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.