# Proving Laplace expansion using exterior algebra

Let $$A = (a_{ij})$$ be an $$n\times n$$ matrix with entries in a ring $$R$$, $$M$$ a free $$R$$-module of rank $$n$$ with an ordered basis $$(e_i)_{i \leq n}$$ and $$\phi\colon M\to M$$ is an endomorphism which $$A$$ represents (such that $$\phi(e_j) = \sum_{i = 1}^n a_{ij}e_i$$ fop all $$j$$). Then $$\bigwedge^n M$$ has rank $$1$$ with a basis $$\{e_1\wedge ... \wedge e_n\}$$ and there is a unique scalar $$r \in R$$ such that $$\left(\bigwedge^n \phi\right)(e_1\wedge ... \wedge e_n) = \phi(e_1)\wedge ... \wedge \phi(e_n) = r(e_1\wedge ... \wedge e_n).$$ This scalar is precisely the determinant $$\det(A)$$ of $$A$$. It can be easily shown that the value of $$\det(A)$$ doesn't depend on the choice of $$M$$. I take this as the definition of a determinant. From this other definitions can be deduced as theorems, including the formula $$\det(A) = \sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)a_{\sigma(1)1}...a_{\sigma(n)n}.$$

I understand that there is a shorter proof of the Laplace expansion of a determinant using the definition of a determinant in question. However, a book I consulted has a serious gap in the proof. I will state what I want to prove:

Let $$A = (a_{ij})$$ be an $$n\times n$$ matrix with entries in a commutative ring $$R$$. Denote by $$A_{ij}$$ the $$(n-1)\times(n-1)$$ matrix obtained from $$A$$ by deleting the $$i$$-th row and the $$j$$-th column of $$A$$. Then, for all $$i$$, $$\det(A) = \sum_{j = 1}^n (-1)^{i + j} a_{ij}\det(A_{ij}).$$

## 2 Answers

$$\newcommand{\bw}{\bigwedge} \newcommand{\w}{\wedge}$$ There is a simple proof (although tedious if we write down all the computations, which is what I did), but for that we need to interpret the quantity $$\det(A_{ij})$$ in terms of maps.

My notations will be the following : $$\iota_k : R^{n-1}\to R^n$$ is the map that sends basis vectors to basis vectors in linear order, but not touching the $$k$$th one in $$R^n$$, $$p_k : R^n\to R^{n-1}$$ will be the map that forgets about the $$k$$th coordinate, $$\rho_k = \iota_k\circ p_k : R^n\to R^n$$ is the map that sets the $$k$$th coordinate to $$0$$, and finally $$\pi_j : R^n \to R^n$$ is projection onto the $$j$$th coordinate (that is, it's $$id_{R^n} - \rho_k$$)

Then, identifying matrices with linear maps, we have the following commutative square :

$$\require{AMScd} \begin{CD}R^n @>A>> R^n \\ @A\iota_jAA @Vp_iVV\\ R^{n-1} @>A_{ij}>>R^{n-1}\end{CD}$$

This will be important later, because it lets us interpret $$\det(A_{ij})$$ : indeed take $$\bigwedge^{n-1}$$ of this diagram and you get :

$$\require{AMScd} \begin{CD}\bigwedge^{n-1}R^n @>\bigwedge^{n-1}A>> \bigwedge^{n-1}R^n \\ @A\bigwedge^{n-1}\iota_jAA @V\bigwedge^{n-1}p_iVV\\ \bigwedge^{n-1}R^{n-1} @>\det(A_{ij})>>\bigwedge^{n-1}R^{n-1}\end{CD}$$

Right, now let $$e_1,...,e_n$$ denote the standard basis of $$R^n$$ (I will let $$b_1,...,b_{n-1}$$ denote the one of $$R^{n-1}$$), and take any $$i$$; we have

$$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = (-1)^i \bw^n\phi (e_i \w e_1 \w... \w \hat{e_i} \w ... \w e_n) = (-1)^i \phi(e_i) \w \bw^{n-1}\phi(e_1 \w ... \w \hat{e_i} \w ... \w e_n)$$

where as usual, $$\hat{e_i}$$ means we omit $$e_i$$ from the term. Now any element of $$R^n$$ is a sum of its projections, which implies that $$\phi = \sum_j \pi_j \circ \phi$$. It follows that

$$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = (-1)^i \phi(e_i) \w \sum_j \bw^{n-1}(\pi_j \circ \phi)(e_1\w ... \w \hat{e_i} \w...\w e_n)$$

Moreover, $$(e_1,... \hat{e_i}, ..., e_n) = (\iota_i(b_1), ..., \iota_i(b_{n-1}))$$ so that

$$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = (-1)^i \phi(e_i) \w \sum_j \bw^{n-1}(\pi_j \circ \phi)\circ \bw^{n-1}\iota_i(b_1\w...\w b_{n-1}) = (-1)^i \phi(e_i) \w \sum_j \bw^{n-1}(\pi_j \circ \phi\circ \iota_i)(b_1\w...\w b_{n-1})$$

Now, note that $$\phi(e_i) = \sum_k a_{ki} e_k$$

Therefore $$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = \sum_{k,j} (-1)^i a_{ki} e_k\w \bw^{n-1}(\pi_j \circ \phi\circ \iota_i)(b_1\w...\w b_{n-1})$$

$$\pi_j$$ of anything is colinear to $$e_j$$, therefore if $$j=k$$, the term in the sum is $$0$$. So we may remove all the $$j=k$$ terms. Now note that for fixed $$k$$, $$\sum_{j\neq k} \pi_j = \rho_k$$.

