The Determinant and Alternating Forms

Introduction

The exterior powers of the companion article Exterior Powers carry, in their top degree, a one-dimensional module attached to a free module of finite rank, and an endomorphism acts on it by a scalar. That scalar is the determinant. The determinant is therefore not an ad hoc formula in the entries of a matrix but the unique scalar invariant of a linear transformation, obtained by transport along the top exterior power. The same construction in every intermediate degree produces the minors, and the multiplicativity of the determinant is a special case of the functoriality of $\Lambda^n$, while the formula relating the minors of a composite to those of its factors, known as the Cauchy–Binet formula, is the degreewise statement that $\Lambda^n(g \circ f) = \Lambda^n g \circ \Lambda^n f$.

The linear functionals on the exterior powers are the alternating forms, and the exterior algebra of the dual module is the algebraic prototype of the algebra of differential forms of Part III, which is named only and not used; the operations of the calculus — the wedge product, the pullback, and the exterior derivative — are visible already at the level of a single free module over a commutative ring, with no smooth structure required.

Throughout, $R$ denotes a commutative ring with identity $1 \neq 0$ and $M$, $N$, $P$ are $R$-modules; over a commutative ring left and right modules coincide, so no side is specified. The field is written $K$, and $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$ when only those two are meant. The determinant is defined over an arbitrary commutative ring, and the place where a field, a characteristic assumption, or the invertibility of $2$ is needed is flagged; in particular the sign-based expansions below require no division, while the Pfaffian and the identification of alternating with skew-symmetric forms require care in characteristic $2$. No physics is invoked.

Induced Maps and Minors

The Induced Map on Exterior Powers

Let $f : M \to N$ be an $R$-linear map and let $\Lambda^n f : \Lambda^n M \to \Lambda^n N$ be the induced map of Exterior Powers, characterised on elementary wedges by

$$ \Lambda^n f(x_1 \wedge \cdots \wedge x_n) = f(x_1) \wedge \cdots \wedge f(x_n). $$

Functoriality gives $\Lambda^n(\mathrm{id}_M) = \mathrm{id}_{\Lambda^n M}$ and $\Lambda^n(g \circ f) = \Lambda^n g \circ \Lambda^n f$, and these two identities are the source of the multiplicativity properties recorded below. In degree $1$ the induced map is $f$ itself, and in degree $0$ it is the identity of $R$.

Minors and the Matrix of $\Lambda^n f$

Suppose now that $M$ and $N$ are free with bases $e_1, \ldots, e_m$ and $f_1, \ldots, f_r$, and let $A = (a_{ij})$ be the matrix of $f$ in these bases, so that $f(e_j) = \sum_i a_{ij} f_i$. The induced bases of $\Lambda^n M$ and $\Lambda^n N$ consist of the increasing wedges, and the matrix of $\Lambda^n f$ in them is described by the minors of $A$.

Definition. Let $I \subseteq \{1, \ldots, r\}$ and $J \subseteq \{1, \ldots, m\}$ have the same cardinality $n$, say $I = (i_1 < \cdots < i_n)$ and $J = (j_1 < \cdots < j_n)$. The minor of $A$ with row set $I$ and column set $J$ is

$$ A_{I,J} = \det\bigl(a_{i_p, j_q}\bigr)_{p,q=1}^{n}, $$

the determinant of the $n \times n$ submatrix formed by those rows and columns. A minor is therefore an alternating function of the rows and of the columns separately.

Proposition. In the induced bases, the matrix entry of $\Lambda^n f$ from the basis wedge $e_J = e_{j_1} \wedge \cdots \wedge e_{j_n}$ to the basis wedge $f_I = f_{i_1} \wedge \cdots \wedge f_{i_n}$ is the minor $A_{I,J}$:

$$ \Lambda^n f(e_J) = \sum_{|I| = n} A_{I,J}\, f_I. $$

Proof. Expand each $f(e_{j_q}) = \sum_i a_{i,j_q} f_i$ and use multilinearity of the wedge: the coefficient of $f_{i_1} \wedge \cdots \wedge f_{i_n}$ is the alternating sum $\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_q a_{i_q, j_{\sigma(q)}}$, which is the Leibniz expansion of the minor.

