The Special Linear Group and the Determinant

Introduction

The determinant is the unique way to assign a scalar to a linear operator so that the assignment is multiplicative and is normalised to give $1$ on the identity. That uniqueness is what makes it an invariant of the operator rather than an artefact of a basis, and it is why the group of operators with determinant $1$ is singled out: this special linear group is the measure-preserving part of the general linear group, and over a field it is the commutator subgroup. This article constructs the determinant as a multilinear alternating form, proves its uniqueness and multiplicativity, identifies its geometric meaning as an oriented volume, and works out the structure of the group it defines.

Throughout, $F$ is a field, $V$ is a finite-dimensional $F$-vector space of dimension $n$, and the notation of the preceding article on the general linear group is used: $\operatorname{GL}(V)$, its centre $F^{\times}\operatorname{id}$, and the homomorphism $\det:\operatorname{GL}(V)\to F^{\times}$ whose kernel is $\operatorname{SL}(V)$. What the preceding article took as given — that the determinant exists, is multiplicative, and is unique — is proved here. The elementary matrices $E_{ij}(\lambda)=I+\lambda e_{ij}$ and the generation statement for $\operatorname{SL}$ are used, with the details of the transvections themselves of the category.

Two results are the aim. The determinant is the unique normalised alternating multilinear function of the columns, and it is the unique polynomial group homomorphism $\operatorname{GL}_n(F)\to F^{\times}$ up to powers: the scalar invariants of a linear operator are exhausted by the determinant. And $\operatorname{SL}_n(F)$ is generated by elementary matrices for $n \ge 2$ and equals the commutator subgroup $[\operatorname{GL}_n(F),\operatorname{GL}_n(F)]$ except in the two exceptional cases $n=2$ with $F=\mathbb{F}_2$ and $n=2$ with $F=\mathbb{F}_3$.

The Determinant

Alternating Multilinear Forms

Definition. A map $D:V^n \to F$ is multilinear if it is linear in each argument separately, and alternating if $D(v_1,\dots,v_n)=0$ whenever $v_i=v_j$ for some $i \neq j$. It is normalised if $D(e_1,\dots,e_n)=1$ for a chosen basis $e_1,\dots,e_n$.

Alternating implies that the interchange of two arguments changes the sign:

$$ D(v_1,\dots,v_i,\dots,v_j,\dots,v_n)=-D(v_1,\dots,v_j,\dots,v_i,\dots,v_n), $$

because multilinearity expands $D(\dots,v_i+v_j,\dots,v_i+v_j,\dots)=0$ into the four terms, of which the two diagonal terms vanish.

Theorem (existence and uniqueness). For each $n \ge 1$ there is exactly one normalised alternating multilinear form $D:V^n \to F$, and in a basis $\mathcal{B}$ with coordinates $v_j=\sum_i a_{ij}e_i$ it is given by the Leibniz formula

$$ D(v_1,\dots,v_n)=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)\prod_{j=1}^{n}a_{\sigma(j)j}. $$

Proof. Uniqueness: by multilinearity it suffices to evaluate $D$ on basis vectors. Writing each $v_j$ in the basis, expanding by multilinearity expresses $D(v_1,\dots,v_n)$ as a sum of terms $D(e_{i_1},\dots,e_{i_n})$ times coefficients; an alternating form vanishes unless $i_1,\dots,i_n$ are distinct, and then the term is $\operatorname{sgn}(\sigma)D(e_1,\dots,e_n)$ for the permutation $\sigma$ with $i_j=\sigma(j)$. So $D$ is determined by $D(e_1,\dots,e_n)$, and normalisation fixes it. Existence: the Leibniz formula is readily checked to be multilinear, to vanish on repeated arguments, and to equal $1$ on the basis.

Definition. The determinant of the matrix $A=(a_{ij})$ with columns $v_j$ is $\det A=D(v_1,\dots,v_n)$ with $D$ the normalised form; equivalently the Leibniz sum above. The determinant of an operator $T \in \operatorname{End}_F(V)$ is $\det T=\det[T]_{\mathcal{B}}^{\mathcal{B}}$ for any basis $\mathcal{B}$.

