Not all has been said about this question that is worth saying -- at
the very least, someone could have written down the version without the
absolute values; but more importantly, there are various other equally good proofs.
Notations and statement
Let me first state the result with proper signs and no absolute values.
Standing assumptions. The following notations will be used throughout this post:
Let $\mathbb{K}$ be a commutative ring. All matrices that appear in the
following are matrices over $\mathbb{K}$.
Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $.
For every $n\in\mathbb{N}$, we let $\left[ n\right] $ denote the set
$\left\{ 1,2,\ldots,n\right\} $.
Fix $n\in\mathbb{N}$.
Let $S_n$ denote the $n$-th symmetric group (i.e., the group of
permutations of $\left[ n\right] $).
If $A\in\mathbb{K}^{n\times m}$ is an $n\times m$-matrix, if $I$ is a
subset of $\left[ n \right]$, and if $J$ is a subset of $\left[ m \right]$,
then $A_J^I$ is the $\left| I\right| \times\left| J\right| $-matrix
defined as follows: Write
$A$ in the form $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq m}$;
write the set $I$ in the form $I = \left\{ i_1 < i_2 < \cdots < i_u \right\}$;
write the set $J$ in the form $J = \left\{ j_1 < j_2 < \cdots < j_v \right\}$.
Then, set $A_J^I = \left( a_{i_x, j_y} \right) _{1\leq x\leq u,\ 1\leq
y\leq v}$. (Thus, roughly speaking, $A_J^I$ is the $\left|
I\right| \times\left| J\right| $-matrix obtained from $A$ by
removing all rows whose indices do not belong to $I$, and removing all columns
whose indices do not belong to $J$.)
If $K$ is a subset of $\left[ n\right] $, then:
Now, we claim the following:
Theorem 1 (Jacobi's complementary minor formula). Let $A\in\mathbb{K} ^{n\times n}$ be an invertible $n\times n$-matrix. Let $I$ and $J$ be two subsets of $\left[ n\right] $ such that $\left| I\right| =\left| J\right| $. Then,
\begin{align}
\det\left( A_{J}^{I}\right) =\left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right) .
\end{align}
Three references
Here are three references to proofs of Theorem 1:
Theorem 1 is Lemma A.1 (e) in Sergio Caracciolo, Alan D. Sokal, Andrea
Sportiello, Algebraic/combinatorial proofs of Cayley-type identities for
derivatives of determinants and pfaffians, arXiv:1105.6270v2 (published in:
Advances in Applied Mathematics 50, 474--594 (2013)). In the paragraph
following Theorem A.16, a proof is given using what the authors call
"Grassmann-Berezin integration" (despite its name, a purely algebraic
mock-calculus on the exterior
algebra of a vector space).
Theorem 1 is (1) in Pierre Lalonde, A non-commutative version of Jacobi's
equality on the cofactors of a matrix, Discrete Mathematics 158 (1996), pp.
161--172. The goal of the paper is to generalize it to a (mildly)
noncommutative setting.
Theorem 1 is Exercise 6.56 in my Notes on the combinatorial
fundamentals of algebra, version of 10 January 2019. The first proof
I give is fairly classical and similar to the ones given by Federico Poloni
and Denis Serre, but requires no WLOG assumptions (instead of using the Schur
complement, I use a cheap generalization of it which is Exercise 6.38 in the
notes). The formal bookkeeping that leads to the sign $\left( -1\right)
^{\sum I+\sum J}$ takes a huge lot of space, although it is so easy to
convince yourself of it with some handwaving that you might not notice that it
requires any proof at all. The second proof is an expansion of an argument
briefly outlined in
D. Laksov, A. Lascoux, P. Pragacz, and A. Thorup, The LLPT Notes,
old (2001) version
(Chapter SCHUR, proof of (1.9)).
Note that every source uses different notations. What I call $A_J^I$ above
is called $A_{IJ}$ in the paper by Caracciolo, Sokal and Sportiello, is
called $A\left[ I,J\right] $ in Lalonde's paper, and is called
$\operatorname*{sub}\nolimits_{w\left( I\right) }^{w\left( J\right) }A$ in
my notes. Also, the $I$ and $J$ in the paper by Caracciolo, Sokal and
Sportiello correspond to the $\widetilde{I}$ and $\widetilde{J}$ in
Theorem 1 above.
