A linear change of coordinates should not lose information. If a [vector space](/page/Vector%20Space) is being used to encode solutions of equations, symmetries of a geometric object, or states of a system, then a useful coordinate transformation must be reversible: it should move vectors around while preserving the possibility of recovering the original vector. The general linear group packages exactly these reversible linear transformations into a single algebraic object. The point is not merely that invertible matrices can be multiplied. The point is that invertible linear maps form the natural symmetry group of a vector space. Once we remember this, the determinant becomes more than a number attached to a matrix: in finite dimensions it detects whether a [linear map](/page/Linear%20Map) belongs to the group and measures volume scaling, while over an oriented real finite-dimensional vector space its sign separates orientation-preserving transformations from orientation-reversing ones. [example: A Singular Matrix Loses a Direction] Here $M_2(\mathbb{R})$ denotes the set of all $2\times 2$ matrices with real entries. Let $A \in M_2(\mathbb{R})$ have entries $A_{11}=1$, $A_{12}=1$, $A_{21}=2$, and $A_{22}=2$. The associated linear map $T_A:\mathbb{R}^2\to\mathbb{R}^2$ is \begin{align*} T_A(x_1,x_2)=(x_1+x_2,2x_1+2x_2). \end{align*} Evaluating this formula at $(1,-1)$ gives \begin{align*} T_A(1,-1)=(1+(-1),2\cdot 1+2\cdot(-1)). \end{align*} Since $1+(-1)=0$ and $2\cdot 1+2\cdot(-1)=2-2=0$, this becomes \begin{align*} T_A(1,-1)=(0,0). \end{align*} Thus the nonzero vector $(1,-1)$ is sent to the zero vector. For an arbitrary vector $(x_1,x_2)\in\mathbb{R}^2$, put $t=x_1+x_2$. Then \begin{align*} T_A(x_1,x_2)=(x_1+x_2,2x_1+2x_2)=(t,2t). \end{align*} Therefore every vector in the image of $T_A$ lies on the line $\{(t,2t):t\in\mathbb{R}\}$. Since, for example, $(0,1)$ is not of the form $(t,2t)$ because $t=0$ would force $2t=0\ne 1$, the map is not onto $\mathbb{R}^2$. This matrix is a linear transformation, but it is not a symmetry of $\mathbb{R}^2$ because it collapses a direction and loses information. [/example] This example shows the first boundary of the subject: all matrices form a ring under addition and multiplication, but only the invertible matrices form a group under multiplication. The general linear group is the part of matrix algebra where composition is reversible. ## Definition Before giving the definition, we need to fix the ambient linear algebra. The same group can be described either as invertible matrices or as invertible linear maps. The matrix version depends on a chosen basis, while the linear-map version is intrinsic to the vector space. The intrinsic definition answers the coordinate-free question: what are all reversible linear symmetries of $V$? [definition: General Linear Group of a Vector Space] Let $k$ be a field and let $V$ be a vector space over $k$. The general linear group of $V$ is the group \begin{align*} GL(V)=\{T:V\to V : T \text{ is a } k\text{-linear isomorphism}\}, \end{align*} with group operation given by composition of maps. [/definition] This is a child of the notion of a [group](/page/Group): the set is not just a collection of maps, but a group because identities, inverses, and associativity come from composition. The identity element is $\operatorname{id}_V$, and the inverse of $T \in GL(V)$ is its inverse linear isomorphism $T^{-1}:V\to V$. ## Coordinate Form For computations, we usually choose a basis and replace maps by matrices. This does not change the underlying symmetry idea, but it gives a concrete model whose entries can be manipulated directly. The matrix version records the same reversible transformations after the dimension and scalar field have been fixed. [definition: Matrix General Linear Group] Let $k$ be a field and let $n \in \mathbb{N}$. The matrix general linear group of degree $n$ over $k$ is \begin{align*} GL_n(k)=\{A \in M_n(k): A \text{ is invertible}\}, \end{align*} with group operation given by matrix multiplication. [/definition] The subscript $n$ records the dimension, and the field $k$ records the scalars. Thus $GL_2(\mathbb{R})$ is a real group of invertible $2\times 2$ real matrices, while $GL_n(\mathbb{F}_q)$ is a finite group when $\mathbb{F}_q$ is a finite field. The remaining question is whether these two definitions describe the same mathematics once a basis is chosen. [quotetheorem:9732] This theorem explains why it is safe to compute with matrices after choosing coordinates. It also warns us that a matrix group is often a coordinate shadow of a