The Determinant

Definition for Endomorphisms

Let $M$ be free of rank $m$ with basis $(e_1, \ldots, e_m)$. The top exterior power $\Lambda^m M$ is free of rank $1$, with the single basis element $e_1 \wedge \cdots \wedge e_m$, so every endomorphism of it is multiplication by an element of $R$. For $f \in \operatorname{End}_R(M)$, the induced map $\Lambda^m f$ is such an endomorphism, and one names its scalar.

Definition. Let $M$ be a free $R$-module of rank $m$ and let $f : M \to M$ be $R$-linear. The determinant of $f$ is the unique scalar $\det(f) \in R$ with

$$ \Lambda^m f(e_1 \wedge \cdots \wedge e_m) = \det(f)\, e_1 \wedge \cdots \wedge e_m $$

for a basis $(e_1, \ldots, e_m)$. By the proposition above it is independent of the choice of basis, and it is the alternating $m$-multilinear form

$$ \det(f) = \sum_{\sigma \in S_m} \operatorname{sgn}(\sigma) \prod_{q=1}^{m} a_{q, \sigma(q)}, $$

where $(a_{ij})$ is the matrix of $f$ in the basis.

The Determinant of a Matrix

Writing $\det(A)$ for the right-hand side, the determinant of a square matrix is

$$ \det(A) = \sum_{\sigma \in S_m} \operatorname{sgn}(\sigma)\, a_{1,\sigma(1)} a_{2,\sigma(2)} \cdots a_{m,\sigma(m)}. $$

This is the Leibniz formula. It is a polynomial with integer coefficients in the entries, hence defined over every commutative ring, and it is alternating in the columns and in the rows.

Proposition (basic expansions). For an $m \times m$ matrix $A$ over a commutative ring:

(a) $\det(A)$ is $R$-multilinear and alternating in the columns of $A$;

(b) $\det(I) = 1$ and $\det(A) = 0$ if two columns of $A$ are equal;

(c) $\det(A^{\mathsf{T}}) = \det(A)$.

Proof. (a) Each term of the Leibniz formula contains exactly one entry from each column, so the formula is linear in each column; exchanging two columns permutes the summation index $\sigma$ by a transposition and multiplies each term by $-1$. (b) is immediate from the formula. (c) The transpose relabels $\sigma$ by $\sigma^{-1}$, and $\operatorname{sgn}(\sigma^{-1}) = \operatorname{sgn}(\sigma)$.

Multiplicativity and Invertibility

Theorem. Let $M$ be free of rank $m$ and let $f, g \in \operatorname{End}_R(M)$. Then

$$ \det(g \circ f) = \det(g)\, \det(f). $$

Moreover $f$ is invertible if and only if $\det(f)$ is a unit of $R$, and then $\det(f^{-1}) = \det(f)^{-1}$.

Proof. Functoriality of the top exterior power gives $\Lambda^m(g \circ f) = \Lambda^m g \circ \Lambda^m f$, and on the rank-one module $\Lambda^m M$ these are multiplications by $\det(g \circ f)$, $\det(g)$, and $\det(f)$ respectively, so the scalars multiply. If $f$ is invertible, applying the first part to $g = f^{-1}$ and $g \circ f = \mathrm{id}$ gives $\det(f)\det(f^{-1}) = \det(\mathrm{id}) = 1$. Conversely, if $\det(f)$ is a unit, then the adjugate identity $f \circ \operatorname{adj}(f) = \det(f)\,\mathrm{id}_M$ exhibits an inverse: $\operatorname{adj}(f)$ is the endomorphism whose matrix entries are the $(m-1)$-minors of the matrix of $f$, with alternating signs, and it is defined over any commutative ring.

Remark. Over a general commutative ring one cannot conclude that a non-invertible endomorphism has $\det(f) = 0$: only that $\det(f)$ is a non-unit. For example, over $R = \mathbb{Z}$ the map of $\mathbb{Z}$ given by multiplication by $2$ has determinant $2$, which is neither a unit nor zero. This is the point at which the determinant over a ring differs from the determinant over a field.

The Characteristic Polynomial

Proposition. Let $M$ be free of rank $m$ and let $f \in \operatorname{End}_R(M)$. Then

$$ \det(\mathrm{id}_M + t f) = \sum_{k=0}^{m} \operatorname{tr}(\Lambda^k f)\, t^k, $$

where $\operatorname{tr}(\Lambda^k f)$ denotes the trace of the induced endomorphism of $\Lambda^k M$ and $\operatorname{tr}(\Lambda^0 f) = 1$.

