(\emptyset)=1$. A matrix $A$ is irreducible if there is no permutation matrix $P$ so that

$$ P^{-1} A P = \begin{bmatrix} E & G \\ 0 & F \end{bmatrix} $$

where $E$ and $F$ are square.

Is there a invertible, irreducible matrix $A \in \mathbb{C}^{4 \times 4}$ with $\textrm{det } A(\alpha) = \textrm{det } \overline{A}^{-1}(\alpha)$ for all $\alpha \subseteq \{1,2,3,4\}$ and $\textrm{rank } A([1,2],[3,4]) = 1$?

I'm searching for rather specific counter-example.

Some notation: $A(\alpha|\beta)$ is the sub matrix of $A$ with with rows $\alpha$ and columns $\beta$. $\textrm{det } A(\alpha|\alpha) =: \textrm{det } A(\alpha)$ are the principal minors of $A$. We define $\textrm{det } A(\emptyset)=1$. A matrix $A$ is irreducible if there is no permutation matrix $P$ so that

$$ P^{-1} A P = \begin{bmatrix} E & G \\ 0 & F \end{bmatrix} $$

where $E$ and $F$ are square.

Is there a invertible, irreducible matrix $A \in \mathbb{C}^{4 \times 4}$ with $\textrm{det } A(\alpha) = \textrm{det } ((\overline{A}^{-1})(\alpha))$ for all $\alpha \subseteq \{1,2,3,4\}$ and $\textrm{rank } A([1,2],[3,4]) = 1$?

I'm searching for rather specific counter-example.

Some notation: $A(\alpha|\beta)$ is the sub matrix of $A$ with with rows $\alpha$ and columns $\beta$. $\textrm{det } A(\alpha|\alpha) =: \textrm{det } A(\alpha)$ are the principal minors of $A$. A matrix $A$ is irreducible if there is no permutation matrix $P$ so that

$$ P^{-1} A P = \begin{bmatrix} E & G \\ 0 & F \end{bmatrix} $$

where $E$ and $F$ are square.

Is there a invertible, irreducible matrix $A \in \mathbb{C}^{4 \times 4}$ with $\textrm{det } A(\alpha) = \textrm{det } \overline{A}^{-1}(\alpha)$ for all $\alpha$ and $\textrm{rank } A([1,2],[3,4]) = 1$?

I'm searching for rather specific counter-example.

Some notation: $A(\alpha|\beta)$ is the sub matrix of $A$ with with rows $\alpha$ and columns $\beta$. $\textrm{det } A(\alpha|\alpha) =: \textrm{det } A(\alpha)$ are the principal minors of $A$. We define $\textrm{det } A(\emptyset)=1$. A matrix $A$ is irreducible if there is no permutation matrix $P$ so that

$$ P^{-1} A P = \begin{bmatrix} E & G \\ 0 & F \end{bmatrix} $$

where $E$ and $F$ are square.

Is there a invertible, irreducible matrix $A \in \mathbb{C}^{4 \times 4}$ with $\textrm{det } A(\alpha) = \textrm{det } \overline{A}^{-1}(\alpha)$ for all $\alpha \subseteq \{1,2,3,4\}$ and $\textrm{rank } A([1,2],[3,4]) = 1$?

I'm searching for rather specific counter-example.

Some notation: $A(\alpha|\beta)$ is the sub matrix of $A$ with with rows $\alpha$ and columns $\beta$. $\textrm{det } A(\alpha|\alpha) =: \textrm{det } A(\alpha)$ are the principal minors of $A$.

Is there a invertible, irreducible matrix $A \in \mathbb{C}^{4 \times 4}$ with $\textrm{det } A(\alpha) = \textrm{det } \overline{A}^{-1}(\alpha)$ for all $\alpha$ and $\textrm{rank } A([1,2],[3,4]) = 1$?

I'm searching for rather specific counter-example.

Some notation: $A(\alpha|\beta)$ is the sub matrix of $A$ with with rows $\alpha$ and columns $\beta$. $\textrm{det } A(\alpha|\alpha) =: \textrm{det } A(\alpha)$ are the principal minors of $A$. A matrix $A$ is irreducible if there is no permutation matrix $P$ so that

$$ P^{-1} A P = \begin{bmatrix} E & G \\ 0 & F \end{bmatrix} $$

where $E$ and $F$ are square.

Is there a invertible, irreducible matrix $A \in \mathbb{C}^{4 \times 4}$ with $\textrm{det } A(\alpha) = \textrm{det } \overline{A}^{-1}(\alpha)$ for all $\alpha$ and $\textrm{rank } A([1,2],[3,4]) = 1$?