basis-free object. [example: Change of Basis Conjugates Matrices] Let $V$ be finite-dimensional over $k$, let $\mathcal{B}$ and $\mathcal{C}$ be ordered bases of $V$, and let $P_{\mathcal{C}\leftarrow\mathcal{B}}$ satisfy $[x]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}[x]_{\mathcal{B}}$ for every $x\in V$. Since the coordinate map from $\mathcal{B}$-coordinates to $\mathcal{C}$-coordinates is bijective, $P_{\mathcal{C}\leftarrow\mathcal{B}}$ is invertible, and therefore \begin{align*} [x]_{\mathcal{B}}=P_{\mathcal{C}\leftarrow\mathcal{B}}^{-1}[x]_{\mathcal{C}}. \end{align*} For $T\in GL(V)$ and $x\in V$, the definition of the matrix of $T$ in the basis $\mathcal{B}$ gives \begin{align*} [T(x)]_{\mathcal{B}}=[T]_{\mathcal{B}}[x]_{\mathcal{B}}. \end{align*} Changing the coordinates of $T(x)$ from $\mathcal{B}$ to $\mathcal{C}$ gives \begin{align*} [T(x)]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}[T(x)]_{\mathcal{B}}. \end{align*} Substituting the previous two identities into this expression gives \begin{align*} [T(x)]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}[T]_{\mathcal{B}}P_{\mathcal{C}\leftarrow\mathcal{B}}^{-1}[x]_{\mathcal{C}}. \end{align*} But the definition of the matrix of $T$ in the basis $\mathcal{C}$ also gives \begin{align*} [T(x)]_{\mathcal{C}}=[T]_{\mathcal{C}}[x]_{\mathcal{C}}. \end{align*} Because this holds for every coordinate vector $[x]_{\mathcal{C}}\in k^n$, the two matrices acting on $[x]_{\mathcal{C}}$ are equal: \begin{align*} [T]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}[T]_{\mathcal{B}}P_{\mathcal{C}\leftarrow\mathcal{B}}^{-1}. \end{align*} Thus changing basis replaces the representing matrix by a conjugate matrix, so matrix conjugacy records the same linear automorphism written in different coordinate systems. [/example] ## Invertibility and Determinants ### Tests for Membership The central membership question for $GL_n(k)$ is practical: given a matrix, how do we decide whether it belongs to the group? In finite dimensions, many conditions are equivalent. The determinant is the most compact test, but it sits alongside kernel, image, rank, and the existence of an inverse. The theorem below is the reason these tests may be used interchangeably. [quotetheorem:9735] The determinant condition turns the general linear group into the complement of a hypersurface inside the [affine space](/page/Affine%20Space) of all matrices. In equations, singular matrices are those satisfying $\det A=0$, and invertible matrices are the remaining points. [example: Membership in $GL_2(\mathbb{R})$] Let $A\in M_2(\mathbb{R})$ have entries $A_{11}=a$, $A_{12}=b$, $A_{21}=c$, and $A_{22}=d$. Put $\Delta=ad-bc$. We show that $A\in GL_2(\mathbb{R})$ exactly when $\Delta\ne 0$. First suppose $\Delta\ne 0$. Define $B\in M_2(\mathbb{R})$ by \begin{align*} B_{11}=\frac{d}{\Delta},\quad B_{12}=\frac{-b}{\Delta},\quad B_{21}=\frac{-c}{\Delta},\quad B_{22}=\frac{a}{\Delta}. \end{align*} The entries of $AB$ are \begin{align*} (AB)_{11}=a\frac{d}{\Delta}+b\frac{-c}{\Delta}=\frac{ad-bc}{\Delta}=1. \end{align*} \begin{align*} (AB)_{12}=a\frac{-b}{\Delta}+b\frac{a}{\Delta}=\frac{-ab+ab}{\Delta}=0. \end{align*} \begin{align*} (AB)_{21}=c\frac{d}{\Delta}+d\frac{-c}{\Delta}=\frac{cd-dc}{\Delta}=0. \end{align*} \begin{align*} (AB)_{22}=c\frac{-b}{\Delta}+d\frac{a}{\Delta}=\frac{-cb+ad}{\Delta}=\frac{ad-bc}{\Delta}=1. \end{align*} Similarly, the entries of $BA$ are \begin{align*} (BA)_{11}=\frac{d}{\Delta}a+\frac{-b}{\Delta}c=\frac{ad-bc}{\Delta}=1. \end{align*} \begin{align*} (BA)_{12}=\frac{d}{\Delta}b+\frac{-b}{\Delta}d=\frac{db-bd}{\Delta}=0. \end{align*} \begin{align*} (BA)_{21}=\frac{-c}{\Delta}a+\frac{a}{\Delta}c=\frac{-ca+ac}{\Delta}=0. \end{align*} \begin{align*} (BA)_{22}=\frac{-c}{\Delta}b+\frac{a}{\Delta}d=\frac{-cb+ad}{\Delta}=1. \end{align*} Thus $AB=BA=I_2$, so $B=A^{-1}$ and $A\in GL_2(\mathbb{R})$. Conversely suppose $\Delta=0$. If $(a,b)\ne(0,0)$, then $(b,-a)$ is nonzero and \begin{align*} A(b,-a)=(ab-ab,cb-da). \end{align*} Since $\Delta=ad-bc=0$, we have $cb-da=-(ad-bc)=0$, so $A(b,-a)=(0,0)$. If instead $(a,b)=(0,0)$ and $(c,d)\ne(0,0)$, then $(d,-c)$ is nonzero and \begin{align*} A(d,-c)=(0,cd-dc)=(0,0). \end{align*} If all four entries are zero, then $A(1,0)=(0,0)$. In every case, $A$ sends a nonzero vector to zero, so it cannot have an inverse. Therefore \begin{align*} A\in GL_2(\mathbb{R}) \quad \Longleftrightarrow \quad ad-bc\ne 0. \end{align*} For the shear matrix $S$ with entries $S_{11}=1$, $S_{12}=t$, $S_{21}=0$, and $S_{22}=1$, its determinant is \begin{align*} S_{11}S_{22}-S_{12}S_{21}=1\cdot 