Proof. Both sides are polynomials with integer coefficients in the matrix entries of $f$. It therefore suffices to prove the identity after base change to a field over which the matrix has a full set of eigenvalues, for instance over the algebraic closure of the fraction field of the polynomial ring in the entries. There $f$ is triangularisable, with eigenvalues $\lambda_1, \ldots, \lambda_m$; the eigenvalues of $\Lambda^k f$ are the products $\lambda_{i_1} \cdots \lambda_{i_k}$ over increasing $k$-tuples, so $\operatorname{tr}(\Lambda^k f)$ is the elementary symmetric function $e_k(\lambda)$ of the eigenvalues and $\det(\mathrm{id} + tf) = \prod_{i=1}^{m}(1 + t\lambda_i) = \sum_{k=0}^{m} e_k(\lambda) t^k$.

Setting $t = -1/s$ and clearing denominators recovers the characteristic polynomial $\det(s\,\mathrm{id} - f) = s^m \sum_k \operatorname{tr}(\Lambda^k f)(-1/s)^k$; the coefficients are the traces of the induced maps on the exterior powers.

Alternating Forms

Forms and the Exterior Algebra of the Dual

Definition. An alternating form of degree $n$ on $M$ with values in $R$ is an alternating $n$-multilinear map $M^n \to R$. The $R$-module of such forms is

$$ \operatorname{Alt}^n(M; R) \cong (\Lambda^n M)^*, $$

the last identification being the universal property of the exterior power. Assembling the degrees,

$$ \operatorname{Alt}^\bullet(M; R) = \bigoplus_{n \geq 0} \operatorname{Alt}^n(M; R) $$

is the algebra of alternating forms on $M$.

Proposition. Let $M$ be free of finite rank. Then for each $n$ there is a natural isomorphism $\operatorname{Alt}^n(M; R) \cong \Lambda^n(M^*)$, and these assemble into an isomorphism of graded algebras

$$ \operatorname{Alt}^\bullet(M; R) \cong \Lambda(M^*) $$

between the algebra of alternating forms under the wedge product and the exterior algebra of the dual module.

Proof. A form of degree $n$ corresponds to a linear functional on $\Lambda^n M$, hence to an element of $(\Lambda^n M)^*$; over a free module of finite rank the natural map $\Lambda^n(M^*) \to (\Lambda^n M)^*$ sending $\varphi_1 \wedge \cdots \wedge \varphi_n$ to the form $(x_1, \ldots, x_n) \mapsto \det(\varphi_p(x_q))$ is an isomorphism, by the basis theorem of Exterior Powers applied to the dual basis. Multiplicativity of the correspondence is the definition of the wedge product of forms.

Definition. The wedge product of forms $\alpha \wedge \beta \in \operatorname{Alt}^{p+q}(M; R)$ of forms $\alpha$ of degree $p$ and $\beta$ of degree $q$ is the form

$$ (\alpha \wedge \beta)(x_1, \ldots, x_{p+q}) = \sum_{\sigma} \operatorname{sgn}(\sigma)\, \alpha(x_{\sigma(1)}, \ldots, x_{\sigma(p)})\, \beta(x_{\sigma(p+1)}, \ldots, x_{\sigma(p+q)}), $$

the sum over the $(p, q)$-shuffles of $\{1, \ldots, p+q\}$. It is graded-commutative, $\alpha \wedge \beta = (-1)^{pq}\beta \wedge \alpha$, and associative.

Pullback

Definition. Let $f : M \to N$ be $R$-linear. The pullback along $f$ is the algebra homomorphism

$$ f^* : \operatorname{Alt}^\bullet(N; R) \to \operatorname{Alt}^\bullet(M; R), \qquad (f^*\omega)(x_1, \ldots, x_n) = \omega(f(x_1), \ldots, f(x_n)). $$

Proposition. The pullback is contravariantly functorial: $\mathrm{id}^* = \mathrm{id}$ and $(g \circ f)^* = f^* \circ g^*$.

Proof. Both are immediate from the definition; the reversal of order is the usual contravariance of a hom functor.