The definition of $\det T$ is independent of the basis, because changing the basis conjugates the matrix and conjugate matrices have equal determinants, as the linear-maps article shows. So the determinant is a function of the operator.

Multiplicativity

Theorem. For all $A,B \in M_n(F)$ one has $\det(AB)=\det A \det B$, and for invertible $A$, $\det(A^{-1})=(\det A)^{-1}$.

Proof. Fix $A$ and consider the function $(v_1,\dots,v_n)\mapsto D(Av_1,\dots,Av_n)$ where $D$ is the normalised determinant. It is multilinear and alternating in the $v_j$, since $A$ is linear and $D$ is, so by uniqueness it equals $c(A)D(v_1,\dots,v_n)$ for a scalar $c(A)$ depending on $A$. Evaluating on the standard basis gives $c(A)=D(Ae_1,\dots,Ae_n)=\det A$. Hence $D(A(Bv_1),\dots)=\det A\cdot D(Bv_1,\dots)=\det A\det B\cdot D(v_1,\dots)$ for all $v_j$, and evaluating on the basis gives $\det(AB)=\det A\det B$. The inverse statement follows by applying $\det$ to $AA^{-1}=I$.

This is the reason the determinant is a group homomorphism on $\operatorname{GL}(V)$: multiplicativity is exactly the homomorphism property.

Expansion and Computation

Theorem (Laplace expansion). For $A=(a_{ij})$, expanding along the $i$-th row,

$$ \det A=\sum_{j=1}^{n}(-1)^{i+j}a_{ij}M_{ij}, $$

where $M_{ij}$ is the determinant of the $(n-1)\times(n-1)$ matrix obtained by deleting row $i$ and column $j$.

Proof. Expand the Leibniz sum according to which index is paired with $i$. Each choice of $j$ contributes $a_{ij}$ times a sum over permutations of the remaining indices, which is the determinant of the minor with the sign $(-1)^{i+j}$.

Corollary. (i) If $A$ is upper or lower triangular, or block triangular $\begin{pmatrix}A_1&B\\0&A_2\end{pmatrix}$, then $\det A$ is the product of the diagonal entries, respectively $\det A_1\det A_2$. (ii) If two rows or two columns of $A$ are equal, or one is a scalar multiple of the other, then $\det A=0$. (iii) Adding a multiple of one row to another leaves $\det$ unchanged. (iv) The Vandermonde matrix $V(x_1,\dots,x_n)$ with entries $x_i^{j-1}$ has $\det V=\prod_{i

Proof. (i) In the Leibniz sum only the identity permutation survives for a triangular matrix; for the block case expand successively or note the same. (ii) is alternation. (iii) Multilinearity and (ii). (iv) The determinant is an alternating polynomial in the $x_i$ of degree $\binom{n}{2}$; it vanishes when $x_i=x_j$, so it is divisible by $\prod_{i

Definition. The adjugate $A^{\mathrm{adj}}$ has entries $(-1)^{i+j}M_{ji}$, and $A A^{\mathrm{adj}}=(\det A)I$. Hence over a commutative ring, $A$ is invertible if and only if $\det A$ is a unit, with $A^{-1}=(\det A)^{-1}A^{\mathrm{adj}}$; over a field this is the criterion $\det A \neq 0$.

Cramer's Rule

Theorem (Cramer). Let $A$ be invertible and let $b \in F^n$. Then the unique solution of $Ax=b$ is given by

$$ x_i=\frac{\det A_i}{\det A}, $$

where $A_i$ is the matrix obtained from $A$ by replacing column $i$ by $b$.

Proof. From $AA^{\mathrm{adj}}=(\det A)I$ one gets $x=A^{-1}b=(\det A)^{-1}A^{\mathrm{adj}}b$. The $i$-th coordinate of $A^{\mathrm{adj}}b$ is $\sum_j(-1)^{i+j}M_{ji}b_j$, and this is exactly the Laplace expansion along the $i$-th column of the matrix $A_i$, whose $i$-th column is $b$ and whose other columns agree with those of $A$.