A fourth proof
Let me now give a fourth proof, using exterior algebra. The proof is probably
not new (the method is definitely not), but I find it instructive.
This proof would become a lot shorter if I didn't care for the signs
and would only prove the weaker claim that
$\det\left( A_J^I \right)
= \pm \det A\cdot \det\left(
\left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right) $
for some value of $\pm$. But this weaker claim is not as useful as
Theorem 1 in its full version (in particular, it would not suffice to
fill the gap in Macdonald's book that has motivated this question).
The permutation $w\left( K\right) $
Let us first introduce some more notations:
If $K$ is a subset of $\left[ n\right] $, and if $k = \left|K\right|$,
then we let $w\left( K\right) $
be the (unique) permutation $\sigma\in S_n$ whose first $k$ values
$\sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left(
k\right) $ are the elements of $K$ in increasing order, and whose next $n-k$
values $\sigma\left( k+1\right) ,\sigma\left( k+2\right) ,\ldots
,\sigma\left( n\right) $ are the elements of $\widetilde{K}$ in increasing order.
The first important property of $w\left( K\right) $ is the following fact:
Lemma 2. Let $K$ be a subset of $\left[ n\right] $. Then, $\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left| K\right| \right) }$.
You don't need to prove Lemma 2 if you only care about the weaker
version of Theorem 1 with the $\pm$ sign.
Proof of Lemma 2. Let $k=\left| K\right| $. Let $a_{1},a_{2}
,\ldots,a_{k}$ be the $k$ elements of $K$ in increasing order (with no
repetitions). Let $b_{1},b_{2},\ldots,b_{n-k}$ be the $n-k$ elements of
$\widetilde{K}$ in increasing order (with no repetitions). Let $\gamma
=w\left( K\right) $. Then, the definition of $w\left( K\right) $ shows
that the first $k$ values $\gamma\left( 1\right) ,\gamma\left( 2\right)
,\ldots,\gamma\left( k\right) $ of $\gamma$ are the elements of $K$ in
increasing order (that is, $a_{1},a_{2},\ldots,a_{k}$), and the next $n-k$
values $\gamma\left( k+1\right) ,\gamma\left( k+2\right) ,\ldots
,\gamma\left( n\right) $ of $\gamma$ are the elements of $\widetilde{K}$ in
increasing order (that is, $b_{1},b_{2},\ldots,b_{n-k}$). In other words,
\begin{align}
\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots
,\gamma\left( n\right) \right) =\left( a_{1},a_{2},\ldots,a_{k}
,b_{1},b_{2},\ldots,b_{n-k}\right) .
\end{align}
Now, you can obtain the list $\left( \gamma\left( 1\right) ,\gamma\left(
2\right) ,\ldots,\gamma\left( n\right) \right) $ from the list $\left(
1,2,\ldots,n\right) $ by successively switching adjacent entries, as follows:
First, move the element $a_{1}$ to the front of the list, by successively
switching it with each of the $a_{1}-1$ entries smaller than it.
Then, move the element $a_{2}$ to the second position, by successively
switching it with each of the $a_{2}-2$ entries (other than $a_{1}$) smaller
than it.
Then, move the element $a_{3}$ to the third position, by successively
switching it with each of the $a_{3}-3$ entries (other than $a_{1}$ and
$a_{2}$) smaller than it.
And so on, until you finally move the element $a_{k}$ to the $k$-th position.
More formally, you are iterating over all $i\in\left\{ 1,2,\ldots,k\right\}
$ (in increasing order), each time moving the element $a_{i}$ to the $i$-th
position in the list, by successively switching it with each of the $a_{i}-i$
entries (other than $a_{1},a_{2},\ldots,a_{i-1}$) smaller than it.