The Cauchy–Binet Formula

The multiplicativity of the determinant in the square case is the $n = m$ case of a formula valid in every degree.

Theorem (Cauchy–Binet). Let $A$ be an $r \times m$ matrix and $B$ an $m \times s$ matrix over a commutative ring $R$. For increasing index sets $I \subseteq \{1, \ldots, r\}$ and $J \subseteq \{1, \ldots, s\}$ of size $n$,

$$ \det\bigl((AB)_{I,J}\bigr) = \sum_{K} \det(A_{I,K})\, \det(B_{K,J}), $$

the sum over the increasing $n$-element subsets $K \subseteq \{1, \ldots, m\}$.

Proof. Let $f : R^s \to R^m$ and $g : R^m \to R^r$ be the linear maps with matrices $B$ and $A$, so that $AB$ is the matrix of $g \circ f$. By the minor proposition, the matrix entries of $\Lambda^n(g \circ f)$ are the minors of $AB$, and by functoriality $\Lambda^n(g \circ f) = \Lambda^n g \circ \Lambda^n f$, whose matrix entries are sums of products of the minors of $A$ and of $B$ indexed by the intermediate basis wedges. Equating the two expressions gives the formula.

Corollary. If $m = r = s = n$, Cauchy–Binet reduces to $\det(AB) = \det(A)\det(B)$.

Corollary (Jacobi's complementary minor identity). Let $A$ be an invertible $m \times m$ matrix over a commutative ring, with inverse $A^{-1}$. For increasing index sets $I, J \subseteq \{1, \ldots, m\}$ of size $n$, written $I^c$ and $J^c$ for their complements,

$$ \det\bigl((A^{-1})_{I,J}\bigr) = (-1)^{\textstyle\sum I + \sum J}\, \det(A_{J^c, I^c}) \cdot \det(A)^{-1}, $$

where $\sum I$ denotes the sum of the entries of $I$. Thus the $n \times n$ minors of the inverse are the complementary $(m-n) \times (m-n)$ minors of $A$, up to a sign and the scalar $\det(A)^{-1}$.

This is the classical complementary-minor theorem of Jacobi; it is the statement that the compound matrix of $A^{-1}$ is the inverse of the compound matrix of $A$, combined with the description of the inverse of a compound matrix through complementary minors. When $n = m$ it reduces to $\det(A^{-1}) = \det(A)^{-1}$.

The Top Power, Volume and the Pfaffian

Definition. Let $V$ be a free $K$-module of finite rank $m$. A volume form on $V$ is a nonzero element of the one-dimensional space $\Lambda^m V$; a choice of volume form is a choice of basis of the determinant line $\Lambda^m V$, of which it records the direction but not the scale, and the determinant of an automorphism is the factor by which it scales the volume form.

Proposition (the determinant as the volume form). Let $\omega$ be a nonzero alternating $m$-form on $V$ and let $(e_1, \ldots, e_m)$ be a basis with $\omega(e_1, \ldots, e_m) = 1$. Then for every endomorphism $f$ of $V$,

$$ \omega(f(e_1), \ldots, f(e_m)) = \det(f). $$

Proof. The form $\omega$ is a basis of the one-dimensional space of alternating $m$-forms; the pullback $f^*\omega = \det(f)\omega$ by definition of the determinant, and evaluating at the basis gives the claim.

Example (alternating forms of degree $2$). An alternating $2$-form on a free module with basis $(e_1, \ldots, e_m)$ is determined by its values $\omega_{ij} = \omega(e_i, e_j)$ on pairs, and these satisfy $\omega_{ij} = -\omega_{ji}$ and $\omega_{ii} = 0$. In matrix form $\Omega = (\omega_{ij})$ is alternating, $\Omega^{\mathsf{T}} = -\Omega$, and the form is

$$ \omega(x, y) = x^{\mathsf{T}} \Omega\, y. $$

Such a form is the matrix object studied. When $m = 2n$ is even and $\Omega$ is alternating and invertible, its determinant is a square, $\det(\Omega) = \operatorname{Pf}(\Omega)^2$, where the Pfaffian $\operatorname{Pf}(\Omega)$ is the polynomial

$$ \operatorname{Pf}(\Omega) = \frac{1}{2^n n!} \sum_{\sigma \in S_{2n}} \operatorname{sgn}(\sigma) \prod_{i=1}^{n} \Omega_{\sigma(2i-1), \sigma(2i)}. $$

