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Let $A$ be a square matrix of order $n$, say with complex coefficients, and let $M$ be the plain matrix of minors of $A$ of order $n-1$ (no transpose, no sing changes). Let $I$ and $J$ be $r$-subsets of the index set $[n]$. Then, apparently $$\det M_{I\times J}=\det A_{([n]\setminus I)\times ([n]\setminus J)}\det(A)^{r-1}.$$ (One can also write the identity for the classical adjoint of $A$ instead of $M$, or also assume $A$ invertible, and express the identity in terms of $A^{-1}$).

So for $r=1$ this is just the definition of $M_{ij}$. For small values of $n$, it is possible to check the identity, keeping track of the terms of the expansion of the determinant. But is there a more synthetic proof, and an interpretation of it?

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    $\begingroup$ The case when M is the classical cofactor matrix was discussed here. $\endgroup$
    – Ivan Izmestiev
    Commented Dec 19, 2017 at 16:08
  • $\begingroup$ "Plain matrix of minors" = plain matrix of $\left(n-1\right)\times\left(n-1\right)$-minors? $\endgroup$
    – darij grinberg
    Commented Dec 19, 2017 at 17:01
  • $\begingroup$ @darij grinberg: Yep. Btw, I like very much your exterior algebra duality interpretation, which is what I wished! $\endgroup$
    – Pietro Majer
    Commented Dec 19, 2017 at 17:14

2 Answers 2

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For me, a standard book to look for such things is Prasolov's linear algebra. This is Theorem 1.2.6.1. Here is the Russian edition, but the proof is essentially a formula.

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    $\begingroup$ Thank you, the proof is very clear from the two formulas in the proof! (The formula in the statement should be an equality between determinants, and so should be the last one, right?) $\endgroup$
    – Pietro Majer
    Commented Dec 19, 2017 at 16:04
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    $\begingroup$ English translation of the 1994 edition. $\endgroup$
    – darij grinberg
    Commented Dec 19, 2017 at 17:02
  • $\begingroup$ Yes, for determinants. Maybe, your theorem is rather 1.2.6.3 (1.2.6.1 is for angular minors). $\endgroup$
    – Fedor Petrov
    Commented Dec 19, 2017 at 17:04
  • $\begingroup$ (there was a typo, which in the English translation has been fixed) $\endgroup$
    – Pietro Majer
    Commented Dec 19, 2017 at 17:23
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    $\begingroup$ strange: the Russian version is much more fresh $\endgroup$
    – Fedor Petrov
    Commented Dec 19, 2017 at 17:34
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You might try using the very interesting combinatorial approach of Doron Zeilberger in A Combinatorial Approach to Matrix Algebra, Discrete Mathematics 56 (1985), 61-72. There he gives short proofs of various matrix results such as the Cayley-Hamilton Theorem and the Matrix-Tree Theorem as combinatorial identities, which for me is the "right" way to understand these, rather than appealing to, for example, properties of complex matrices and eigenvalues.

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    $\begingroup$ This may work, but probably not as easily as the proofs in Zeilberger's paper. The comment by André Camargo in terrytao.wordpress.com/2017/08/28/… (from 30 August, 2017 at 7:25 pm) sketches how the $r = n$ case can be obtained from Sylvester's identity (which is combinatorially proven in a paper by Berliner and Brualdi). I suspect the same argument goes through for all $r$, but have not checked. $\endgroup$
    – darij grinberg
    Commented Dec 19, 2017 at 17:07

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