At the end, the first $k$ positions of the list are filled with $a_{1}
,a_{2},\ldots,a_{k}$ (in this order), whereas the remaining $n-k$ positions
are filled with the remaining entries $b_{1},b_{2},\ldots,b_{n-k}$ (again, in
this order, because the switches have never disrupted their
strictly-increasing relative order). Thus, at the end, your list is precisely
$\left( a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{n-k}\right) =\left(
\gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left(
n\right) \right) $. You have used a total of
\begin{align}
& \left( a_{1}-1\right) +\left( a_{2}-2\right) +\cdots+\left(
a_{k}-k\right) \\
& = \underbrace{\left( a_{1}+a_{2}+\cdots+a_{k}\right) }_{\substack{=\sum
K\\\text{(by the definition of }a_{1},a_{2},\ldots,a_{k}\text{)}
}}-\underbrace{\left( 1+2+\cdots+k\right) }_{\substack{=1+2+\cdots
+\left| K\right| \\\text{(since }k=\left| K\right| \text{)}}} \\
& =\sum K-\left( 1+2+\cdots+\left| K\right| \right)
\end{align}
switches. Thus, you have obtained the list $\left( \gamma\left( 1\right)
,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right) $ from the
list $\left( 1,2,\ldots,n\right) $ by $\sum K-\left( 1+2+\cdots+\left|
K\right| \right) $ switches of adjacent entries. In other words, the
permutation $\gamma$ is a composition of $\sum K-\left( 1+2+\cdots+\left|
K\right| \right) $ simple transpositions (where a "simple
transposition" means a transposition switching $u$ with $u+1$ for some
$u$). Hence, it has sign $\left( -1\right) ^{\sum K-\left(
1+2+\cdots+\left| K\right| \right) }$. This proves Lemma 2.
$\blacksquare$
Exterior algebras
Now, let's introduce some more notations and state some well-known properties
concerning exterior algebras.
For any $\mathbb{K}$-module $V$, we let $\wedge V$ denote the exterior algebra
of $V$. The multiplication in this exterior algebra will be written as
juxtaposition (i.e., we will write $ab$ for the product of two elements $a$
and $b$ of $\wedge V$) or as multiplication (i.e., we will write $a\cdot b$
for this product).
If $k\in\mathbb{N}$ and if $V$ is a $\mathbb{K}$-module, then $\wedge^{k}V$
shall mean the $k$-th exterior power of $V$. If $k\in\mathbb{N}$, if $V$ and
$W$ are two $\mathbb{K}$-modules, and if $f:V\rightarrow W$ is a $\mathbb{K}
$-linear map, then the $\mathbb{K}$-linear map $\wedge^{k}V\rightarrow
\wedge^{k}W$ canonically induced by $f$ will be denoted by $\wedge^{k}f$. It
is well-known that if $V$ and $W$ are two $\mathbb{K}$-modules, if
$f:V\rightarrow W$ is a $\mathbb{K}$-linear map, then
\begin{align}
\left( \wedge^{k}f\right) \left( a\right) \cdot\left(
\wedge^{\ell}f\right) \left( b\right) =\left( \wedge^{k+\ell}f\right)
\left( ab\right)
\label{darij1.eq1}
\tag{1}
\end{align}
for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $a\in\wedge^{k}V$ and
$b\in\wedge^{\ell}V$.
If $V$ is a $\mathbb{K}$-module, then
\begin{align}
uv=\left( -1\right) ^{k\ell}vu
\label{darij1.eq2}
\tag{2}
\end{align}
for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $u\in\wedge^{k}V$ and
$v\in\wedge^{\ell}V$.
For any $u\in\mathbb{N}$, we consider $\mathbb{K}^{u}$ as the $\mathbb{K}
$-module of column vectors with $u$ entries.
For any $u\in\mathbb{N}$ and $v\in\mathbb{N}$, and any $v\times u$-matrix
$B\in\mathbb{K}^{v\times u}$, we define $f_{B}$ to be the $\mathbb{K}$-linear
map $\mathbb{K}^{u}\rightarrow\mathbb{K}^{v}$ sending each $x\in\mathbb{K}
^{u}$ to $Bx\in\mathbb{K}^{v}$. This $\mathbb{K}$-linear map $f_{B}$ satisfies
$\det\left( f_{B}\right) =\det B$, and is often identified with the matrix
$B$ (though we will not identify it with $B$ here).
