# Contents

## Idea

The determinant is the (essentially unique) universal alternating multilinear map.

## Definition

### Preliminaries on exterior algebra

Let Vect${}_k$ be the category of vector spaces over a field $k$, and assume for the moment that the characteristic $char(k) \neq 2$. For each $j \geq 0$, let

$sgn_j \colon S_j \to \hom(k, k)$

be the 1-dimensional sign representation? on the symmetric group $S_j$, taking each transposition $(i j)$ to $-1 \in k^\ast$. We may linearly extend the sign action of $S_j$, so that $sgn$ names a (right) $k S_j$-module with underlying vector space $k$. At the same time, $S_j$ acts on the $j^{th}$ tensor product of a vector space $V$ by permuting tensor factors, giving a left $k S_j$-module structure on $V^{\otimes j}$. We define the Schur functor

$\Lambda^j \colon Vect_k \to Vect_k$

by the formula

$\Lambda^j(V) = sgn_j \otimes_{k S_j} V^{\otimes j}.$

It is called the $j^{th}$ alternating power (of $V$).

###### Proposition

If $V$ is $n$-dimensional, then $\Lambda^j(V)$ has dimension $\binom{n}{j}$. In particular, $\Lambda^n(V)$ is 1-dimensional.

###### Proof

If $e_1, \ldots, e_n$ is a basis for $V$, then expressions of the form $e_{n_1} \otimes \ldots \otimes e_{n_j}$ form a basis for $V^{\otimes j}$. Let $e_{n_1} \wedge \ldots \wedge e_{n_j}$ denote the image of this element under the quotient map $V^{\otimes j} \to \Lambda^j(V)$. We have

$e_{n_1} \wedge \ldots \wedge e_{n_i} \wedge e_{n_{i+1}} \wedge \ldots \wedge e_{n_j} = -e_{n_1} \wedge \ldots \wedge e_{n_{i+1}} \wedge e_{n_i} \wedge \ldots \wedge e_{n_j}$

(consider the transposition in $S_j$ which swaps $i$ and $i+1$) and so we may take only such expressions on the left where $n_1 \lt \ldots \lt n_j$ as forming a spanning set for $\Lambda^j(V)$, and indeed these form a basis. The number of such expressions is $\binom{n}{j}$.

###### Remark

In the case where $char(k) = 2$, the same development may be carried out by simply decreeing that $e_{n_1} \wedge \ldots \wedge e_{n_j} = 0$ whenever $n_i = n_{i'}$ for some pair of distinct indices $i$, $i'$.

Now let $V$ be an $n$-dimensional space, and let $f \colon V \to V$ be a linear map. By the proposition, the map

$\Lambda^n(f) \colon \Lambda^n(V) \to \Lambda^n(V),$

being an endomorphism on a 1-dimensional space, is given by multiplying by a scalar $D(f) \in k$. It is manifestly functorial since $\Lambda^n$ is, i.e., $D(f g) = D(f) D(g)$. The quantity $D(f)$ is called the determinant of $f$.

### Determinant of a matrix

We see then that if $V$ is of dimension $n$,

$\det \colon End(V) \to k$

is a homomorphism of multiplicative monoids; by commutativity of multiplication in $k$, we infer that

$\det(U A U^{-1}) = \det(A)$

for each invertible linear map $U \in GL(V)$.

If we choose a basis of $V$ so that we have an identification $GL(V) \cong Mat_n(k)$, then the determinant gives a function

$\det \colon Mat_n(k) \to k$

that takes products of $n \times n$ matrices to products in $k$. The determinant however is of course independent of choice of basis, since any two choices are related by a change-of-basis matrix $U$, where $A$ and its transform $U A U^{-1}$ have the same determinant.

By following the definitions above, we can give an explicit formula:

$\det(A) = \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i = 1}^n a_{i \sigma(i)}.$

## Properties

We work over fields of arbitrary characteristic. The determinant satisfies the following properties, which taken together uniquely characterize the determinant. Write a square matrix $A$ as a row of column vectors $(v_1, \ldots, v_n)$.

1. $\det$ is separately linear in each column vector:

$\det(v_1, \ldots, a v + b w, \ldots, v_n) = a\det(v_1, \ldots, v, \ldots, v_n) + b\det(v_1, \ldots, w, \ldots, v_n)$
2. $\det(v_1, \ldots, v_n) = 0$ whenever $v_i = v_j$ for distinct $i, j$.