Cramer's rule is an existence and uniqueness statement and a formal solution formula; it is not an efficient algorithm, since it evaluates $n+1$ determinants, and Gaussian elimination is used in practice.

The Cauchy–Binet Formula

For rectangular matrices the multiplicativity of the determinant is replaced by a sum over maximal minors.

Theorem (Cauchy–Binet). Let $A$ be $m \times n$ and $B$ an $n \times m$ matrix with $m \le n$. For a subset $S \subseteq \{1,\dots,n\}$ of size $m$ let $A_{\cdot S}$ be the $m \times m$ matrix formed by the columns of $A$ indexed by $S$, and $B_{S \cdot}$ the $m \times m$ matrix formed by the rows of $B$ indexed by $S$. Then

$$ \det(AB)=\sum_{|S|=m}\det(A_{\cdot S})\det(B_{S\cdot}). $$

Proof. The $m$-th alternating power is a functor: for linear maps $F^n \xrightarrow{B} F^n \xrightarrow{A} F^m$ one has $\wedge^m(AB)=(\wedge^m A)(\wedge^m B)$. In the basis $\{e_S\}$ of $\wedge^m F^n$ indexed by the $m$-element subsets $S$, the entries of the matrix of $\wedge^m A$ are the maximal minors $\det(A_{\cdot S})$, and similarly for $B$. Taking determinants of both sides of the functorial identity gives the formula.

The special case $m=1$ is the identity $a\cdot b=\sum_i a_ib_i$, and the case $n=m$ reduces to $\det(AB)=\det A\det B$.

Unimodularity and Volume

The determinant is not merely an algebraic device: it measures volume.

Proposition. Let $V=\mathbb{R}^n$ with the standard inner product. For vectors $v_1,\dots,v_n$, the Euclidean volume of the parallelepiped $\{\sum t_jv_j : 0 \le t_j \le 1\}$ is $|\det(v_1,\dots,v_n)|$, and the determinant is positive precisely when the ordered basis $v_1,\dots,v_n$ is positively oriented.

Proof. An orthogonal change of coordinates in the domain preserves both the volume of the parallelepiped and the absolute value of the determinant, by multiplicativity and the fact that orthogonal matrices have determinant $\pm1$; a diagonal matrix rescales the $j$-th edge by $|d_j|$ and the volume by $\prod|d_j|=|\det|$; a shear, which is an elementary matrix of the form $E_{ij}(\lambda)$, has determinant $1$ and does not change the volume, since it moves one edge parallel to the others. Every invertible matrix is a product of orthogonal, diagonal and shear factors (the $QR$ and $LU$ decompositions, or equivalently singular value decomposition), and the formula follows factor by factor.

Definition. A linear operator $T$ on a real vector space is unimodular if $|\det T|=1$, so that it preserves volume; it is orientation-preserving if $\det T=1$ and orientation-reversing if $\det T=-1$. Over an arbitrary field the analogue is $\det T=1$, which is the condition of membership in $\operatorname{SL}(V)$.

Thus $\operatorname{SL}(V)$ is exactly the group of volume-preserving and orientation-preserving linear maps of a real vector space with a fixed volume form, and $\operatorname{GL}(V)$ is the group of all volume-scaling maps, the scale factor being $\det T$. Over a finite field the same statement is read as a count: $\operatorname{SL}_n(\mathbb{F}_q)$ has index $q-1$ in $\operatorname{GL}_n(\mathbb{F}_q)$, since $\det:\operatorname{GL}_n(\mathbb{F}_q)\to\mathbb{F}_q^{\times}$ is surjective, so $|\operatorname{SL}_n(\mathbb{F}_q)|=|\operatorname{GL}_n(\mathbb{F}_q)|/(q-1)$. The full count is in the applications article on vector spaces over finite fields.

The Gram Determinant

Proposition. Let $v_1,\dots,v_k$ be vectors in a real inner product space and let $G=(v_i \cdot v_j)$ be their Gram matrix. Then $\det G \ge 0$, with $\det G=0$ exactly when $v_1,\dots,v_k$ are linearly dependent, and $\det G$ is the square of the $k$-dimensional volume of the parallelepiped spanned by the $v_i$.

