[proofplan] We use the power-series definition of the matrix exponential. Absolute convergence, supplied by [citetheorem:8777], allows us to multiply the two exponential series by the Cauchy product and regroup terms by total degree. Since both matrices are scalar multiples of the same matrix $X$, their powers commute and the coefficient of $X^k$ is the usual [binomial coefficient](/page/Binomial%20Coefficient) sum. The inverse formula follows by applying the exponential law to the parameters $t$ and $-t$. [/proofplan]

[step:Expand both exponentials as absolutely convergent matrix series] Fix $s,t \in \mathbb C$. Let $\|\cdot\|$ be any submultiplicative matrix norm on $M(n,\mathbb C)$. By [citetheorem:8777], the matrix exponential is defined by the absolutely convergent series \begin{align*} \exp(A)=\sum_{m=0}^{\infty}\frac{A^m}{m!} \end{align*} for every $A \in M(n,\mathbb C)$. Applying this to $A=sX$ and $A=tX$ gives \begin{align*} \exp(sX)=\sum_{m=0}^{\infty}\frac{(sX)^m}{m!} \end{align*} and \begin{align*} \exp(tX)=\sum_{\ell=0}^{\infty}\frac{(tX)^\ell}{\ell!}. \end{align*} [/step]

[step:Multiply the absolutely convergent series by the Cauchy product]Define sequences $(A_m)_{m\geq 0}$ and $(B_\ell)_{\ell\geq 0}$ in $M(n,\mathbb C)$ by \begin{align*} A_m=\frac{(sX)^m}{m!} \end{align*} and \begin{align*} B_\ell=\frac{(tX)^\ell}{\ell!}. \end{align*} The two series $\sum_{m=0}^{\infty}A_m$ and $\sum_{\ell=0}^{\infty}B_\ell$ are absolutely convergent. Since the norm is submultiplicative, \begin{align*} \sum_{m=0}^{\infty}\sum_{\ell=0}^{\infty}\|A_mB_\ell\|\leq \left(\sum_{m=0}^{\infty}\|A_m\|\right)\left(\sum_{\ell=0}^{\infty}\|B_\ell\|\right)<\infty. \end{align*} Hence the double series $\sum_{m,\ell\geq 0}A_mB_\ell$ is absolutely convergent, so it may be regrouped by the total degree $k=m+\ell$. Therefore \begin{align*} \exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\sum_{m=0}^{k}\frac{(sX)^m(tX)^{k-m}}{m!(k-m)!}. \end{align*}[/step]

[guided]We need to justify the passage from the product of two infinite matrix series to a single series indexed by total degree. This is the only convergence point in the proof. Define \begin{align*} A_m=\frac{(sX)^m}{m!} \end{align*} and \begin{align*} B_\ell=\frac{(tX)^\ell}{\ell!}. \end{align*} By [citetheorem:8777], both $\sum_{m=0}^{\infty}A_m$ and $\sum_{\ell=0}^{\infty}B_\ell$ converge absolutely in the matrix norm $\|\cdot\|$. The submultiplicativity hypothesis on the norm gives $\|A_mB_\ell\|\leq \|A_m\|\|B_\ell\|$ for all $m,\ell\geq 0$. Thus \begin{align*} \sum_{m=0}^{\infty}\sum_{\ell=0}^{\infty}\|A_mB_\ell\|\leq \left(\sum_{m=0}^{\infty}\|A_m\|\right)\left(\sum_{\ell=0}^{\infty}\|B_\ell\|\right)<\infty. \end{align*} So the double series of products is absolutely convergent. Absolute convergence is exactly what permits the Cauchy product regrouping: the product of the two sums equals the sum over all pairs $(m,\ell)$, and this absolutely convergent double sum may be collected along the finite diagonals $m+\ell=k$. Therefore \begin{align*} \exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\sum_{m=0}^{k}\frac{(sX)^m(tX)^{k-m}}{m!(k-m)!}. \end{align*}[/guided]

[step:Use commutativity of scalar multiples of one matrix to identify the coefficients] For each $k\geq 0$ and each $m\in\{0,\dots,k\}$, scalar multiplication commutes with matrix multiplication and powers of the same matrix multiply by adding exponents. Hence \begin{align*} (sX)^m(tX)^{k-m}=s^m t^{k-m}X^mX^{k-m}=s^m t^{k-m}X^k. \end{align*} Substituting this into the Cauchy product gives \begin{align*} \exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\left(\sum_{m=0}^{k}\frac{s^m t^{k-m}}{m!(k-m)!}\right)X^k. \end{align*} For each $k\geq 0$, the binomial identity gives \begin{align*} \sum_{m=0}^{k}\frac{s^m t^{k-m}}{m!(k-m)!}=\frac{(s+t)^k}{k!}. \end{align*} Therefore \begin{align*} \exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\frac{((s+t)X)^k}{k!}=\exp((s+t)X). \end{align*} [/step]

[step:Apply the exponential law with opposite parameters to obtain the inverse] Let $t\in\mathbb C$. Applying the identity already proved with parameters $t$ and $-t$ yields \begin{align*} \exp(tX)\exp(-tX)=\exp(0X)=I. \end{align*} Applying the same identity with parameters $-t$ and $t$ yields \begin{align*} \exp(-tX)\exp(tX)=\exp(0X)=I. \end{align*} Thus $\exp(-tX)$ is both a right inverse and a left inverse for $\exp(tX)$. Hence $\exp(tX)$ is invertible, so $\exp(tX)\in GL(n,\mathbb C)$, and its inverse is \begin{align*} \exp(tX)^{-1}=\exp(-tX). \end{align*} This proves both assertions. [/step]