# Jacobi's equality between complementary minors of inverse matrices

What's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse?

Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then

$$|\det A[I,J]| = | (\det A) \det A^{-1}[J^c,I^c]|,$$ where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$.

I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(

Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.

• Wow - I did not know that algebra proofs could have a "sexist bias". I am too curious to let it pass --- what do you mean exactly? Feb 8, 2012 at 10:19
• I was just referring to the preface, where he said the book is suitable for undergraduate students, or boys in their last years of school. Maybe the word "boy" has a gender neutral meaning back then? Feb 8, 2012 at 19:39

The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be $$M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix},$$ where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.

You may verify the following identity ("magic wand Schur complement formula") $$\begin{bmatrix}A & B\\\\ C & D\end{bmatrix} = \begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix} \begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix} \begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1}$$ By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).

Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.

• I guess for any thesis involving minors the Schur complement formula would be among the first things to try. Feb 8, 2012 at 10:30

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:

• we use $$\widetilde{K}$$ to denote the complement $$\left[ n\right] \setminus K$$ of this subset in $$\left[ n\right]$$.

• we use $$\sum K$$ to denote the sum of the elements of $$K$$.

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:

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} .

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...

• If we agree to see a $\mathbb{K}$-matrix as a function $S\times S\to\mathbb{K}$ , with any finite index set $S$ (not just some $[m]$), then, according to Proposition 3, the matrix of the linear map $\Lambda^k(f_B)$ in the basis $\{e_I\}_{I\in\mathcal{P}_k([n])}$ is $[\det(B^I_J)]_{(I\times J)\in\mathcal{P}_k([n])\times \mathcal{P}_k([n])}$. Dec 19, 2017 at 18:48
• So the functoriality of $\Lambda^k$ reads as the (generalized) Cauchy-Binet formula: en.wikipedia.org/wiki/Cauchy–Binet_formula#Generalization . In the same spirit, I'd like to see Theorem 1 as a statement on the inverse matrix of the linear map $\Lambda^k(f_A)$, or better, since it also refers to $\tilde I$ and $\tilde J$, of the linear map $\Lambda^k(f_A)^*\sim\Lambda^{n-k}(f_A)$. Can this produce a proof to Theorem 1 consisting in a plain translation of facts about exterior algebra in the language of matrices? Dec 19, 2017 at 18:48
• @PietroMajer: This is what I was trying to achieve when I started writing the proof. Unfortunately, it's not directly clear to me how this works. Dec 19, 2017 at 19:54
• I feel it is quite similar to what we wrote extending Jacobi identity to matrices with non-commuting entries (Manin matrices) arxiv.org/abs/0901.0235 Jan 13, 2019 at 8:24

This is nothing but the Schur complement formula. See my book Matrices; Theory and Applications, 2nd ed., Springer-Verlag GTM 216, page 41.

Up to some permutation of rows and columns, we may assume that $I=J=[1,p]$. Let us write blockwise $$A=\begin{pmatrix} W & X \\\\ Y & Z \end{pmatrix}.$$ Assume WLOG that $W$ is invertible. On the one hand, we have (Schur C.F) $$\det A=\det W\cdot\det(Z-YW^{-1}X).$$ Finally, we have $$A^{-1}=\begin{pmatrix} \cdot & \cdot \\\\ \cdot & (Z-YW^{-1}X)^{-1} \end{pmatrix},$$which gives the desired result.

These formulas are obtained by factorizing $A$ into $LU$ (namely, $L= \begin{pmatrix} I_* & 0 \\ YW^{-1} & I_* \end{pmatrix}$ and $U = \begin{pmatrix} W & X \\ 0 & Z-YW^{-1}X \end{pmatrix}$, with the $I_*$ being identity matrices of appropriate size).

• Your answer is equally valid. Thanks! Feb 8, 2012 at 10:27

Here is a short proof inspired by darij grinberg's answer and not using Schur complements (in particular, no invertibility of sub-matrices or continuity assumptions are needed).

WLOG let $I=J=[k]$. Let $A_i$ denote the $i$th column of a matrix $A$. Consider the product $$A \left[ \begin{array}{c|c|c|c|c|c} e_1 & \cdots & e_k & A^{-1}_{k+1} & \cdots & A^{-1}_n \end{array} \right] = \left[ \begin{array}{c|c|c|c|c|c} A_1 & \cdots & A_k & e_{k+1} & \cdots & e_n \end{array} \right]$$ which can also be written $$A \begin{pmatrix} I_k & A^{-1}[J,I^c] \\ 0 & A^{-1}[J^c,I^c] \end{pmatrix} = \begin{pmatrix} A[I,J] & 0 \\ A[I^c,J] & I_{n-k} \end{pmatrix}$$ Taking the determinant of both sides yields $$(\det A) (\det A^{-1}[J^c,I^c]) = \det A[I,J]$$.

• Really nice... although a nontrivial part of my answer is justifying the "WLOG" in the first sentence. Still, this simplifies a lot. Feb 17, 2017 at 1:19
• I believe the justification is not too bad. Assume the result is true for I=K, J=K, K=[k]. Then for any other I,J of cardinality k, we have $\det(A[I,J])=\det((PAQ)[K,K])=\det(PAQ)\det(Q^{-1}A^{-1}P^{-1}[K^c,K^c]) = \det(PAQ)\det(A^{-1}[J^c,I^c])$, where $P$ is the permutation matrix taking $I$ to $1,\dots,k$ and $I^c$ to $k+1,\dots,n$ (I believe you call this permutation $w(I)$), and $Q$ is defined similarly. So we incur a factor of $\text{sign}(P)\text{sign}(Q)$ which is $(-1)^{\sum I + \sum J}$ by a counting argument, as you proved. Feb 17, 2017 at 15:05
• I suspect that your $Q$ is not "defined similarly" but rather is the inverse of the matrix defined similarly. But yes, this is essentially how it's done. Feb 18, 2017 at 5:42