Proof. Let $V$ be the matrix whose columns are the $v_i$, so that $G=V^{\mathsf{T}}V$. If $\sum c_iv_i=0$ with $c \neq 0$ then $Gc=0$ and $\det G=0$; conversely $\det G=0$ makes $G$ singular, so $Gc=0$ for some nonzero $c$ by the rank–nullity theorem, and then $0=c^{\mathsf{T}}Gc=c^{\mathsf{T}}V^{\mathsf{T}}Vc=\|Vc\|^2$, so $Vc=0$. If the $v_i$ are independent, Gram–Schmidt factors $V=QR$ with $Q$ having orthonormal columns and $R$ upper triangular invertible, so $G=R^{\mathsf{T}}R$ and $\det G=(\det R)^2$. The volume of the parallelepiped is $|\det R|$, because the columns of $Q$ span the same space with pairwise orthogonal unit edges.

The Gram determinant converts a question about volumes, which needs an inner product, into a determinant of a symmetric matrix, and it is the sense in which the determinant is available without a metric while the volume is not: the abstract determinant of a linear operator needs only the vector-space structure, while the identification of $|\det|$ with a Euclidean volume requires the inner product.

Worked Examples

Example. The determinant of $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is $ad-bc$, and of the $3 \times 3$ matrix $\begin{pmatrix}1&2&3\\4&5&6\\7&8&10\end{pmatrix}$ it is $1\cdot(5\cdot10-6\cdot8)-2\cdot(4\cdot10-6\cdot7)+3\cdot(4\cdot8-5\cdot7)=2-2\cdot(-2)+3\cdot(-3)=2+4-9=-3$.

Example. The Vandermonde determinant at $x_1=1$, $x_2=2$, $x_3=3$ is $\det\begin{pmatrix}1&1&1\\1&2&4\\1&3&9\end{pmatrix}=(2-1)(3-1)(3-2)=2$, in agreement with the general formula $\det V=\prod_{i

Example. For an upper triangular matrix the determinant is the product of the diagonal entries, so $\det E_{ij}(\lambda)=1$ for every elementary matrix and every field: the shears are exactly the elementary operations that leave volume and determinant unchanged.

The Special Linear Group

Definition and Basic Structure

Definition. $\operatorname{SL}(V)=\{T \in \operatorname{GL}(V) : \det T=1\}=\ker\det$, the special linear group of $V$, and $\operatorname{SL}_n(F)=\ker(\det:\operatorname{GL}_n(F)\to F^{\times})$.

Proposition. $\operatorname{SL}(V)$ is a normal subgroup of $\operatorname{GL}(V)$, the quotient is $\operatorname{GL}(V)/\operatorname{SL}(V)\cong F^{\times}$, and $Z(\operatorname{SL}(V))=Z(\operatorname{GL}(V))\cap\operatorname{SL}(V)=\mu_n$, the group of $n$-th roots of unity in $F$.

Proof. Normality and the quotient are the first isomorphism theorem for $\det$. For the centre: a scalar $\lambda\operatorname{id}$ lies in $\operatorname{SL}(V)$ exactly when $\lambda^n=1$, and scalars are central in $\operatorname{GL}(V)$ hence in $\operatorname{SL}(V)$; conversely an element of $\operatorname{SL}(V)$ central in $\operatorname{SL}(V)$ commutes with all elementary matrices, which generate $\operatorname{SL}(V)$ for $n \ge 2$, and a short computation with $E_{ij}(\lambda)$ shows it must be scalar.

Generation by Elementary Matrices

Theorem. For $n \ge 2$ and any field $F$, $\operatorname{SL}_n(F)$ is generated by the elementary matrices $E_{ij}(\lambda)=I+\lambda e_{ij}$, $i \neq j$, $\lambda \in F$.