Therefore our sum simplifies to

$$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = \sum_k (-1)^i a_{ki}e_k \w \bw^{n-1}(\rho_k\circ \phi \circ \iota_i) (b_1\w...\w b_{n-1})$$

We're almost there : $$\rho_k = \iota_k\circ p_k$$ as we mentioned earlier, so that

$$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = \sum_k (-1)^i a_{ki}e_k \w \bw^{n-1}\iota_k \circ \bw^{n-1}(p_k \circ \phi \circ \iota_i)(b_1\w...\w b_{n-1})$$

Our interpretation above yields that $$\bw^{n-1}(p_k \circ \phi \circ \iota_i)(b_1\w...\w b_{n-1}) = \det(A_{ki}) b_1\w...\w b_{n-1}$$; and $$\bw^{n-1}\iota_k (b_1\w...\w b_{n-1}) = e_1\w...\w \hat{e_k} \w ... \w e_n$$ so that

$$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = \sum_k (-1)^i a_{ki}\det(A_{ki}) e_k \w e_1\w...\w \hat{e_k} \w ... \w e_n = \sum_k (-1)^i a_{ki}\det(A_{ki}) (-1)^k e_1 \w... \w e_n$$

All in all :

$$\bigwedge^n\phi(e_1\wedge ... \wedge e_n) = \sum_k (-1)^{i+k} a_{ki} \det(A_{ki}) e_1\w ... \w e_n$$

and in particular, $$\det(A) = \sum_k (-1)^{i+k} a_{ki} \det(A_{ki})$$

What worries me a tad is that I get $$ji$$ instead of your $$ij$$. Now of course this isn't fundamentally a problem, as $$\det(A) = \det(A^T)$$, so we do get your formula in the end, but that's not the one you get if you just follow through the given proof. Hopefully I didn't mix things up along the way

As a passing (not so useful) comment, you can see how this is an instance of categorification : we have a completely concrete formula in terms of elements of $$R$$, and we interpret it as saying something about various maps (if you look at it closely, all the equalities I wrote can be interpreted as saying something about maps) instead of elements, and then the computations become more straightforward - to get back the concrete thing you decategorify at the end.

• Nicely done! The proof is correct, no mistake noticed from my side. Oct 8, 2019 at 0:57
• As for categorification, yeah, it seems like a nice example. I have personally found category theory to simply abstract algebra. For instance, the fact that adjoint functors preserve (co)limits allows us to prove that the basis of a tensor product of free modules is a tensor product of bases (tensor-hom adjunction) and how to compute the basis of an exterior power of a free module ((with a little work on the side, which is important on its own). Oct 8, 2019 at 1:03

I would like to present here an alternative categorification.

Step 1: defining the general adjugation process. We start by taking any free $$R$$-module $$M$$ of finite rank $$n$$, and observing that the natural map

$$\bigwedge^kM\otimes_R\bigwedge^{n-k}M\xrightarrow{\wedge}\bigwedge^nM$$

is a perfect pairing. We thus have a canonical isomorphism

$$\bigwedge^kM\xrightarrow{\phi}\text{Hom}\left(\bigwedge^{n-k}M,\bigwedge^nM\right).$$

So for any $$\alpha\in\bigwedge^kM$$ we have $$\phi(\alpha)=\phi_\alpha$$ is the morphism taking $$\beta\in\bigwedge^{n-k}M$$ and sending it to $$\alpha\wedge\beta$$:

$$\phi_\alpha(\beta)=\alpha\wedge\beta.$$

This enables us to define adjugate morphisms, i.e. for any endomorphism $$f:\bigwedge^{n-k}M\longrightarrow\bigwedge^{n-k}M$$ one has a morphism $$f^\dagger:\bigwedge^kM\longrightarrow\bigwedge^kM$$ verifying

$$\forall\alpha\in\bigwedge^kM,\beta\in\bigwedge^{n-k}M,\quad \alpha\wedge f(\beta)=f^\dagger(\alpha)\wedge\beta.$$

Step 2: fundamental lemma concerning general adjugates. We prove the following for any endomorphism $$f:M\longrightarrow M$$ and any $$k=0,\ldots,n$$

$$\left(\bigwedge^{n-k}f\right)^\dagger\circ\bigwedge^kf= \det(f)\cdot\text{Id}_{\wedge^kM}.$$

Indeed, write $$F$$ for the above composition and take any $$\alpha\in\bigwedge^kM,\beta\in\bigwedge^{n-k}M$$:

$$F(\alpha)\wedge\beta=\left(\bigwedge^kf\right)(\alpha)\wedge \left(\bigwedge^{n-k}f\right)(\beta)= \left(\bigwedge^nf\right)(\alpha\wedge\beta)=\det(f)\alpha\wedge\beta= \left(\det(f)\alpha\right)\wedge\beta,$$

thus, considering this equality is made for arbitrary $$\beta\in\bigwedge^{n-k}M$$, we have $$F(\alpha)=\det(f)\alpha$$.

