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Given an invertible matrix $A \in \mathbb{R}^{n \times n}$, and index set $\langle n\rangle = \{ 1, \dots, n \}$, and the submatrix $A(\alpha)$ with the columns and rows of $A$ with indices $\alpha \subset \langle n \rangle$.

Is there are characterisation of all matrices $A$ with

$\textrm{det } A(\alpha) = \textrm{det } A^{-1}(\alpha) \qquad \forall \alpha \subset \langle n \rangle \qquad ?$

Example: Sufficient conditions are involutory ($A^{-1} = A$) or orthogonal ($A^{-1} = A^T$) matrices.

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Sebastian, I changed deeply my answer, because there was a mistake in calculations. It is still nteresting, I hope, but in a different way. – Denis Serre May 14 '13 at 15:41
Denis, I find your structural findings very interesting. Although my pure algebra knowledge might be too restricted to create an answer from this. I was coming more from a point of diagonally similar (with transpose) matrices, which do have the same corresponding principal minors under certain conditions. (Like it was investigated by D.J. Hartfiel and R. Leowy in On matrices having equal corresponding principal minors) So your answer went far off, what I expected and hopefully someone else can help out on this. – Sebastian Schlecht May 14 '13 at 16:14
I eventually delete my answer. It seems that I described the set of involutory matrices! Fortunately, this was not a doctoral dissertation; see the MO question about urban legends... – Denis Serre May 14 '13 at 16:29
........... :-) – Sebastian Schlecht May 14 '13 at 17:05
What gives the case $n=2$ with $A=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$ ? – Dietrich Burde May 16 '13 at 18:11

Let $A=(a_{ij})$. Then we obtain a system of polynomial equations in the variables $a_{ij}$, given by determinant conditions $\det A(\alpha)=\det A^{-1}(\alpha)$. If one determines a Groebner basis for such a system with a given $n$, we see that the dimension of the variety given by the polynomial equations is rather high. We see that there are always many other solutions for $A$, which do not satisfy $A=A^{-1}$ or $A^{-1}=A^T$. One reason is, that many determinants of the submatrices of size less than $n$ can be chosen to be zero, for $A$ and $A^{-1}$.

I know this does not answer the question how to give a nice characterization for this variety other than by determinants. Therefore, let me just give a typical example for $n=3$, with $a,b\in \mathbb{R}$ and $b\neq 0$:

$$ A=\begin{pmatrix} 0 & 0 & -1/b \cr a & 1 & 0 \cr b & 0 & 0 \end{pmatrix}, \quad A^{-1}=\begin{pmatrix} 0 & 0 & 1/b \cr 0 & 1 & -a/b \cr -b & 0 & 0 \end{pmatrix}. $$

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