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 2


$\newcommand{\bw}{\bigwedge} \newcommand{\w}{\wedge}$ 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.

  • $\begingroup$ Nicely done! The proof is correct, no mistake noticed from my side. $\endgroup$
    – Jxt921
    Oct 8, 2019 at 0:57
  • 1
    $\begingroup$ 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). $\endgroup$
    – Jxt921
    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


is a perfect pairing. We thus have a canonical isomorphism


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 $\alpha\wedge\beta$:


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 $M=\bigwedge^1M$:


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:


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


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.

  • 1
    $\begingroup$ 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 $A$ $\endgroup$ Mar 24, 2021 at 11:56
  • $\begingroup$ Oh yes you are right, since $(-)^\dagger$ is self-dual. How silly of me. I'll edit it right away. $\endgroup$
    – Andrei.B
    Mar 24, 2021 at 13:20
  • 1
    $\begingroup$ It is true, but you don't even need the self-duality :D $\endgroup$ Mar 24, 2021 at 13:21

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