Matrix exponential as the Lie-group exponential

Suppose first that $n\ge 1$ and put $\parallel A{\parallel }_{\mathrm{\infty }}={\max }_{i,j}\mathrm{\mid }{a}_{ij}\mathrm{\mid }$. The row-by-column formula [F4] gives $\parallel AB{\parallel }_{\mathrm{\infty }}\le n\parallel A{\parallel }_{\mathrm{\infty }}\parallel B{\parallel }_{\mathrm{\infty }}$, and hence $\parallel X^k{\parallel }_{\mathrm{\infty }}\le n^{k-1}\parallel X{\parallel }_{\mathrm{\infty }}^k$ for $k\ge 1$. Thus [F2] dominates the matrix series and its termwise derivative uniformly on every compact $t$-interval by scalar exponential series. Consequently, for $E(t)={\sum }_{k\ge 0}{t}^k{X}^k\mathrm{/}k!$, termwise differentiation is valid for every real $t$, and [F4] gives $E^{\mathrm{\prime }}(t)=XE(t)=E(t)X$ and $E(0)=I$. When $n=0$ this is the constant identity curve directly.

Let ${\gamma }_X:\mathbb{R}\to G$ be the subgroup from [F1], viewed as a matrix curve. Its velocity at $t$ is the left translate of $X$, so ${\gamma }_X^{\mathrm{\prime }}(t)={\gamma }_X(t)X$ and ${\gamma }_X(0)=I$. Transposing gives $({\gamma }_X^T{)}^{\mathrm{\prime }}=X^T{\gamma }_X^T$. Step 1.1 likewise gives $(E^T{)}^{\mathrm{\prime }}=X^T{E}^T$ with $E(0{)}^T=I$. On every compact interval containing zero, [F3] makes these transposed solutions equal. Thus ${\gamma }_X(t)=E(t)$ for all $t\in \mathbb{R}$.

Evaluating step 2.1 at $t=1$ proves the displayed formula and, because ${\gamma }_X(1)\in G$, also proves $e^X\in G$. For $n=0$ both sides are the identity of the one-point group, and for $X=0$ both equal $I$; no invertibility or spectral hypothesis on $X$ is used. The parameter is global, so $t=1$ is not an endpoint issue. This is an equality, not an iff claim. ${\mathrm{A}\mathrm{C}}_{\omega }$ is used exactly through [F1]; the series and ODE comparison add no choice.