Proof (sketch). Let $A \in \operatorname{SL}_n(F)$. Gaussian elimination with elementary row operations of the form "add a multiple of row $i$ to row $j$" reduces $A$ to a matrix with a single nonzero entry in the first row and column unless $F=\mathbb{F}_2$, where the two-dimensional case needs a separate argument. The surviving scalar is $1$, because left multiplication by an elementary matrix has determinant $1$ and $\det A=1$; induction on $n$ then expresses $A$ as a product of elementary matrices. The exception $\operatorname{SL}_2(\mathbb{F}_2)=S_3$ is still generated by elementary matrices: over $\mathbb{F}_2$ the matrices $E_{12}(1)=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $E_{21}(1)=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ generate a group of order $6$.

The elementary matrices are products of transvections, the shear maps $x\mapsto x+f(x)v$ with $f(v)=0$; another article of this category develops them and gives the full generation proof, including the role of row and column operations. The generation statement is the algebraic reason the determinant controls $\operatorname{SL}$: the group is generated by operations each of which visibly preserves the determinant.

The Commutator Subgroup

Theorem (Dieudonné). For $n \ge 2$,

$$ [\operatorname{GL}_n(F),\operatorname{GL}_n(F)]=\operatorname{SL}_n(F) $$

except for $n=2$ with $F=\mathbb{F}_2$ or $F=\mathbb{F}_3$. For $F=\mathbb{F}_2$ one has $\operatorname{GL}_2(\mathbb{F}_2)=\operatorname{SL}_2(\mathbb{F}_2)=S_3$ and the commutator subgroup is $A_3$, the cyclic group of order $3$, of index $2$ in $\operatorname{SL}_2(\mathbb{F}_2)$; for $F=\mathbb{F}_3$ the commutator subgroup is the quaternion group $Q_8$, of index $3$ in $\operatorname{SL}_2(\mathbb{F}_3)$ and index $6$ in $\operatorname{GL}_2(\mathbb{F}_3)$. For all other $n$ and $F$ the abelianisation of $\operatorname{GL}_n(F)$ is $F^{\times}$, induced by the determinant.

Proof. The elementary matrix $E_{ij}(\lambda)$ with $i,j,k$ distinct is a commutator, $E_{ij}(\lambda)=[E_{ik}(\lambda),E_{kj}(1)]$, so for $n \ge 3$ the elementary matrices lie in $[\operatorname{GL}_n(F),\operatorname{GL}_n(F)]$ and generation gives the inclusion $\operatorname{SL}_n(F)\subseteq[\operatorname{GL}_n(F),\operatorname{GL}_n(F)]$. For $n=2$ and $|F|>3$ choose $t \in F^{\times}$ with $t^2 \neq 1$; then

$$ [\operatorname{diag}(t,t^{-1}),E_{12}(\lambda)]=\operatorname{diag}(t,t^{-1})E_{12}(\lambda)\operatorname{diag}(t^{-1},t)E_{12}(-\lambda)=E_{12}\bigl((t^2-1)\lambda\bigr), $$

and as $\lambda$ runs over $F$ so does $(t^2-1)\lambda$, so the elementary matrices lie in the commutator subgroup and again generation gives the inclusion. The reverse inclusion holds because $\operatorname{GL}_n(F)/\operatorname{SL}_n(F)\cong F^{\times}$ is abelian, so $[\operatorname{GL}_n(F),\operatorname{GL}_n(F)]\subseteq\operatorname{SL}_n(F)$; the two inclusions together give equality whenever the elementary matrices generate $\operatorname{SL}_n(F)$, that is for $n \ge 3$ and for $n=2$ with $|F|>3$. The two remaining cases $n=2$ with $F=\mathbb{F}_2$ and $F=\mathbb{F}_3$ are exceptional, and in each of them the commutator subgroup is proper in $\operatorname{SL}_2(F)$: $A_3$ in $S_3$ and $Q_8$ in the binary tetrahedral group.

Theorem (Dieudonné, perfection). $\operatorname{SL}_n(F)$ is perfect — equal to its own commutator subgroup — for $n \ge 3$, and for $n=2$ whenever $|F|>3$. The exceptions are $\operatorname{SL}_2(\mathbb{F}_2)=S_3$, whose commutator subgroup is the cyclic group of order $3$, and $\operatorname{SL}_2(\mathbb{F}_3)$, whose commutator subgroup is the quaternion group $Q_8$ of order $8$, of index $3$.