1-t\cdot 0=1. \end{align*} Since $1\ne 0$, the criterion above gives $S\in GL_2(\mathbb{R})$ for every $t\in\mathbb{R}$. Its inverse has entries $1,-t,0,1$, so a shear moves points reversibly rather than collapsing a direction. [/example] ### Determinant as a Homomorphism The determinant is compatible with multiplication, so it should be viewed as a map between groups rather than only as a scalar-valued test. To state this group map, we need the target group consisting of nonzero scalars. This target records the possible determinant values of invertible matrices. [definition: Multiplicative Group of a Field] Let $k$ be a field. The multiplicative group of $k$ is \begin{align*} k^\times=k\setminus\{0\}, \end{align*} with group operation given by multiplication in $k$. [/definition] This group is the natural target of the determinant because determinants multiply under matrix multiplication and are nonzero exactly on invertible matrices. The remaining issue is not just that determinant detects invertibility, but that it must respect the group operation: multiplying two invertible matrices should correspond to multiplying their determinant scalars. Without this compatibility, determinant would be only a test on individual matrices, not a structural map from $GL_n(k)$ to $k^\times$. [quotetheorem:395] This multiplicativity is the point that promotes determinant from a numerical invariant to a [group homomorphism](/page/Group%20Homomorphism). For $A,B\in GL_n(k)$, the product $AB$ again has nonzero determinant because $\det(AB)=\det(A)\det(B)$ lies in $k^\times$. Thus determinant preserves the group operation and gives a way to organize $GL_n(k)$ by determinant value. The most important fibre for the structure of $GL_n(k)$ is the fibre over the identity element $1\in k^\times$. ### The Determinant-One Subgroup A homomorphism is often understood through its fibres, and the determinant fibre over $1$ is the one that keeps volume unchanged. This asks for a named subgroup of matrices whose determinant scaling has been removed. [definition: Special Linear Group] Let $k$ be a field and let $n\in\mathbb{N}$. Here $k^n$ denotes the coordinate vector space of $n$-tuples over $k$, $M_n(k)$ denotes the set of all $n\times n$ matrices with entries in $k$, and $GL_n(k)$ denotes the group of invertible matrices in $M_n(k)$. The special linear group of degree $n$ over $k$ is \begin{align*} SL_n(k)=\{A\in GL_n(k):\det A=1\}, \end{align*} with group operation inherited from $GL_n(k)$. [/definition] After isolating determinant-one matrices, the structural question is how this subgroup sits inside the larger general linear group. Kernel language answers that question and connects the construction to the standard normal-subgroup machinery of group theory. [quotetheorem:9740] This kernel description is more than a convenient way to restate the definition. It shows that $SL_n(k)$ is a [normal subgroup](/page/Normal%20Subgroup) of $GL_n(k)$, so determinant-one matrices are stable under conjugation by arbitrary invertible changes of coordinates. That matters because many later constructions compare linear transformations only up to a [change of basis](/page/Change%20Of%20Basis). The limitation is equally important: the determinant remembers only the total volume-scaling factor, not the full geometry of a matrix. Passing to the kernel removes that one scalar invariant while leaving the richer internal structure of $SL_n(k)$ to be studied separately. ## Structure from Bases and Frames The general linear group is the symmetry group of a vector space, but it also acts on the ways of coordinatizing that vector space. A basis is a coordinate frame. An invertible linear transformation sends bases to bases, so $GL(V)$ acts on the set of all frames. Naming these frames lets us express the action without hiding it inside matrix notation. [definition: Ordered Basis Frame] Let $k$ be a field and let $V$ be an $n$-dimensional vector space over $k$. An ordered basis frame of $V$ is an ordered tuple \begin{align*} \mathcal{F}=(v_1,\dots,v_n) \end{align*} whose entries form a basis of $V$. [/definition] Once frames have been named, the next question is how the symmetry group moves them. The relevant construction should send a frame to the frame obtained by applying the same invertible linear map to each vector. [definition: Frame Set] Let $k$ be a field and let $V$ be an $n$-dimensional vector space over $k$. The frame set of $V$ is the set \begin{align*} \operatorname{Fr}(V)=\{(v_1,\dots,v_n): (v_1,\dots,v_n) \text{ is an ordered basis frame of } V\}. \end{align*} [/definition] With the set of frames named, we can ask whether $GL(V)$ really accounts for every possible change of coordinates. To answer that, the group must act on frames themselves: a linear automorphism should transform a coordinate frame by moving each basis vector to its image. [definition: Natural Action of $GL(V)$ on Frames] Let $k$ be a field and let $V$ be an $n$-dimensional vector space over $k$. The natural action of $GL(V)$ on ordered basis frames is the map \begin{align*} GL(V)\times \operatorname{Fr}(V)\to \operatorname{Fr}(V) \end{align*} that sends $(T,(v_1,\dots,v_n))$ to $(T(v_1),\dots,T(v_n))$. [/definition] Having defined the action, we need to measure its size: can every frame be reached, and can two different linear automorphisms send one fixed frame to the same target frame? The answer is the strongest possible form of transitivity. [quotetheorem:9745] Simple transitivity is a useful slogan: $GL(V)$ is the space of all choices of ordered basis, once one reference basis has been fixed. This explains why change-of-basis matrices are themselves elements of a general linear group. [example: Every Ordered Basis is a Matrix in $GL_n(k)$] Take the standard basis $(e_1,\dots,e_n)$ of $k^n$, and let $C\in M_n(k)$ be the matrix whose $i$-th column is $v_i$. For each $i$ and each coordinate index $j$, the $j$-th coordinate of $Ce_i$ is \begin{align*} (Ce_i)_j=\sum_{\ell=1}^n C_{j\ell}(e_i)_\ell=C_{ji}. \end{align*} Since the $i$-th column of $C$ is $v_i$, we have $C_{ji}=(v_i)_j$ for every $j$, so $Ce_i=v_i$. If $(v_1,\dots,v_n)$ is a basis, then every $y\in k^n$ has a unique expression $y=\alpha_1v_1+\cdots+\alpha_nv_n$. Define $B(y)=(\alpha_1,\dots,\alpha_n)$. Then \begin{align*} B(Ce_i)=B(v_i)=e_i. \end{align*} Also, if $y=\alpha_1v_1+\cdots+\alpha_nv_n$, then \begin{align*} CB(y)=C(\alpha_1e_1+\cdots+\alpha_ne_n)=\alpha_1v_1+\cdots+\alpha_nv_n=y. \end{align*} Thus $BC=CB=I_n$, so $C$ is invertible and hence $C\in GL_n(k)$. Conversely, suppose $C\in GL_n(k)$. If $y\in k^n$, write $x=C^{-1}y$ and express $x=\alpha_1e_1+\cdots+\alpha_ne_n$. Then \begin{align*} y=Cx=C(\alpha_1e_1+\cdots+\alpha_ne_n)=\alpha_1v_1+\cdots+\alpha_nv_n, \end{align*} so the vectors $v_1,\dots,v_n$ span $k^n$. If \begin{align*} \alpha_1v_1+\cdots+\alpha_nv_n=0, \end{align*} then \begin{align*} C(\alpha_1e_1+\cdots+\alpha_ne_n)=0. \end{align*} Applying $C^{-1}$ gives \begin{align*} \alpha_1e_1+\cdots+\alpha_ne_n=0, \end{align*} so $\alpha_1=\cdots=\alpha_n=0$ because the standard basis is linearly independent. Therefore $(v_1,\dots,v_n)$ is a basis. Finally, if $T\in GL_n(k)$ satisfies $T(e_i)=v_i$ for every $i$, then for $x=\alpha_1e_1+\cdots+\alpha_ne_n$, \begin{align*} T(x)=\alpha_1v_1+\cdots+\alpha_nv_n=Cx. \end{align*} Thus $T=C$, so the unique element of $GL_n(k)$ sending each $e_i$ to $v_i$ is exactly the matrix with columns $v_1,\dots,v_n$. [/example] ## Subgroups and Linear Symmetries Many familiar groups arise by imposing extra conditions on invertible matrices. This is how the general linear group acts as a universal ambient group for finite-dimensional linear symmetry: orthogonal, symplectic, diagonal, triangular, and permutation groups are all found inside some $GL_n(k)$. The first useful restriction is to allow independent scaling along the chosen coordinate axes. [definition: Diagonal Subgroup] Let $k$ be a field and let $n\in\mathbb{N}$. The diagonal subgroup of $GL_n(k)$ is \begin{align*} D_n(k)=\{\operatorname{diag}(a_1,\dots,a_n): a_i\in k^\times \text{ for }1\le i\le n\}. \end{align*} [/definition] Diagonal matrices scale the coordinate axes independently. They are the simplest invertible transformations that do not mix the chosen coordinates, and their multiplication reflects only multiplication of scalars. [example: The Diagonal Subgroup is Abelian] Let $A=\operatorname{diag}(a_1,\dots,a_n)$ and $B=\operatorname{diag}(b_1,\dots,b_n)$ be elements of $D_n(k)$. Thus $A_{ii}=a_i$, $B_{ii}=b_i$, and $A_{ij}=B_{ij}=0$ whenever $i\ne j$. For each pair of indices $i,j$, the matrix product formula gives \begin{align*} (AB)_{ij}=\sum_{\ell=1}^n A_{i\ell}B_{\ell j}. \end{align*} If $i=j$, every term is zero except possibly the term $\ell=i$, so \begin{align*} (AB)_{ii}=A_{ii}B_{ii}=a_i b_i. \end{align*} If $i\ne j$, then for each $\ell$ either $\ell\ne i$, so $A_{i\ell}=0$, or $\ell=i$, in which case $B_{ij}=0$; hence every summand is zero and \begin{align*} (AB)_{ij}=0. \end{align*} Therefore \begin{align*} AB=\operatorname{diag}(a_1b_1,\dots,a_nb_n). \end{align*} The same entrywise calculation gives \begin{align*} (BA)_{ii}=B_{ii}A_{ii}=b_i