3. $\det(I) = 1$, where $I$ is the identity matrix.

Other properties may be worked out, starting from the explicit formula or otherwise:

• If $A$ is a diagonal matrix, then $\det(A)$ is the product of its diagonal entries.

• More generally, if $A$ is an upper (or lower) triangular matrix, then $\det(A)$ is the product of the diagonal entries.

• If $E/k$ is an extension field and $f$ is a $k$-linear map $V \to V$, then $\det(f) = \det(E \otimes_k f)$. Using the preceding properties and the Jordan normal form? of a matrix, this means that $\det(f)$ is the product of its eigenvalues (counted with multiplicity), as computed in the algebraic closure of $k$.

• If $A^t$ is the transpose of $A$, then $\det(A^t) = \det(A)$.

### Cramer’s rule

A simple observation which flows from these basic properties is

###### Proposition

(Cramer’s Rule)

Let $v_1, \ldots, v_n$ be column vectors of dimension $n$, and suppose

$w = \sum_j a_j v_j.$

Then for each $i$ we have

$a_j \det(v_1, \ldots, v_i, \ldots, v_n) = \det(v_1, \ldots, w, \ldots, v_n)$

where $w$ occurs as the $i^{th}$ column vector on the right.

###### Proof

This follows straightforwardly from properties 1 and 2 above.

For instance, given a square matrix $A$ such that $\det(A) \neq 0$, and writing $A = (v_1, \ldots, v_n)$, this allows us to solve the equation

$A \cdot a = w$

and we easily conclude that $A$ is invertible if $\det(A) \neq 0$.

###### Remark

This holds true even if we replace the field $k$ by an arbitrary commutative ring $R$, and we replace the condition $\det(A) \neq 0$ by the condition that $\det(A)$ is a unit. (The entire development given above goes through, mutatis mutandis.)

### Characteristic polynomial and Cayley-Hamilton theorem

Given a linear endomorphism $f: M\to M$ of a finite rank free unital module over a commutative unital ring, one can consider the zeros of the characteristic polynomial $\det(t \cdot 1_V - f)$. The coefficients of the polynomial are the concern of the Cayley-Hamilton theorem.

### Over the real numbers: volume and orientation

A useful intuition to have for determinants of real matrices is that they measure change of volume. That is, an $n \times n$ matrix with real entries will map a standard unit cube in $\mathbb{R}^n$ to a parallelpiped in $\mathbb{R}^n$ (quashed to lie in a hyperplane if the matrix is singular), and the determinant is, up to sign, the volume of the parallelpiped. It is easy to convince oneself of this in the planar case by a simple dissection of a parallelogram, rearranging the dissected pieces in the style of Euclid to form a rectangle. In algebraic terms, the dissection and rearrangement amount to applying shearing or elementary column operations to the matrix which, by the properties discussed earlier, leave the determinant unchanged. These operations transform the matrix into a diagonal matrix whose determinant is the area of the corresponding rectangle. This procedure easily generalizes to $n$ dimensions.

The sign itself is a matter of interest. An invertible transformation $f \colon V \to V$ is said to be orientation-preserving if $\det(f)$ is positive, and orientation-reversing if $\det(f)$ is negative. Orientations play an important role throughout geometry and algebraic topology, for example in the study of orientable manifolds (where the tangent bundle as $GL(n)$-bundle can be lifted to a $GL_+(n)$-bundle structure, $GL_+(n) \hookrightarrow GL(n)$ being the subgroup of matrices of positive determinant). See also KO-theory.

Finally, we include one more property of determinants which pertains to matrices with real coefficients (which works slightly more generally for matrices with coefficients in a local field):

• If $A$ is an $n \times n$ matrix, then $\det(\exp(A)) =$ $\exp($trace$(A))$

## In terms of Berezinian integrals

see Pfaffian for the moment

## References

The proof of Proposition 3 on surjective endomorphisms of finitely generated modules was extracted from

• Stacks Project, Commutative Algebra, section 13 (pdf)

Revised on August 25, 2014 02:26:57 by Urs Schreiber (82.113.121.148)