Here is another known fact:
Proposition 2a. Let $f:\mathbb{K}^{n}\rightarrow\mathbb{K}^{n}$ be a $\mathbb{K}$-linear map. The map $\wedge^{n}f:\wedge^{n}\left( \mathbb{K} ^{n}\right) \rightarrow\wedge^{n}\left( \mathbb{K}^{n}\right) $ is multiplication by $\det f$. In other words, every $z\in\wedge^{n}\left( \mathbb{K}^{n}\right) $ satisfies
\begin{align}
\left( \wedge^{n}f\right) \left( z\right) =\left( \det f\right) z .
\label{darij1.eq3}
\tag{3}
\end{align}
Let $\left( e_{1},e_{2},\ldots,e_{n}\right) $ be the standard basis of the
$\mathbb{K}$-module $\mathbb{K}^{n}$. (Thus, $e_i$ is the column vector
whose $i$-th entry is $1$ and whose all other entries are $0$.)
For every subset $K$ of $\left[ n\right] $, we define $e_K\in
\wedge^{\left| K\right| }\left( \mathbb{K}^{n}\right) $ to be the
element $e_{k_{1}}\wedge e_{k_{2}}\wedge\cdots\wedge e_{k_{\left|
K\right| }}$, where $K$ is written in the form $K=\left\{ k_{1}
<k_{2}<\cdots<k_{\left| K\right| }\right\} $.
For every $k\in\mathbb{N}$ and every set $S$, we let $\mathcal{P}_{k}\left( S
\right) $ denote the set of all $k$-element subsets of $S$.
It is well-known that, for every $k\in\mathbb{N}$, the family $\left(
e_K\right) _{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ is a
basis of the $\mathbb{K}$-module $\wedge^{k}\left( \mathbb{K}^{n}\right) $.
Applying this to $k=n$, we conclude that the family $\left( e_K\right)
_{K\in\mathcal{P}_{n}\left( \left[ n\right] \right) }$ is a basis of the
$\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right) $. Since this
family $\left( e_K\right) _{K\in\mathcal{P}_{n}\left( \left[ n\right]
\right) }$ is the one-element family $\left( e_{\left[ n\right] }\right)
$ (because the only $K\in\mathcal{P}_{n}\left( \left[ n\right] \right) $
is the set $\left[ n\right] $), this rewrites as follows: The one-element
family $\left( e_{\left[ n\right] }\right) $ is a basis of the
$\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right) $.
If $B$ is an $n\times n$-matrix and $k\in\mathbb{N}$, then evaluating the map
$\wedge^{k}f_{B}$ on the elements of the basis $\left( e_K\right)
_{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ of $\wedge
^{k}\left( \mathbb{K}^{n}\right) $, and expanding the results again in this
basis gives rise to coefficients which are the $k\times k$-minors of $B$. More precisely:
Proposition 3. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right) $. Then,
\begin{align}
\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) = \sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J} ^{I}\right) e_{I} .
\end{align}
(This can be generalized:
If $u\in\mathbb{N}$, $v \in \mathbb{N}$,
$B\in\mathbb{K}^{u\times v}$,
$k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ v\right] \right) $,
then $\left( \wedge^{k}f_{B}\right) \left( e_{J}\right)
=
\sum\limits_{I\in\mathcal{P}_{k}\left( \left[ u\right] \right) }\det\left( B_{J}
^{I}\right) e_{I}$, where the elements $e_{J}\in\wedge^{k}\left(
\mathbb{K}^{v}\right) $ and $e_{I}\in\wedge^{k}\left( \mathbb{K}^{u}\right)
$ are defined as before but with $v$ and $u$ instead of $n$.)
Extracting minors from the exterior algebra
Now, we shall need a simple lemma:
Lemma 4. Let $K$ be a subset of $\left[ n\right] $. Then,
\begin{align}
e_K e_{\widetilde{K}}=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left| K\right| \right) }e_{\left[ n\right] } .
\end{align}
Proof of Lemma 4. Let $k = \left|K\right|$.