a_i \end{align*} and, for $i\ne j$, \begin{align*} (BA)_{ij}=0. \end{align*} Thus \begin{align*} BA=\operatorname{diag}(b_1a_1,\dots,b_na_n). \end{align*} Since $k$ is a field, multiplication in $k$ is commutative, so $a_i b_i=b_i a_i$ for every $i$. The diagonal entries of $AB$ and $BA$ are equal, and their off-diagonal entries are all zero, so $AB=BA$. Hence any two elements of $D_n(k)$ commute, which means $D_n(k)$ is abelian. [/example] The diagonal subgroup does not allow one coordinate to influence another. The next problem is to model transformations that preserve a hierarchy of coordinate subspaces while still permitting controlled mixing, and this leads to the upper triangular subgroup. [definition: Upper Triangular Subgroup] Let $k$ be a field and let $n\in\mathbb{N}$. The upper triangular subgroup of $GL_n(k)$ is \begin{align*} B_n(k)=\{A\in GL_n(k): A_{ij}=0 \text{ whenever } i>j\}. \end{align*} [/definition] Upper triangular matrices preserve the flag of coordinate subspaces $\operatorname{span}(e_1)\subsetneq\operatorname{span}(e_1,e_2)\subsetneq\cdots\subsetneq k^n$. The membership question for this subgroup should be simpler than the general determinant test, because triangular shape concentrates invertibility on the diagonal. [quotetheorem:7912] For an upper triangular matrix, this theorem says that the [characteristic polynomial](/page/Characteristic%20Polynomial) has the form \begin{align*} \prod_{i=1}^n (x-A_{ii}). \end{align*} In particular, the constant term is nonzero exactly when every diagonal entry $A_{ii}$ is nonzero, which is the triangular version of the determinant criterion for invertibility. Thus $B_n(k)$ can be read concretely as the invertible upper triangular matrices, or equivalently as the upper triangular matrices whose diagonal entries all lie in $k^\times$. The triangular subgroup preserves a flag, while another basic subgroup preserves only the set of coordinate axes and permutes them. This asks for a matrix model of a permutation, turning the operation of rearranging labels into a linear automorphism. [definition: Permutation Matrix] Let $k$ be a field, let $n\in\mathbb{N}$, and let $\sigma\in S_n$. The permutation matrix associated to $\sigma$ is the matrix $P_\sigma\in GL_n(k)$ determined by $P_\sigma e_i=e_{\sigma(i)}$ for every $1\le i\le n$. [/definition] This construction turns each permutation into a linear map, but it is not yet clear that it preserves the whole group structure. The possible obstruction is order of composition: if matrix multiplication encoded permutations in the wrong order, or if two different permutations produced the same matrix, then this would be only a convenient notation rather than a subgroup of $GL_n(k)$. The required check is that the assignment $\sigma\mapsto P_\sigma$ is injective and respects multiplication. [quotetheorem:9749] This embedding is the first hint of a broader idea: studying a group by letting it act linearly on a vector space. ## Finite Fields and Counting When the field is finite, the general linear group becomes a finite group with a beautiful counting formula. The count is not obtained by expanding determinants; it comes from the frame interpretation. To choose an invertible matrix, choose its columns as an ordered basis. This finite case deserves its own named object because it supplies central examples throughout finite group theory. [definition: General Linear Group over a Finite Field] Let $\mathbb{F}_q$ be a finite field with $q$ elements and let $n\in\mathbb{N}$. The finite general linear group of degree $n$ over $\mathbb{F}_q$ is \begin{align*} GL_n(\mathbb{F}_q)=\{A\in M_n(\mathbb{F}_q):\det A\ne 0\}. \end{align*} [/definition] After naming the finite group, the natural question is how large it is. Counting by columns turns invertibility into the condition that the chosen vectors remain independent at each step. [quotetheorem:9753] This formula is one of the cleanest demonstrations that the general linear group is the group of frames. Each factor counts the number of choices for the next column after avoiding the span of the previous columns. [example: Counting $GL_2(\mathbb{F}_3)$] For $n=2$ and $q=3$, the order formula gives \begin{align*} |GL_2(\mathbb{F}_3)|=(3^2-1)(3^2-3). \end{align*} Since $3^2=9$, this becomes \begin{align*} |GL_2(\mathbb{F}_3)|=(9-1)(9-3)=8\cdot 6=48. \end{align*} The same number comes from counting possible columns. The vector space $\mathbb{F}_3^2$ has $3\cdot 3=9$ vectors, and the first column of an invertible matrix may be any nonzero vector, so there are $9-1=8$ choices. After