Finite matrix exponential series and its identities

If $A$ has size $0\times 0$, the matrix space has one element, which is both $0$ and $I_0$; every partial sum in the definition is this same matrix. Thus $e^{tA}=I_0$ for every $t$, its derivative is the zero matrix, and the derivative identities, initial value, and commuting-product identity all hold because every product and sum is the unique empty matrix. Hence assume below that $A$ has size $n\times n$ with $n\ge 1$.

Put $M=\mathrm{\parallel }A{\mathrm{\parallel }}_1$. By induction using [F1] and [F4], $\mathrm{\parallel }{A}^k{\mathrm{\parallel }}_1\le M^k$ for every $k\ge 0$. For $\mathrm{\mid }t\mathrm{\mid }\le T$, [F5] then bounds each entry of $t^k{A}^k\mathrm{/}k!$ by $(TM{)}^k\mathrm{/}k!$. The scalar majorant converges by [F2], so each entry series converges absolutely and uniformly on $[-T,T]$; its entrywise sums define a matrix $E(t)$. Each column has only $n$ entries, so the maximum-column-sum formula [F5] bounds the norm of a matrix tail by $n$ times the scalar majorant tail. Therefore the matrix series converges in induced $1$-norm uniformly on $[-T,T]$. Since $T$ was arbitrary, it converges locally uniformly and defines $e^{tA}=E(t)$.

Suppose $A,B\in M_n(\mathbb{R})$ commute. For each $r\ge 0$ set $D_r:={\sum }_{m=0}^r{A}^m{B}^{r-m}\mathrm{/}(m!(r-m)!)$. Induction from $AB=BA$ and [F3] gives $B^{s}A=A{B}^s$ for every $s\ge 0$. Distributing $D_r(A+B)$ and using this commutation, the coefficient of $A^j{B}^{r+1-j}$ is $1\mathrm{/}(j!(r-j)!)+1\mathrm{/}((j-1)!(r+1-j)!)$ for $1\le j\le r$, which equals $(r+1)\mathrm{/}(j!(r+1-j)!)$ by [F6]; for $j=0$ and $j=r+1$ the single endpoint coefficient has that same value by $(r+1)!=(r+1)r!$. Thus $D_r(A+B)=(r+1)D_{r+1}$. Since $D_0=I_n$, induction, the recursive power law [F4], and the nonzero factorial images in [F6] give $D_r=(A+B{)}^r\mathrm{/}r!$ for every $r$.

Fix $c\in \mathbb{R}$ and choose $R>\mathrm{\mid }c\mathrm{\mid }$. If $\mathrm{\mid }h\mathrm{\mid }<R-\mathrm{\mid }c\mathrm{\mid }$, then $\mathrm{\mid }c\mathrm{\mid },\mathrm{\mid }c+h\mathrm{\mid }<R$; for $k\ge 1$, the factorization $(c+h{)}^k-c^k=h{\sum }_{j=0}^{k-1}(c+h{)}^{k-1-j}{c}^j$ bounds the norm of the $k$th matrix-series difference quotient by $M^k{R}^{k-1}\mathrm{/}(k-1)!$, and the same bound holds for the $k$th proposed derivative term $c^{k-1}{A}^k\mathrm{/}(k-1)!$. The sum of these bounds is $M{\sum }_{j\ge 0}(MR{)}^j\mathrm{/}j!<\mathrm{\infty }$ by [F2]. For each finite partial sum, the factorization also shows that its difference quotient tends to the finite sum of the proposed derivative terms, using $(k+1)!=(k+1)k!$ from [F6]. For each fixed nonzero $h$, the full difference quotient is the limit of those of the finite partial sums by [step 1.2]; the two uniform tail bounds let us first choose a large partial sum and then take $h\to 0$. Consequently $E$ is differentiable at $c$ and $E^{\mathrm{\prime }}(c)={\sum }_{k\ge 1}{c}^{k-1}{A}^k\mathrm{/}(k-1)!$.

Apply the power bound and convergence argument of step 1.2 to $B$ and $A+B$ as well. Put $a_m=\mathrm{\parallel }A{\mathrm{\parallel }}_1^m\mathrm{/}m!$ and $b_n=\mathrm{\parallel }B{\mathrm{\parallel }}_1^n\mathrm{/}n!$; both scalar series converge by [F2]. The finite product $E_N(A)E_N(B)$ expands by [F3] as the square sum over $0\le m,n\le N$ of $A^m{B}^n\mathrm{/}(m!n!)$, whose term norms are at most $a_m{b}_n$ by [F1] and [F4]. Its difference from the triangular sum over $m+n\le N$ has norm at most $({\sum }_{m>N\mathrm{/}2}{a}_m)({\sum }_n{b}_n)+({\sum }_m{a}_m)({\sum }_{n>N\mathrm{/}2}{b}_n)$, which tends to zero; every omitted pair has $m>N\mathrm{/}2$ or $n>N\mathrm{/}2$. By step 1.2, $E_N(A)E_N(B)\to e^A{e}^B$ (matrix multiplication is continuous by [F1]); by step 1.3 the triangular sum is ${\sum }_{r=0}^N(A+B{)}^r\mathrm{/}r!$, which converges to $e^{A+B}$. Therefore $e^A{e}^B=e^{A+B}$.

From [F3] and the recursive power definition [F4], induction gives $A^k=A{A}^{k-1}=A^{k-1}A$ for every $k\ge 1$. Hence the derivative series in step 2.1 is the limit of both $A{\sum }_{j=0}^N{c}^j{A}^j\mathrm{/}j!$ and $({\sum }_{j=0}^N{c}^j{A}^j\mathrm{/}j!)A$ as $N\to \mathrm{\infty }$. Submultiplicativity [F1] makes multiplication by fixed $A$ continuous, so these limits are $AE(c)$ and $E(c)A$. At $c=0$, all positive-degree terms vanish and the constant term is $A^0=I_n$. Thus $E^{\mathrm{\prime }}(c)=AE(c)=E(c)A$ and $E(0)=I_n$.

Step 1.1 covers the zero-sized matrix case. For $n\ge 1$, steps 1.2 and 2.1 establish convergence and differentiation, step 3.1 gives the derivative identities and initial value, and steps 1.3 and 2.2 prove the commuting-product identity. Thus every claim in the Statement holds.