Let $\sigma$ be the permutation $w\left( K\right) \in
S_n$ defined above. Its first $k$ values $\sigma\left( 1\right)
,\sigma\left( 2\right) ,\ldots,\sigma\left( k\right) $ are the elements of
$K$ in increasing order; thus, $e_{\sigma\left( 1\right) }\wedge
e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right)
}=e_K$. Its next $n-k$ values $\sigma\left( k+1\right) ,\sigma\left(
k+2\right) ,\ldots,\sigma\left( n\right) $ are the elements of
$\widetilde{K}$ in increasing order; thus, $e_{\sigma\left( k+1\right)
}\wedge e_{\sigma\left( k+2\right) }\wedge\cdots\wedge e_{\sigma\left(
n\right) }=e_{\widetilde{K}}$.
From $\sigma=w\left( K\right) $, we obtain $\left( -1\right) ^{\sigma
}=\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum
K-\left( 1+2+\cdots+\left| K\right| \right) }$ (by Lemma 2).
Now, it is well-known that
\begin{align}
e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge
\cdots\wedge e_{\sigma\left( n\right) }=\left( -1\right) ^{\sigma
}\underbrace{e_{1}\wedge e_{2}\wedge\cdots\wedge e_{n}}_{=e_{\left[ n\right]
}}=\left( -1\right) ^{\sigma}e_{\left[ n\right] } .
\end{align}
Hence,
\begin{align}
\left( -1\right) ^{\sigma}e_{\left[ n\right] }
& = e_{\sigma\left( 1\right)
}\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left(
n\right) } \\
& = \underbrace{\left( e_{\sigma\left( 1\right) }\wedge e_{\sigma\left(
2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right) }\right) }_{=e_K
}\underbrace{\left( e_{\sigma\left( k+1\right) }\wedge e_{\sigma\left(
k+2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }\right)
}_{=e_{\widetilde{K}}} \\
& = e_K e_{\widetilde{K}} .
\end{align}
Since $\left( -1\right) ^{\sigma}=\left( -1\right) ^{\sum K-\left(
1+2+\cdots+\left| K\right| \right) }$, this rewrites as $\left(
-1\right) ^{\sum K-\left( 1+2+\cdots+\left| K\right| \right)
}e_{\left[ n\right] }= e_K e_{\widetilde{K}}$. This proves Lemma 4.
$\blacksquare$
We can combine Proposition 3 and Lemma 4 to obtain the following fact:
Corollary 5. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right) $. Then, every $K\in\mathcal{P}_{k}\left( \left[ n\right] \right) $ satisfies
\begin{align}
\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K} }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] } .
\end{align}
Proof of Corollary 5. Let $K \in \mathcal{P}_{k}\left( \left[ n\right] \right) $.
Let $I\in\mathcal{P}_{k}\left( \left[ n\right]
\right) $ be such that $I\neq K$. Then, $I\not \subseteq K$ (since the sets
$I$ and $K$ have the same size $k$). Hence, there exists some $z\in I$ such
that $z\notin K$. Consider this $z$. We have $z\in I$ and $z\in\widetilde{K}$
(since $z\notin K$). Hence, both $e_{I}$ and $e_{\widetilde{K}}$ are "wedge
products" containing the factor $e_{z}$; therefore, the product $e_{I}
e_{\widetilde{K}}$ is a "wedge product" containing this factor twice. Thus,
$e_{I}e_{\widetilde{K}}=0$.
Now, forget that we fixed $I$. We thus have proven that
\begin{align}
e_{I}e_{\widetilde{K}}=0
\text{ for every } I\in\mathcal{P}_{k}\left( \left[
n\right] \right)
\text{ satisfying } I\neq K .
\end{align}
Hence,
\begin{align}
\sum\limits_{\substack{I\in\mathcal{P}_{k}\left( \left[ n\right] \right)
;\\I\neq K}}\det\left( B_{J}^{I}\right) \underbrace{e_{I}e_{\widetilde{K}}
}_{=0}=0 .
\label{darij1.eq4}
\tag{4}
\end{align}
Proposition 3 yields
\begin{align}
\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) =\sum\limits_{I\in
\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J}
^{I}\right) e_{I} .