choosing a nonzero first column $v$, its span is \begin{align*} \operatorname{span}(v)=\{0,v,2v\}. \end{align*} These three vectors are distinct: if $v=0$ then the first column was zero, and if $2v=0$ then multiplying by $2^{-1}=2$ in $\mathbb{F}_3$ gives $v=0$. Therefore the span of the first column contains exactly $3$ vectors. The second column must not lie in $\operatorname{span}(v)$, because otherwise the two columns would be linearly dependent. Hence it has $9-3=6$ possible choices. Thus the total number of ordered column pairs forming a basis of $\mathbb{F}_3^2$ is \begin{align*} 8\cdot 6=48. \end{align*} So $GL_2(\mathbb{F}_3)$ has $48$ elements: each one is exactly an ordered basis of $\mathbb{F}_3^2$ written as the two columns of a matrix. [/example] The same counting question can be asked after imposing determinant $1$. For $n\ge 1$, every nonzero determinant value is attained by a diagonal matrix, so the determinant map spreads $GL_n(\mathbb{F}_q)$ evenly over the nonzero elements of $\mathbb{F}_q$. [quotetheorem:9755] The formula divides the size of $GL_n(\mathbb{F}_q)$ by $q-1$ because the determinant takes values in the nonzero multiplicative group $\mathbb{F}_q^\times$, which has exactly $q-1$ elements. Each determinant fibre has the same size: multiplying one row, or equivalently using a suitable diagonal matrix, moves matrices from one nonzero determinant value to another. For example, when $q=3$ and $n=2$, the earlier count $|GL_2(\mathbb{F}_3)|=48$ gives $|SL_2(\mathbb{F}_3)|=48/(3-1)=24$. This fibre-counting argument depends on working over a finite field and on restricting to invertible matrices, where the determinant is nonzero. The result will be used as the determinant-one analogue of the general linear group count, especially when comparing full linear symmetry with volume-preserving linear symmetry. ## Representations and Linearization The general linear group is not only an object of study; it is also the target for representing other groups. A representation turns an abstract group element into an invertible linear transformation. This is one of the main ways group theory meets linear algebra, because it lets algebraic multiplication be studied through composition of maps. [definition: Linear Representation] Let $G$ be a group, let $k$ be a field, and let $V$ be a vector space over $k$. A linear representation of $G$ on $V$ is a group homomorphism \begin{align*} \rho:G\to GL(V). \end{align*} [/definition] The definition says that every $g\in G$ acts as an invertible linear map, and that multiplication in $G$ is respected by composition of maps. The next problem is computational: after choosing coordinates on $V$, the same action should be recorded as matrices in some $GL_n(k)$. [definition: Matrix Representation] Let $G$ be a group, let $k$ be a field, and let $n\in\mathbb{N}$. A matrix representation of $G$ of degree $n$ over $k$ is a group homomorphism \begin{align*} \rho:G\to GL_n(k). \end{align*} [/definition] Matrix representations appear once a basis of $V$ is chosen. Different bases usually give conjugate matrix representations, echoing the change-of-basis phenomenon from the definition section. [example: Cyclic Groups Acting by Rotations] Let $C_m=\langle r\rangle$ with $r^m=e$, and put $\theta=2\pi/m$. For each real number $\alpha$, let $R_\alpha$ denote the real $2\times 2$ matrix with entries $(R_\alpha)_{11}=\cos\alpha$, $(R_\alpha)_{12}=-\sin\alpha$, $(R_\alpha)_{21}=\sin\alpha$, and $(R_\alpha)_{22}=\cos\alpha$. Its determinant is \begin{align*} (R_\alpha)_{11}(R_\alpha)_{22}-(R_\alpha)_{12}(R_\alpha)_{21}=\cos^2\alpha+\sin^2\alpha=1. \end{align*} Thus $R_\alpha\in GL_2(\mathbb{R})$ for every $\alpha$. We compute the product $R_\alpha R_\beta$ entry by entry. The $(1,1)$-entry is \begin{align*} \cos\alpha\cos\beta+(-\sin\alpha)\sin\beta=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta). \end{align*} The $(1,2)$-entry is \begin{align*} \cos\alpha(-\sin\beta)+(-\sin\alpha)\cos\beta=-(\cos\alpha\sin\beta+\sin\alpha\cos\beta)=-\sin(\alpha+\beta). \end{align*} The $(2,1)$-entry is \begin{align*} \sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta). \end{align*} The $(2,2)$-entry is \begin{align*} \sin\alpha(-\sin\beta)+\cos\alpha\cos\beta=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta). \end{align*} Therefore $R_\alpha R_\beta=R_{\alpha+\beta}$. Define $\rho:C_m\to GL_2(\mathbb{R})$ by $\rho(r^j)=R_{j\theta}$. If $r^i=r^j$, then $i-j=qm$ for some integer $q$, so \begin{align*} i\theta-j\theta=qm\frac{2\pi}{m}=2\pi q. \end{align*} Since sine and cosine are $2\pi$-periodic, $R_{i\theta}=R_{j\theta}$, so $\rho$ is well-defined. For powers $r^i$ and $r^j$, \begin{align*} \rho(r^i r^j)=\rho(r^{i+j})=R_{(i+j)\theta}=R_{i\theta+j\theta}=R_{i\theta}R_{j\theta}=\rho(r^i)\rho(r^j). \end{align*} Hence $\rho$ is a group homomorphism. Also, \begin{align*} \rho(r)^m=R_\theta^m=R_{m\theta}=R_{2\pi}=I_2. \end{align*} This representation realizes the abstract [cyclic group](/page/Cyclic%20Group) as the subgroup of $GL_2(\mathbb{R})$ generated by rotation through angle $2\pi/m$. [/example] A representation may still lose information if distinct group elements act by the same linear map. To distinguish groups that merely act linearly from groups that are genuinely embedded in a general linear group, we need the injective case. [definition: Faithful Representation] Let $G$ be a group, let $k$ be a field, and let $V$ be a vector space over $k$. A representation $\rho:G\to GL(V)$ is faithful if $\rho$ is injective. [/definition] A faithful representation embeds $G$ as a subgroup of a general linear group. The natural obstruction is that an abstract finite group may come with no visible vectors, matrices, or coordinates at all. To put it inside some $GL_n(k)$, one must first manufacture a vector space on which the group acts without identifying distinct elements. The finite case has a universal way to do this: let the group move a basis indexed by its own elements. [quotetheorem:846] [Cayley's theorem](/theorems/795) first embeds $G$ into a [symmetric group](/page/Symmetric%20Group) by left multiplication, and the permutation-matrix construction then turns that symmetric group into a subgroup of $GL_n(k)$. If $|G|=n$, this gives a faithful $n$-dimensional representation over any field: the basis vectors are indexed by the elements of $G$, and left multiplication permutes those basis vectors without collapsing distinct group elements. The construction is universal but not economical. It usually gives a large representation, and after one chooses an ordering of the group elements the resulting matrices depend on that choice rather than on a canonical coordinate system. ## Real and Complex General Linear Groups Over $\mathbb{R}$ and $\mathbb{C}$, the general linear group carries topology as well as algebra. As before, $M_n(k)$ denotes the set of all $n\times n$ matrices with entries in a field $k$, and $GL_n(k)$ denotes the matrices in $M_n(k)$ with nonzero determinant. Since $GL_n(k)$ is described by the condition $\det A\ne 0$, it is an open subset of $M_n(k)$ for $k=\mathbb{R}$ or $\mathbb{C}$. This gives it the structure of a manifold and, with multiplication and inversion, a Lie group. The real case is worth naming separately because determinant sign has geometric meaning. [definition: Real General Linear Group] Let $n\in\mathbb{N}$. The real general linear group is \begin{align*} GL_n(\mathbb{R})=\{A\in M_n(\mathbb{R}):\det A\ne 0\}. \end{align*} [/definition] Once the real group is viewed topologically, the question becomes whether all invertible real matrices can be connected by continuous paths. The determinant sign gives an obstruction, and the theorem says this is the only obstruction. [quotetheorem:9756] This real theorem is not just a classification of components; it explains why determinant sign is a topological invariant of real invertible matrices. For instance, a real matrix with positive determinant can be continuously deformed toward the identity inside $GL_n(\mathbb{R})$, while a matrix with negative determinant cannot cross into that component without passing through determinant zero. Thus orientation is the visible obstruction, and the theorem says there is no more subtle obstruction hiding behind it. The corresponding complex group should be separated from the real one because determinants now live in $\mathbb{C}^\times$ rather than in ordered real scalars. Without a positive-negative split, the topology changes even though the algebraic formula looks the same. This motivates giving the complex group its own definition before stating its connectedness behavior. [definition: Complex General Linear Group] Let $n\in\mathbb{N}$. The complex general linear group is \begin{align*} GL_n(\mathbb{C})=\{A\in M_n(\mathbb{C}):\det A\ne 0\}. \end{align*} [/definition] The complex case has no determinant-sign split. The natural topological question is therefore whether any hidden obstruction remains. In finite dimensions, there is none: every invertible complex matrix can be connected to the identity by a continuous path inside $GL_n(\mathbb{C})$. One way to see this is to factor a matrix into elementary operations