Determinant

2020 Mathematics Subject Classification: Primary: 15-XX [MSN][ZBL]

The determinant of a square matrix $A = (a_{ij})$ of order $n$ over a commutative associative ring $R$ with unit 1 is the element of $R$ equal to the sum of all terms of the form

$$(-1)^k a_{1i_1}\cdots a_{ni_n},$$ where $i_1,\dots,i_n$ is a permutation of the numbers $1,\dots,n$ and $k$ is the number of inversions of the permutation $1\mapsto i_1,\dots,n\mapsto i_n$, so that $(-1)^k$ is the signature of this permutation.. The determinant of the matrix

$$A=\begin{pmatrix}a_{11} & \dots & a_{1n}\\ \vdots & \ddots & \vdots\\ a_{n1} & \dots & a_{nn} \end{pmatrix}$$ is written as

$$\begin{vmatrix}a_{11} & \dots & a_{1n}\\ \vdots & \ddots & \vdots\\ a_{n1} & \dots & a_{nn} \end{vmatrix} \textrm{ or } \det A.$$ The determinant of the matrix $A$ contains $n!$ terms. When $n=1$, $\det A = a_{11}$, when $n=2$, $\det A = a_{11}a_{22} - a_{21}a_{12}$. The most important instances in practice are those in which $R$ is a field (especially a number field), a ring of functions (especially a ring of polynomials) or a ring of integers.

From now on, $R$ is a commutative associative ring with 1, $\def\Mn{\textrm{M}_n(R)}\Mn$ is the set of all square matrices of order $n$ over $R$ and $E_n$ is the identity matrix over $R$. Let $A\in\Mn$, while $a_1,\dots,a_n$ are the rows of the matrix $A$. (All that is said from here on is equally true for the columns of $A$.) The determinant of $A$ can be considered as a function of its rows:

$$\det A = D(a_1,\dots,a_n).$$ The mapping

$$d:\Mn\to R\quad(A\mapsto \det A)$$ is subject to the following three conditions:

1) $d$ is a linear function of any row of $A$: $$\def\l{\lambda}\def\m{\mu}D(a_1,\dots,\l a_i+\m b_i,\dots,a_n) = \l D(a_1,\dots,a_i,\dots,a_n) + \m D(a_1,\dots,b_i,\dots,a_n),$$ where $\l,\m\in R$;

2) if the matrix $B$ is obtained from $A$ by replacing a row $a_i$ by a row $a_i+a_j$, $i\ne j$, then $d(A)=d(b)$;

3) $d(E_n) = 1$.

Conditions 1)–3) uniquely define $d$, i.e. if a mapping $h:\Mn\to R$ satisfies conditions 1)–3), then $h(A) = \det(A)$. An axiomatic construction of the theory of determinants is obtained in this way.

Let a mapping $f:\Mn\to R$ satisfy the condition:

$1'$) if $B$ is obtained from $A$ by multiplying one row by $\l\in R$, then $f(B)=\l f(A)$. Clearly 1) implies $1'$). If $R$ is a field, the conditions 1)–3) prove to be equivalent to the conditions $1'$), 2), 3).

The determinant of a diagonal matrix is equal to the product of its diagonal entries. The surjectivity of the mapping $d:\Mn\to R$ follows from this. The determinant of a triangular matrix is also equal to the product of its diagonal entries. For a matrix

$$A=\begin{pmatrix}B&0\\D&C\end{pmatrix}\;,$$ where $B$ and $C$ are square matrices,

$$\det A = \det B \det C.$$ It follows from the properties of transposition that $\det A^t = \det A$, where ${}^t$ denotes transposition. If the matrix $A$ has two identical rows, its determinant equals zero; if two rows of a matrix $A$ change places, then its determinant changes its sign;

$$D(a_1,\dots,a_i+\l a_j,\dots,a_n) = D(a_1,\dots,a_i,\dots,a_n)$$ when $i\ne j$, $\l\in R$; for $A$ and $B$ from $\Mn$,

$$\det (AB) = (\det A)(\det B).$$ Thus, $d$ is an epimorphism of the multiplicative semi-groups $\Mn$ and $R$.