Proof (sketch). For $n \ge 3$ the commutator of two elementary matrices can be made to produce a third, and the generation of $\operatorname{SL}_n(F)$ by elementary matrices then gives perfection. For $n=2$ the same computation with $E_{12}(\lambda)$ and $\operatorname{diag}(t,t^{-1})$ produces $E_{12}((t^2-1)\lambda)$ together with its transpose, and for $|F|>3$ these generate $\operatorname{SL}_2(F)$. The two exceptional groups are small: $\operatorname{SL}_2(\mathbb{F}_2)\cong S_3$ is the symmetry group of the triangle and $\operatorname{SL}_2(\mathbb{F}_3)\cong 2T$ is the binary tetrahedral group of order $24$, whose commutator subgroup is $Q_8$.

The corresponding projective statements are the classical simplicity theorems: $\operatorname{PSL}_n(F)$ is simple for $n \ge 2$ except for $(n,|F|)=(2,2)$ and $(2,3)$, where $\operatorname{PSL}_2(\mathbb{F}_2)=S_3$ and $\operatorname{PSL}_2(\mathbb{F}_3)=A_4$ are not simple. This is quoted as standard; it is the reason the projective special linear groups are the building blocks of the classical finite groups.

The Determinant as the Unique Invariant

Uniqueness Among Multilinear Forms

The uniqueness theorem above already says that, among the alternating multilinear forms, the determinant is the only normalised one: the space of alternating multilinear $n$-forms on an $n$-dimensional space is one-dimensional, spanned by the determinant. This is a statement about forms, not about operators, and it is the source of every other uniqueness statement about the determinant.

Uniqueness Among Polynomial Characters

Theorem. Let $F$ be an infinite field and let $p:M_n(F)\to F$ be a polynomial function, in the matrix entries, that is multiplicative: $p(AB)=p(A)p(B)$ for all $A,B$, and not identically zero. Then $p=\det^k$ for some $k \ge 0$. If in addition $p(A) \neq 0$ exactly when $A$ is invertible, then $p$ restricts to a group homomorphism $\operatorname{GL}_n(F)\to F^{\times}$.

In particular the only polynomial group homomorphisms $\operatorname{GL}_n(F)\to F^{\times}$ are the powers $A\mapsto(\det A)^k$; the determinant is the primitive one. This is the precise sense in which the determinant is the invariant of a linear operator: every multiplicative polynomial invariant is a power of it.

Proof (sketch). A multiplicative polynomial vanishing on all non-invertible matrices is determined on $\operatorname{GL}_n(F)$; restricting to diagonal matrices, multiplicativity forces $p(\operatorname{diag}(t_1,\dots,t_n))=\prod_i t_i^{k_i}$ for some exponents, and conjugating by permutation matrices shows the exponents are equal, $k_i=k$. Then $p$ and $\det^k$ are two multiplicative functions agreeing on all diagonalisable matrices over an algebraically closed field, and they agree everywhere because diagonalisable matrices are Zariski dense in $M_n$.

Corollary (the extreme coefficients). The characteristic polynomial $c_A(x)=\det(xI-A)=x^n-(\operatorname{tr}A)x^{n-1}+\cdots+(-1)^n\det A$ is determined by the sequence of elementary symmetric functions of the eigenvalues, of which $\operatorname{tr}A$ and $\det A$ are the two extreme ones. The determinant is the multiplicative invariant and the trace the additive one, and neither of them alone determines the operator up to similarity.

The Determinant of an Operator and of a Form

Proposition. For $T \in \operatorname{End}_F(V)$ and any basis $\mathcal{B}$, the scalar $\det T$ is independent of $\mathcal{B}$ and equals the factor by which $T$ multiplies every alternating $n$-form on $V$; more precisely $D(Tv_1,\dots,Tv_n)=\det T\cdot D(v_1,\dots,v_n)$ for every alternating multilinear $n$-form $D$.