or use a complex version of Gaussian elimination; unlike over $\mathbb{R}$, no sign obstruction appears because $\mathbb{C}^\times$ itself is connected by paths around the origin. This connectedness statement is meant here only for the finite-dimensional matrix group $GL_n(\mathbb{C})$, not for general spaces of bounded operators. It explains why the complex general linear group has one [connected component](/page/Connected%20Component), in contrast with $GL_n(\mathbb{R})$. The tangent space at the identity of a Lie group is its [Lie algebra](/page/Lie%20Algebra). For the general linear group, no determinant constraint remains at first order, so every matrix occurs as an infinitesimal direction. The bracket then records the first-order failure of matrix multiplication to commute. [definition: General Linear Lie Algebra] Let $k$ be $\mathbb{R}$ or $\mathbb{C}$, and let $n\in\mathbb{N}$. The general linear Lie algebra is \begin{align*} \mathfrak{gl}(n,k)=M_n(k), \end{align*} with Lie bracket \begin{align*} [X,Y]=XY-YX \end{align*} for all $X,Y\in M_n(k)$. [/definition] The Lie algebra records infinitesimal noncommutativity. Even when the underlying vector space is all matrices, the bracket remembers matrix multiplication. [example: A Nonzero Matrix Bracket] Let $X,Y\in M_2(\mathbb{R})$ be given by $X_{11}=0$, $X_{12}=1$, $X_{21}=0$, $X_{22}=0$ and $Y_{11}=0$, $Y_{12}=0$, $Y_{21}=1$, $Y_{22}=0$. We compute the bracket $[X,Y]=XY-YX$ entry by entry. For $XY$, the matrix product formula gives \begin{align*} (XY)_{11}=X_{11}Y_{11}+X_{12}Y_{21}=0\cdot 0+1\cdot 1=1. \end{align*} \begin{align*} (XY)_{12}=X_{11}Y_{12}+X_{12}Y_{22}=0\cdot 0+1\cdot 0=0. \end{align*} \begin{align*} (XY)_{21}=X_{21}Y_{11}+X_{22}Y_{21}=0\cdot 0+0\cdot 1=0. \end{align*} \begin{align*} (XY)_{22}=X_{21}Y_{12}+X_{22}Y_{22}=0\cdot 0+0\cdot 0=0. \end{align*} For $YX$, the same product formula gives \begin{align*} (YX)_{11}=Y_{11}X_{11}+Y_{12}X_{21}=0\cdot 0+0\cdot 0=0. \end{align*} \begin{align*} (YX)_{12}=Y_{11}X_{12}+Y_{12}X_{22}=0\cdot 1+0\cdot 0=0. \end{align*} \begin{align*} (YX)_{21}=Y_{21}X_{11}+Y_{22}X_{21}=1\cdot 0+0\cdot 0=0. \end{align*} \begin{align*} (YX)_{22}=Y_{21}X_{12}+Y_{22}X_{22}=1\cdot 1+0\cdot 0=1. \end{align*} Using the definition $[X,Y]=XY-YX$, its entries are therefore \begin{align*} [X,Y]_{11}=(XY)_{11}-(YX)_{11}=1-0=1. \end{align*} \begin{align*} [X,Y]_{12}=(XY)_{12}-(YX)_{12}=0-0=0. \end{align*} \begin{align*} [X,Y]_{21}=(XY)_{21}-(YX)_{21}=0-0=0. \end{align*} \begin{align*} [X,Y]_{22}=(XY)_{22}-(YX)_{22}=0-1=-1. \end{align*} Thus $[X,Y]\ne 0$, so these two infinitesimal transformations do not commute; the Lie bracket records that failure of commutativity. [/example] ## Beyond and Connected Topics The general linear group sits at the meeting point of group theory, linear algebra, geometry, and representation theory. In basic algebra, it is the main source of concrete nonabelian groups and the natural home for matrix calculations. The page [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra) is the natural place to continue with determinants, bases, change of basis, eigenvalues, and canonical forms. From the group-theoretic side, $GL_n(k)$ supplies many examples of subgroups, normal subgroups, quotient groups, and group actions. The page [Cambridge IA Groups](/page/Cambridge%20IA%20Groups) develops the parent language of groups, homomorphisms, actions, orbits, stabilisers, and normality. The page [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules) continues this story with modules and linear actions over rings. Over finite fields, the groups $GL_n(\mathbb{F}_q)$ and $SL_n(\mathbb{F}_q)$ become central examples in finite group theory and representation theory. Their orders, subgroups, and actions on finite vector spaces lead toward projective linear groups and many families of finite simple groups. In commutative algebra and algebraic geometry, $GL_n$ is also an algebraic group: its coordinate functions satisfy the single open condition $\det\ne 0$. This viewpoint connects invertible matrices to localisation, coordinate rings, and group schemes, themes developed further in [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Over $\mathbb{R}$ and $\mathbb{C}$, $GL_n$ is a Lie group. Its Lie algebra $\mathfrak{gl}(n,k)$ leads to matrix exponentials, flows of linear differential equations, and the structure theory of matrix Lie groups such as $SL_n$, $SO_n$, $U_n$, and symplectic groups. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Dummit and Foote, *Abstract Algebra* (2004). Lang, *Algebra* (2002). Artin, *Algebra* (2011).