Step 3: coming back down to the $$k=1$$ case and decategorification. The classical adjugate is defined for the special case $$k=1$$, where we identify $$M=\bigwedge^1M$$:

$$f^{ad}:=\left(\bigwedge^{n-1}f\right)^\dagger.$$

We now wish to show that going from $$f$$ to $$f^{ad}$$ corresponds at the matrix level, to going from a matrix $$A$$ to its adjugate. For this consider $$M=R^n$$ with canonical basis $$e_1,\ldots,e_n$$ and take $$A$$ the matrix of $$f$$ in this basis:

$$f(e_j)=\sum_{i=1}^nA_{i,j}e_i.$$

By a classic result on exterior products, we know that if we set

$$\epsilon_k=e_1\wedge\ldots\wedge e_{k-1}\wedge\widehat{e_k} \wedge e_{k+1}\wedge\ldots\wedge e_n,$$

then $$\epsilon_1,\ldots,\epsilon_n$$ forms a basis for $$\bigwedge^{n-1}M$$. Remark then that

$$e_k\wedge\epsilon_l=\left\{\begin{array}{ll} (-1)^{k-1}e_1\wedge\ldots\wedge e_n & \text{if }k=l\\ 0 & \text{otherwise} \end{array}\right.$$

We now wish to show how does the map $$\left(\bigwedge^{n-1}f\right)^\dagger$$ act on $$V$$. For this it suffices to show what happens on each base vector $$e_k$$, that we wedge with $$\epsilon_l$$:

\begin{align*} \left(\bigwedge^{n-1}f\right)^\dagger(e_k)\wedge\epsilon_l&=e_k\wedge \left(\bigwedge^{n-1}f\right)(\epsilon_l)\\ &=e_k\wedge f(e_1)\wedge\ldots\wedge f(e_{l-1})\wedge f(e_{l+1}) \wedge\ldots\wedge f(e_n)\\ &=(-1)^{k-1}\det(A^{(k,l)})e_1\wedge\ldots\wedge e_n. \end{align*}

where $$A^{(k,l)}$$ is the matrix obtained from $$A$$ by eliminating the $$k$$-th row and $$l$$-th column (this is just Maxime's second commutative square). So we clearly have

$$\left(\bigwedge^{n-1}f\right)^\dagger(e_k)=\sum_{l=1}^n(-1)^{k+l}\det(A^{(k,l)})e_l$$

thus showing that the matrix of $$f^{ad}$$ in the canonical basis is $$A^{ad}={}^t\left((-1)^{k+l}\det(A^{(k,l)})\right)_{1\leqslant k,l\leqslant n}$$, i.e. the classical adjugate matrix of $$A$$. By step two, we get that $$A^{ad}A=\det(A)I_n$$. This shows the classic Laplace expansion, by considering the computation of every $$(i,i)$$ coefficient in the product.

Remark: Suppose $$\det(A)\in R^\times$$, then we wish to show that $$A$$ is invertible. An application of Nakayama's lemma shows that if an endomorphism of fintely generated modules is surjective, then it is injective, and thus an isomorphism, so $$A^{ad}$$ is invertible considering $$A^{ad}A=\det(A)I_n$$ and since $$\det(A)I_n$$ is invertible. But then $$\det(A)^{-1}A$$ is a right inverse and by a simple computation, we can show that it is also a left inverse to $$A^{ad}$$. Thus we have also shown that $$A$$ is invertible.

Edit: In fact, Nakayama isn't necessary to conclude. Since $$\det(A)\in R^\times$$, then $$\det(A^{ad})\det(A)=\det(A)^n$$, so $$\det(A^{ad})=\det(A)^{n-1}$$ is also invertible, so the identity $$(A^{ad})^{ad}A^{ad}=\det(A^{ad})I_n$$ shows that $$A^{ad}$$ also has a left inverse. A simple computation shows that the two inverses are identical.

Any comments and errata are welcome. I hope this helps.

• That's interesting, in my mind $A^{ad}A = \det(A) I_n$ was a consequence of the Laplace expansion; I didn't realize you could do it the other way around with the perfect pairing ! By the way, wrt your remark : you don't need Nakayama's lemma. Indeed, $A^{ad}A = \det(A) I_n$ shows that $\det(A^{ad})$ is invertible, so by the same formula for $A^{ad}$, you see that $A^{ad}$ is also left invertible, and hence invertible; and hence so is $A$ Mar 24, 2021 at 11:56
• Oh yes you are right, since $(-)^\dagger$ is self-dual. How silly of me. I'll edit it right away. Mar 24, 2021 at 13:20
• It is true, but you don't even need the self-duality :D Mar 24, 2021 at 13:21