Proof. The map $(v_1,\dots,v_n)\mapsto D(Tv_1,\dots,Tv_n)$ is alternating multilinear, hence a scalar multiple of the normalised determinant $D_0$ by the one-dimensionality of the space of such forms; evaluating on a basis identifies the scalar as $\det T$.

This is the intrinsic definition of the determinant of an operator: it is the eigenvalue of $T$ acting on the top exterior power of $V$, and it can be introduced without choosing a basis at all. The matrix Leibniz formula is then its coordinate expression.

Unimodular Matrices and Integral Structure

Definition. A matrix $A \in M_n(\mathbb{Z})$ is unimodular if $\det A=\pm1$; equivalently $A \in \operatorname{GL}_n(\mathbb{Z})$, since an integral matrix is invertible over $\mathbb{Z}$ exactly when its determinant is a unit of $\mathbb{Z}$, that is $\pm1$.

Proposition. $\operatorname{GL}_n(\mathbb{Z})=\{A \in M_n(\mathbb{Z}) : \det A=\pm1\}$ and $\operatorname{SL}_n(\mathbb{Z})=\{A \in M_n(\mathbb{Z}) : \det A=1\}$, with $\operatorname{GL}_n(\mathbb{Z})/\operatorname{SL}_n(\mathbb{Z}) \cong \{\pm1\}$.

Proof. Immediate from the adjugate criterion over the ring $\mathbb{Z}$: $A$ invertible over $\mathbb{Z}$ means $\det A$ a unit, and the units of $\mathbb{Z}$ are $\pm1$.

Theorem. For $n \ge 2$, $\operatorname{SL}_n(\mathbb{Z})$ is generated by the elementary matrices $E_{ij}(1)=I+e_{ij}$, $i \neq j$.

This is the integral form of the generation theorem: the Euclidean algorithm converts the reduction of an integral matrix to elementary row operations. The group $\operatorname{SL}_2(\mathbb{Z})$ is generated by $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $\begin{pmatrix}1&0\\1&1\end{pmatrix}$, and it is the free product of the cyclic subgroups they generate modulo their centres, in the form $\operatorname{PSL}_2(\mathbb{Z}) \cong C_2 * C_3$. The unimodular matrices describe the change-of-basis transformations of lattices, and the connection with lattices, determinant and covolume is developed in the applications article on lattices and the quaternion lattice.

The Determinant over a Commutative Ring

Theorem. Let $R$ be a commutative ring. The Leibniz formula defines a function $\det:M_n(R)\to R$, a polynomial with integer coefficients in the entries, satisfying $\det(AB)=\det A \det B$, $\det I=1$, and $\det A \in R^{\times}$ if and only if $A \in \operatorname{GL}_n(R)$.

Proof. The Leibniz formula is a finite integer polynomial, so it makes sense over any commutative ring; multiplicativity is proved as over a field, since the argument uses only the uniqueness of the normalised alternating multilinear form, which holds formally over $R$. The invertibility criterion is the adjugate identity $AA^{\mathrm{adj}}=(\det A)I$ over $R$: if $\det A$ is a unit then $(\det A)^{-1}A^{\mathrm{adj}}$ is an inverse, and the converse follows by taking determinants.

Proposition (naturality). For a ring homomorphism $\varphi:R \to S$ and $A \in M_n(R)$ one has $\det(\varphi(A))=\varphi(\det A)$, where $\varphi(A)$ has entries $\varphi(a_{ij})$. Consequently $\varphi$ restricts to a group homomorphism $\operatorname{GL}_n(R)\to\operatorname{GL}_n(S)$.

Proof. Apply $\varphi$ to the Leibniz sum, a polynomial identity with integer coefficients.

Naturality is what makes the determinant available over $\mathbb{Z}$ and then transported to $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{F}_p$: the integer determinant of an integer matrix determines all of its reductions. For an endomorphism $T$ of a free $R$-module of rank $n$ with a chosen basis, $\det T$ is the determinant of its matrix and is independent of the basis, by the same conjugation argument as over a field, so the determinant is an invariant of an endomorphism of a free module; this is the form in which it is used for lattices in the applications article on lattices and the quaternion lattice. Over a noncommutative ring no single scalar plays this role, and the correct generalisation is the Dieudonné determinant, with values in the abelianisation of the unit group.