\end{align}
Multiplying both sides of this equality by $e_{\widetilde{K}}$ from the right,
we find
\begin{align}
& \left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K}}
=\sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left(
B_{J}^{I}\right) e_{I}e_{\widetilde{K}} \\
& = \det\left( B_{J}^{K}\right) e_K e_{\widetilde{K}}+\sum\limits_{\substack{I\in
\mathcal{P}_{k}\left( \left[ n\right] \right) ;\\I\neq K}}\det\left(
B_{J}^{I}\right) e_{I}e_{\widetilde{K}} \\
& = \det\left( B_{J}^{K}\right) \underbrace{e_K e_{\widetilde{K}}
}_{\substack{=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left|
K\right| \right) }e_{\left[ n\right] }\\\text{(by Lemma 4)}}}
\qquad \text{(by \eqref{darij1.eq4})} \\
& = \left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left| K\right|
\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] } \\
& = \left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left(
B_{J}^{K}\right) e_{\left[ n\right] }
\end{align}
(since $\left| K\right| =k$). This proves Corollary 5. $\blacksquare$
Corollary 5 is rather helpful when it comes to extracting a specific minor of
a matrix $B$ from the maps $\wedge^{k}f_{B}$.
Proof of Theorem 1
Proof of Theorem 1. Set $k=\left| I\right| =\left| J\right|
$. Notice that $\left| \widetilde{I}\right| =n-k$ (since $\left|
I\right| =k$) and $\left| \widetilde{J}\right| =n-k$ (similarly).
Define $y\in\wedge
^{n-k}\left( \mathbb{K}^{n}\right) $ by $y=\left( \wedge^{n-k}f_{A^{-1}
}\right) \left( e_{\widetilde{I}}\right) $.
The maps $f_{A}$ and $f_{A^{-1}}$ are mutually inverse (since the map
$\mathbb{K}^{n\times n}\rightarrow\operatorname*{End}\left( \mathbb{K}
^{n}\right) ,\ B\mapsto f_{B}$ is a ring homomorphism). Hence, the maps
$\wedge^{n-k}f_{A}$ and $\wedge^{n-k}f_{A^{-1}}$ are mutually inverse (since
$\wedge^{n-k}$ is a functor). Thus, $\left( \wedge^{n-k}f_{A}\right)
\circ\left( \wedge^{n-k}f_{A^{-1}}\right) =\operatorname*{id}$. Now,
$y=\left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) $,
so that
\begin{align}
\left( \wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge
^{n-k}f_{A}\right) \left( \left( \wedge^{n-k}f_{A^{-1}}\right) \left(
e_{\widetilde{I}}\right) \right) =\underbrace{\left( \left( \wedge
^{n-k}f_{A}\right) \circ\left( \wedge^{n-k}f_{A^{-1}}\right) \right)
}_{=\operatorname*{id}}\left( e_{\widetilde{I}}\right) =e_{\widetilde{I}}
.
\end{align}
But \eqref{darij1.eq1} (applied to $V=\mathbb{K}^{n}$, $W=\mathbb{K}^{n}$, $f=f_{A}$,
$\ell=n-k$, $a=e_{J}$ and $b=y$) yields
\begin{align}
\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) \cdot\left(
\wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge^{n}f_{A}\right)
\left( e_{J}y\right) .
\end{align}
Thus,
\begin{align}
& \left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\left( \wedge
^{k}f_{A}\right) \left( e_{J}\right) \cdot\underbrace{\left( \wedge
^{n-k}f_{A}\right) \left( y\right) }_{=e_{\widetilde{I}}} \\
& =\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) e_{\widetilde{I}}
= \left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left(
A_J^I \right) e_{\left[ n\right] }
\end{align}
(by Corollary 5, applied to $B=A$ and $K=I$).
Hence,
\begin{align}
& \left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left(
A_J^I \right) e_{\left[ n\right] } \\
& = \left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\underbrace{\left(
\det f_{A}\right) }_{=\det A}e_{J}y \\
& \qquad \text{(by \eqref{darij1.eq3}, applied to $f=f_{A}$ and $z=e_{J}y$)}
\\
& = \left( \det A\right) e_{J}y .