Let $m\le n$, let $A=(a_{ij})$ be an $(m\times n)$-matrix, let $B=(b_{ij})$ be an $(n\times m)$-matrix over $R$, and let $C=AB$. Then the Binet–Cauchy formula holds:

$$\det C = \sum_{1\le j_1<\cdots<j_m\le n} \begin{vmatrix} a_{ij_1}&\dots&a_{ij_m}\\ \vdots&\ddots&\vdots\\ a_{mj_1}&\dots&a_{mj_m} \end{vmatrix} \begin{vmatrix} b_{j_11}&\dots&a_{j_1m}\\ \vdots&\ddots&\vdots\\ b_{j_m1}&\dots&a_{j_mm} \end{vmatrix} $$ Let $A=(a_{ij})\in \Mn$, and let $A_{ij}$ be the cofactor of the entry $a_{ij}$. The following formulas are then true:

$$\begin{equation}\left.\begin{aligned} \sum_{j=1}^n a_{ij}A_{kj} &= \delta_{ik} \det A\\ \sum_{i=1}^n a_{ij}A_{ik} &= \delta_{jk} \det A \end{aligned}\right\}\label{1}\end{equation}$$ where $\delta_{ij}$ is the Kronecker symbol. Determinants are often calculated by development according to the elements of a row or column, i.e. by the formulas (1), by the Laplace theorem (see Cofactor) and by transformations of $A$ which do not alter the determinant. For a matrix $A$ from $\Mn$, the inverse matrix $A^{-1}$ in $\Mn$ exists if and only if there is an element in $R$ which is the inverse of $\det A$. Consequently, the mapping

$$\def\GL{\textrm{GL}}\GL(n.K)\to K^*\quad (A\mapsto \det A).$$ where $\GL(n,K)$ is the group of all invertible matrices in $\Mn$ (i.e. the general linear group) and where $K^*$ is the group of invertible elements in $K$, is an epimorphism of these groups.

A square matrix over a field is invertible if and only if its determinant is not zero. The $n$-dimensional vectors $a_1,\dots,a_n$ over a field $F$ are linearly dependent if and only if

$$D(a_1,\dots,a_n) = 0.$$ The determinant of a matrix $A$ of order $N>1$ over a field is equal to 1 if and only if $A$ is the product of elementary matrices of the form

$$x_{ij}(\l) = E_n+\l e_{ij},$$ where $i\ne j$, while $e_{ij}$ is a matrix with its only non-zero entries equal to 1 and positioned at $(i,j)$.

The theory of determinants was developed in relation to the problem of solving systems of linear equations:

$$\begin{equation}\left.\begin{aligned} a_{11}x_1+\cdots+a_{1n}x_n &=b_1\\ \cdots &\\ a_{n1}x_1+\cdots+a_{nn}x_n &=b_n\\ \end{aligned}\right\}\label{2}\end{equation}$$ where $a_{ij}, b_j$ are elements of the field $R$. If $\det A\ne 0$, where $A=(a_{ij})$ is the matrix of the system (2), then this system has a unique solution, which can be calculated by Cramer's formulas (see Cramer rule). When the system (2) is given over a ring $R$ and $\det A$ is invertible in $R$, the system also has a unique solution, also given by Cramer's formulas.

A theory of determinants has also been constructed for matrices over non-commutative associative skew-fields. The determinant of a matrix over a skew-field $k$ (the Dieudonné determinant) is introduced in the following way. The skew-field $k$ is considered as a semi-group, and its commutative homomorphic image $\bar k$ is formed. $k$ consists of a group, $k^*$, with added zero 0, while the role of $\bar k$ is taken by the group $\overline{k^*}$ with added zero $\bar 0$, where $\overline{k^*}$ is the quotient group of $k^*$ by the commutator subgroup. The epimorphism $k\to \bar k$, $\l \mapsto \bar\l$, is given by the canonical epimorphism of groups $k^*\to \overline{k^*}$ and by the condition $0\to \bar0$. Clearly, $\bar 1$ is the unit of the semi-group $\bar k$.

