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For an integer $c$, repeated multiplicativity of the determinant, with the inverse identity for negative powers, givesFor an arbitrary real $c$, an invertible real matrix does not by itself define a unique real matrix power. For example, the scalar matrix $(-1)$ has no real square root. If a matrix logarithm $L$ has been chosen with $e^L=A$ and $A^c$ is defined as $e^{cL}$, the matrix exponential determinant identity gives the precise formulaThis cannot always be replaced by an independently chosen scalar branch of $(\det A{)}^c$. For instance, $A=-I_2$ has the complex square root $i{I}_2$, whose determinant is $-1$, while the principal scalar square root of $\det A=1$ is $+1$. For a positive-definite symmetric matrix with its usual positive spectral power, all eigenvalues are positive and the familiar formula $(\det A{)}^c$ holds for every real $c$.

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.

The convergent power series defining the matrix exponential gives $e^A=U{e}^D{U}^{†}$, with $e^D=\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})$. Hence the matrix exponential determinant identity isThe diagonalization proof uses the given Hermitian hypothesis; the identity itself holds for every square real or complex matrix.

For any fixed real $n$-by-$n$ matrix $A$, as the real parameter $t$ tends to zero,The big O notation means the remainder is bounded in absolute value by $C{t}^2$ for some constant $C$ near zero. The determinant stays positive for sufficiently small $t$, so its oriented volume factor is also its ordinary volume factor. Thus the trace is the first-order fractional volume change, measured from the identity matrix.

A proof follows directly from multilinearity of the determinant. Each column of $I+tA$ is $e_j+tA{e}_j$. The constant term chooses all the standard basis columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $A{e}_j$; expanding that vector in the standard basis leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\mathrm{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the derivative of the determinant at the identity matrix.

This coefficient has a direct visual meaning. At the point $e_j$, the linear map $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional area or volume change; the off-diagonal entries initially change shape rather than contributing to this first-order total.

For the constant-coefficient linear system of differential equations $x^{\prime }=Ax$, the exact linear flow is $x(t)=e^{tA}x(0)$, using the matrix exponential. Its volume factor isThis is the matrix exponential determinant identity. One can derive it without diagonalization: set $E(t)=e^{tA}$, whose matrix inverse is $E(-t)$, and use $E^{\prime }=AE$ in Jacobi's formula to get $V^{\prime }=\det E\mathrm{tr}(E^{-1}AE)=(\mathrm{tr}A)V$. The scalar differential equation with initial value one gives the displayed exponential function. Thus positive trace causes volume expansion, negative trace causes contraction, and zero trace preserves volume, even when individual directions behave differently.

In two dimensions, compare four generators at $t=0.4$:The first and third diagonal matrices expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged area. For the last nilpotent matrix, $A^2=0$, so its matrix exponential is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a shear mapping. Zero trace allows substantial shape change and does not mean the flow map is stationary.

The first-order statement and the exact flow map should be distinguished. For $A=\mathrm{diag}(1,-1)$, $\det (I+tA)=1-t^2$, whereas $\det (e^{tA})=1$ for all real $t$. Zero trace eliminates the first-order change in the approximate step, and it preserves volume exactly in the matrix exponential flow. The examples use an abstract parameter $t$ and coordinate vectors; no physical units are assumed.