Summary

The determinant is the unique normalised alternating multilinear form on $V^n$, given in coordinates by the Leibniz sum over permutations, and its value on the columns of a matrix defines $\det A$ and, for an operator, $\det T$ independent of any basis. It is multiplicative, $\det(AB)=\det A\det B$, which is exactly the statement that $\det$ is a group homomorphism $\operatorname{GL}(V)\to F^{\times}$; it can be computed by Laplace expansion along a row, it is the product of the diagonal entries for a triangular matrix, and it vanishes exactly when the columns are linearly dependent. The adjugate identity $AA^{\mathrm{adj}}=(\det A)I$ makes the determinant the criterion of invertibility over any commutative ring, and over $\mathbb{R}$ the absolute value of the determinant is the volume of the parallelepiped spanned by the columns, its sign recording orientation.

The special linear group $\operatorname{SL}(V)=\ker\det$ is normal in $\operatorname{GL}(V)$, the quotient is $F^{\times}$, and its centre is the group $\mu_n$ of $n$-th roots of unity in $F$. For $n \ge 2$ over any field it is generated by the elementary matrices, which are transvections, and it coincides with the commutator subgroup of $\operatorname{GL}_n(F)$ except in the two exceptional cases $n=2$, $F=\mathbb{F}_2$, where $\operatorname{GL}_2(\mathbb{F}_2)=S_3$ and the commutator subgroup is $A_3$, and $n=2$, $F=\mathbb{F}_3$, where the commutator subgroup is $Q_8$. The group $\operatorname{SL}_n(F)$ is perfect except for $n=2$ with $F=\mathbb{F}_2$ or $\mathbb{F}_3$, the exceptions having commutator subgroups of orders $3$ and $8$; the projective quotients $\operatorname{PSL}_n(F)$ are simple outside the two exceptional cases.

The determinant is the unique invariant of a linear operator in the following sense: the alternating multilinear $n$-forms form a one-dimensional space, and every multiplicative polynomial function on $M_n(F)$ is a power of the determinant. Among group homomorphisms $\operatorname{GL}_n(F)\to F^{\times}$ the determinant is the primitive one. Over $\mathbb{Z}$ the invertible matrices are exactly the unimodular ones, $\det A=\pm1$, and $\operatorname{SL}_n(\mathbb{Z})$ is generated by the elementary matrices.

Summary of Notation

Symbol	Meaning
$F$	a field
$V$	$F$-vector space of dimension $n$
$D:V^n\to F$	alternating multilinear $n$-form; normalised when $D(e_1,\dots,e_n)=1$
$\det A$, $\det T$	determinant of a matrix, of an operator
$\operatorname{sgn}(\sigma)$	sign of a permutation
$S_n$	symmetric group
$A^{\mathrm{adj}}$	adjugate, $AA^{\mathrm{adj}}=(\det A)I$
$M_{ij}$	minor obtained by deleting row $i$ and column $j$
$V(x_1,\dots,x_n)$	Vandermonde matrix, determinant $\prod_{i
$\operatorname{GL}(V)$, $\operatorname{GL}_n(F)$	general linear group
$\operatorname{SL}(V)=\ker\det$, $\operatorname{SL}_n(F)$	special linear group
$\mu_n$	$n$-th roots of unity, $=\operatorname{SL}(V)\cap F^{\times}\operatorname{id}$
$[G,G]$	commutator subgroup
$E_{ij}(\lambda)=I+\lambda e_{ij}$	elementary matrix
$\operatorname{PSL}_n(F)$	projective special linear group
$Q_8$	quaternion group of order $8$
$\operatorname{GL}_n(\mathbb{Z})$	unimodular integral matrices, $\det=\pm1$
$A_i$, $A_{\cdot S}$, $B_{S\cdot}$	Cramer matrices, maximal column and row selections
$G=(v_i\cdot v_j)$	Gram matrix, $\det G=\mathrm{vol}^2$
$R^{\times}$	unit group of a commutative ring $R$