The theory of determinants over a skew-field is based on the following theorem: There exists a unique mapping

$$\delta:\textrm{M}_n(k)\to \bar k$$ satisfying the following three axioms:

I) if the matrix $B$ is obtained from the matrix $X$ by multiplying one row from the left by $\l \in k$, then $\delta(B) = \bar\l \delta(A)$;

II) if $B$ is obtained from $A$ by replacing a row $a_i$ by a row $a_i+a_j$, where $i\ne j$, then $\delta(B)=\delta(A)$;

III) $\delta(E_n)=\bar1$.

The element $\delta(A)$ is called the determinant of $A$ and is written as $\det A$. For a commutative skew-field, axioms I), II) and III) coincide with conditions $1'$), 2) and 3), respectively, and, consequently, in this instance ordinary determinants over a field are obtained. If $A=\textrm{diag}(a_{11},\dots,a_{nn})$, then $\det A = a_{11}\cdots a_{nn}$; thus, the mapping $\delta:\textrm{M}_n(k)\to \bar k$ is surjective. A matrix $A$ from $\textrm{M}_n(k)$ is invertible if and only if $\det A \ne 0$. The equation $\det AB = (\det A)(\det B)$ holds. As in the commutative case, $\det A$ will not change if a row $a_i$ of $A$ is replaced by a row $a_i+\l a_j$, where $i\ne j$, $\l\in k$. If $n>1$, $\det A = \bar1$ if and only if $A$ is the product of elementary matrices of the form $x_{ij} = E_n+\l e_{ij}$, $i\ne j$, $\l\in k$. If $a\ne 0$, then

$$\begin{vmatrix}a&b\\c&d\end{vmatrix} = \overline{ad-aca^{-1}b},\quad \begin{vmatrix}0&b\\c&d\end{vmatrix} = -\overline{cd}. $$ Unlike the commutative case, $\det A^t$ does not have to coincide with $\det A$. For example, for the matrix

$$A=\begin{pmatrix}i&j\\k&-1\end{pmatrix}$$ over the skew-field of quaternions (cf. Quaternion), $\det A = -\overline{2i}$, while $\det A^t = \bar0$.

Infinite determinants, i.e. determinants of infinite matrices, are defined as the limit towards which the determinant of a finite submatrix converges when its order is growing infinitely. If this limit exists, the determinant is called convergent; in the opposite case it is called divergent.

The concept of a determinant goes back to G. Leibniz (1678). H. Cramer was the first to publish on the subject (1750). The theory of determinants is based on the work of A. Vandermonde, P. Laplace, A.L. Cauchy and C.G.J. Jacobi. The term "determinant" was first coined by C.F. Gauss (1801). The modern meaning was introduced by A. Cayley (1841).

References

[Ar]	E. Artin, "Geometric algebra", Interscience (1957) MR0082463 Zbl 0077.02101
[Bo]	N. Bourbaki, "Elements of mathematics. Algebra: Algebraic structures. Linear algebra", 1, Addison-Wesley (1974) pp. Chapt.1;2 (Translated from French) MR0354207
[Di]	J.A. Dieudonné, "La géométrie des groups classiques", Springer (1955)
[EfRo]	N.V. Efimov, E.R. Rozendorn, "Linear algebra and multi-dimensional geometry", Moscow (1970) (In Russian)
[HoKu]	K. Hoffman, R. Kunze, "Linear algebra", Prentice-Hall (1961) MR0125849
[Ka]	V.F. Kagan, "Foundations of the theory of determinants", Odessa (1922) (In Russian)
[Ko]	A.I. Kostrikin, "Introduction to algebra", Springer (1982) (Translated from Russian) MR0661256 Zbl 0482.00001
[Ko2]	M. Koecher, "Lineare Algebra und analytische Geometrie", Springer (1983) MR0725166 Zbl 0517.15001
[Ku]	A.G. Kurosh, "Higher algebra", MIR (1972) (Translated from Russian) Zbl 0237.13001
[La]	S. Lang, "Linear algebra", Addison-Wesley (1970) Zbl 0216.06001
[TyFe]	R.I. Tyshkevich, A.S. Fedenko, "Linear algebra and analytic geometry", Minsk (1976) (In Russian)