The Pfaffian and its relation to the symplectic form are treated. Over a ring in which $2^n n!$ is invertible, for instance a $\mathbb{Q}$-algebra or a field of characteristic zero, the displayed formula is directly available; the summand is in fact divisible by $2^n n!$ in $\mathbb{Z}[\Omega_{ij}]$, so the Pfaffian is an integral polynomial and is defined over every commutative ring. The identity $\det(\Omega) = \operatorname{Pf}(\Omega)^2$ holds over every commutative ring because both sides are integral polynomials and the identity holds over $\mathbb{Q}$.

Summary

The determinant of an endomorphism of a free module of rank $m$ is the scalar by which it acts on the top exterior power $\Lambda^m M$; in a basis it is the Leibniz sum $\sum_\sigma \operatorname{sgn}(\sigma)\prod_q a_{q,\sigma(q)}$. It is multiplicative, $\det(g \circ f) = \det(g)\det(f)$, it satisfies $\det(f^{\mathsf{T}}) = \det(f)$, and an endomorphism is invertible exactly when its determinant is a unit of the base ring. The characteristic polynomial is governed by the traces of the induced maps on the exterior powers, $\det(\mathrm{id} + tf) = \sum_k \operatorname{tr}(\Lambda^k f)t^k$.

In every degree the induced map $\Lambda^n f$ has matrix the $n \times n$ minors of the matrix of $f$, and the identity $\Lambda^n(g \circ f) = \Lambda^n g \circ \Lambda^n f$ reads, in entries, as the Cauchy–Binet formula

$$ \det\bigl((AB)_{I,J}\bigr) = \sum_K \det(A_{I,K})\det(B_{K,J}). $$

The alternating forms of degree $n$ are the linear functionals on $\Lambda^n M$, equivalently, for free $M$ of finite rank, the elements of $\Lambda^n(M^*)$; they form the graded-commutative algebra $\Lambda(M^*)$, with wedge product and contravariant pullback. This algebra is the algebraic prototype of the algebra of differential forms of Part III, and the determinant is the factor by which a linear map scales a top-degree form. For an alternating matrix of even size the determinant is the square of the Pfaffian.

Summary of Notation

Symbol	Meaning
$R$	Commutative ring with identity $1 \neq 0$; the default base
$K$	Field
$M$, $N$, $P$	$R$-modules, free of finite rank where a determinant is taken
$\Lambda^n M$	$n$-th exterior power
$\Lambda^n f$	Induced map on exterior powers
$e_1, \ldots, e_m$; $f_1, \ldots, f_r$	Bases of $M$ and of $N$ in which matrices and minors are written
$A_{I,J}$	Minor of a matrix on row set $I$ and column set $J$
$\det(f)$, $\det(A)$	Determinant: scalar action on $\Lambda^m M$, Leibniz sum in a basis
$\operatorname{End}_R(M)$	$R$-linear endomorphisms of $M$; the home of $\det$, $\operatorname{tr}$ and $\operatorname{adj}$
$\operatorname{tr}$	Trace; $\operatorname{tr}(\Lambda^k f)$ is the trace on $\Lambda^k M$
$\operatorname{adj}(f)$	Adjugate; $f \circ \operatorname{adj}(f) = \det(f)\,\mathrm{id}_M$
$\det(\mathrm{id} + tf) = \sum_k \operatorname{tr}(\Lambda^k f)t^k$	Characteristic-polynomial identity
$\operatorname{Alt}^n(M; R) \cong (\Lambda^n M)^*$	Alternating $n$-forms on $M$
$\operatorname{Alt}^\bullet(M; R) \cong \Lambda(M^*)$	Algebra of alternating forms, for free $M$ of finite rank
$\alpha \wedge \beta$	Wedge product of forms; graded-commutative
$f^*$	Pullback of forms, contravariant: $(g \circ f)^* = f^* \circ g^*$
$\Omega = (\omega_{ij})$	Matrix of an alternating $2$-form, $\Omega^{\mathsf{T}} = -\Omega$
$\operatorname{Pf}(\Omega)$	Pfaffian of an alternating matrix of even size; $\det(\Omega) = \operatorname{Pf}(\Omega)^2$