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For a finite-dimensional vector space $V$, the affine group consists of invertible affine maps $x\mapsto Ax+b$, where $A$ belongs to the general linear group and $b\in V$. Its product is $(A,b)(A^{\prime },b^{\prime })=(A{A}^{\prime },b+A{b}^{\prime })$, making it the semidirect product of the translation vector space and its general linear group. On the real line the positive-dilation subgroup is the real affine group used in the adjacent article.

The group $\mathrm{GL}(\mathbf{n})={\prod }_i{\mathrm{GL}}_{n_i}(k)$ acts by $(g\cdot f{)}_{\rho }=g_{t(\rho )}{f}_{\rho }{g}_{s(\rho )}^{-1}$. Inverse entries are regular on a general linear group, so this is an algebraic group action. Its stabilizer is the automorphism group of a quiver representation.

Tensoring a rational representation of the general linear group with a determinant power shifts every highest-weight coordinate by the same integer $r$. Positive twists can clear determinant denominators; negative twists supply rational representations that do not extend to singular matrices. Twisting preserves irreducibility and dimension.

An invertible complex Jordan block $\lambda I+N$, with $N^s=0$, has logarithm $\ell I+{\sum }_{k=1}^{s-1}(-1{)}^{k+1}(N/\lambda {)}^k/k$, where $e^{\ell }=\lambda$. Applying this construction blockwise and conjugating back proves surjectivity of the matrix exponential on the complex general linear group. There is no globally single-valued continuous choice of logarithm.

The endomorphisms of a vector space form a Lie algebra under the commutator $[A,B]=AB-BA$. It is the Lie algebra of the general linear group; the Jacobi identity follows by expanding associative products.

A group representation is a group homomorphism from a group to the general linear group of a vector space. It realizes abstract group elements as invertible linear transformations.

Since the differential of the Exponential map of a Lie group at zero is the identity, the inverse function theorem makes it a diffeomorphism on a sufficiently small neighborhood of zero. Its local inverse is a logarithm near the group identity. It need not extend to a global logarithm, even for the real general linear group.

The determinant is a smooth Lie group homomorphism from the real general linear group to ${\mathbb{R}}^{\times }$. Its derivative at the identity is the trace, and the flow-defined exponential on ${\mathbb{R}}^{\times }$ is the ordinary scalar exponential. Naturality of the Lie group exponential therefore proves the identity without diagonalizability assumptions. The series ${\sum }_{k\ge 0}{t}^k{A}^k/k!$ is the flow-defined matrix exponential because it solves $E^{\prime }=EA$, $E(0)=I$, and has inverse $E(-t)$.

A matrix group is a group represented by invertible matrices, equivalently a subgroup of a general linear group.

A matrix Lie group is a subgroup of a general linear group that is also a Lie group for its matrix topology. Near the identity, the matrix exponential and matrix logarithm give inverse coordinate charts whenever they are restricted to sufficiently small neighborhoods in the group and its tangent space.

The Lie algebra of a Lie group $G$ is the vector space $\mathfrak{g}=T_{e}G$ equipped with the bracket transferred from left-invariant vector fields. For $a\in G$, write $L_a(g)=ag$. A field is left-invariant when $(D_g{L}_a)X_g=X_{ag}$ for all $a,g$. Given $v\in T_{e}G$, defineThis is smooth and left-invariant, because $L_a{L}_g=L_{ag}$ and the chain rule applies. Conversely a left-invariant field is determined by its value at $e$, through this formula. Thusis a vector-space isomorphism, whose inverse is evaluation at the identity.

The Lie bracket of vector fields is preserved by diffeomorphisms: on functions this follows by transporting the commutator of derivations. Therefore the bracket of two left-invariant fields is again left-invariant. Define $[v,w]=[X_v,X_w{]}_e$. Bilinearity, antisymmetry and the Jacobi identity follow from the same properties of commutators of derivations. This supplies the asserted Lie algebra structure.

For the general linear group $G{L}_n(\mathbb{R})$, invertible matrices form an open subset of $M_n(\mathbb{R})$, so $T_{I}G{L}_n(\mathbb{R})=M_n(\mathbb{R})$. The left-invariant field determined by $A$ is $X_A(g)=gA$. Its ambient derivative is $D_g{X}_A(H)=HA$. Therefore differentiating left-invariant matrix fields givesHenceThis is the general linear Lie algebra, with the ordinary matrix commutator; right invariance with the same identification would give the opposite sign.

Now let $F:G\to H$ be a smooth Lie group homomorphism. It sends the identity to the identity and induces the linear mapSince $F\circ L_g=L_{F(g)}\circ F$, differentiation givesThese fields are $F$-related; no injectivity or surjectivity of $F$ is needed. For every smooth function $f$ on $H$,Apply this first with $w$ and then with $v$, and subtract the reversed order. The result isAt $e$, smooth functions detect tangent vectors, soThis proves that the differential of a Lie group homomorphism preserves Lie brackets, and hence that $F_{\ast }$ is a Lie algebra homomorphism.

For the exponential identity, let ${\eta }_v$ be the integral curve of $X_v$ through $e$. Left invariance and uniqueness show ${\eta }_v(t+s)={\eta }_v(t){\eta }_v(s)$ for small times: translating the curve through $e$ gives the curve through ${\eta }_v(s)$. Repeating this identity extends the curve to all real times. This proves completeness of left-invariant vector fields without assuming that an arbitrary smooth field is complete. Rescaling the parameter also gives ${\eta }_v(t)={\eta }_{tv}(1)$. By the definition of the Exponential map of a Lie group, ${\exp }_G(v)={\eta }_v(1)$.

The field-related identity shows that $F({\eta }_v(t))$ is an integral curve of $X_{F_{\ast }v}$ starting at $e_H$. By uniqueness it equals ${\eta }_{F_{\ast }v}(t)$ for all times. At time one,This is the naturality of the Lie group exponential.

Apply this to the determinant homomorphism $G{L}_n(\mathbb{R})\to {\mathbb{R}}^{\times }$. The derivative of the determinant at $I$ issince the permutation formula gives $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$. The left-invariant field on ${\mathbb{R}}^{\times }$ with initial tangent $c$ is $y\mapsto cy$; its curve through one solves $y^{\prime }=cy$, hence is $e^{ct}$. The matrix exponential determinant identity therefore follows from naturality:This argument uses only the flow definition of the exponential and the scalar exponential, not an unproved matrix formula.

To identify it explicitly with the usual matrix exponential, the series $E(t)={\sum }_{k\ge 0}{t}^k{A}^k/k!$ converges in operator norm, uniformly with its differentiated series on bounded intervals. Termwise differentiation gives $E^{\prime }=EA$ and $E(0)=I$. The series commutes with $A$, and differentiating $E(t)E(-t)$ gives zero, so $E(t)E(-t)=I$. Thus $E(t)$ stays in $G{L}_n(\mathbb{R})$ and, by uniqueness, is the integral curve of $X_A$ through $I$. The matrix series is exactly the flow-defined exponential, justifying either notation in the determinant identity.