\label{darij1.eq5}
\tag{5}
\end{align}
But \eqref{darij1.eq2} (applied to $\ell=n-k$, $u=e_{J}$ and $v=y$) yields
\begin{align}
& e_{J} y = \left(-1\right)^{k \left(n-k\right)}
\underbrace{y}_{=\left( \wedge^{n-k}f_{A^{-1}}\right) \left(
e_{\widetilde{I}}\right) } \underbrace{e_{J}}
_{=e_{\widetilde{\widetilde{J}}}} \\
& =\left( -1\right) ^{k\left( n-k\right) }\underbrace{\left( \wedge
^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right)
e_{\widetilde{\widetilde{J}}}}_{\substack{=\left( -1\right) ^{\sum
\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left(
\left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[
n\right] }\\\text{(by Corollary 5, applied to }A^{-1}\text{, }n-k\text{,
}\widetilde{I}\text{ and }\widetilde{J}\\\text{instead of }B\text{, }k\text{,
}J\text{ and }K\text{)}}} \\
& =\left( -1\right) ^{k\left( n-k\right) }\left( -1\right) ^{\sum
\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left(
\left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[
n\right] } \\
& =\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left(
1+2+\cdots+\left( n-k\right) \right) }\det\left( \left(
A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right]
} .
\label{darij1.eq6}
\tag{6}
\end{align}
But
\begin{align}
& k\left( n-k\right) +\underbrace{\sum\widetilde{J}}_{=\sum\left\{
1,2,\ldots,n\right\} -\sum J}-\underbrace{\left( 1+2+\cdots+\left(
n-k\right) \right) }_{=\sum\left\{ 1,2,\ldots,n-k\right\} } \\
& = k\left( n-k\right) +\sum\left\{ 1,2,\ldots,n\right\} -\sum
J-\sum\left\{ 1,2,\ldots,n-k\right\} \\
& = \underbrace{\sum\left\{ 1,2,\ldots,n\right\} -\sum\left\{ 1,2,\ldots
,n-k\right\} }_{\substack{=\sum\left\{ n-k+1,n-k+2,\ldots,n\right\}
\\=k\left( n-k\right) +\sum\left\{ 1,2,\ldots,k\right\} \\=k\left(
n-k\right) +\left( 1+2+\cdots+k\right) }}+k\left( n-k\right) -\sum J \\
& = 2k\left( n-k\right) +\left( 1+2+\cdots+k\right) -\sum J \\
& \equiv-\left( 1+2+\cdots+k\right) -\sum J \mod 2 .
\end{align}
Hence,
\begin{align}
\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left(
1+2+\cdots+\left( n-k\right) \right) }=\left( -1\right) ^{-\left(
1+2+\cdots+k\right) -\sum J} .
\end{align}
Thus, \eqref{darij1.eq6} rewrites as
\begin{align}
e_{J}y=\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left(
\left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[
n\right] } .
\end{align}
Hence, \eqref{darij1.eq5} rewrites as
\begin{align}
& \left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left(
A_J^I \right) e_{\left[ n\right] } \\
& = \left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right)
-\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}
}\right) e_{\left[ n\right] } .
\end{align}
We can "cancel" $e_{\left[ n\right] }$ from this equality (because if
$\lambda$ and $\mu$ are two elements of $\mathbb{K}$ satisfying $\lambda
e_{\left[ n\right] }=\mu e_{\left[ n\right] }$, then $\lambda=\mu$), and
thus obtain
\begin{align}
& \left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left(
A_J^I \right) \\
& = \left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right)
-\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}
}\right) .
\end{align}
Dividing this equality by $\left( -1\right) ^{\sum I-\left( 1+2+\cdots
+k\right) }$, we obtain
\begin{align}
& \det\left( A_J^I \right) \\
& = \left(\det A\right) \dfrac{\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}}{\left(
-1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }}\det\left( \left(
A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) \\
& = \left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left(
A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) .
\end{align}
This proves Theorem 1. $\blacksquare$
I have to admit this proof looked far shorter on the scratch paper on
which it was conceived than it has ended up here...
