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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 $\textrm{rank } A([1,2],[3,4]) = 1$?

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  • $\begingroup$ Not sure what irreducible means, but is $\begin{pmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ an example? All principal minors are $1$. $\endgroup$
    – David E Speyer
    Commented Oct 19, 2015 at 22:14
  • $\begingroup$ @DavidSpeyer: Ah, sorry, I will clarify irreducible in the question. $\endgroup$
    – Jiro
    Commented Oct 19, 2015 at 22:15

1 Answer 1

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The matrix $$A=\frac15\left( \begin{array}{cccc} -1 & 4 & -2 & -2 \\ -4 & 1 & 2 & 2 \\ 2 & 2 & -1 & 4 \\ -2 & -2 & -4 & 1 \\ \end{array} \right)$$is orthogonal, i. e. $A=\overline{A}^{-1}$.

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