The exponential Fréchet derivative has an integral and block-matrix formula

For compatible matrices $U,V$, Cauchy–Schwarz applied to each entry gives $\mathrm{\mid }(UV{)}_{ij}{\mathrm{\mid }}^2\le ({\sum }_k\mathrm{\mid }{u}_{ik}{\mathrm{\mid }}^2)({\sum }_k\mathrm{\mid }{v}_{kj}{\mathrm{\mid }}^2)$. Summing over $i,j$ proves $\mathrm{\parallel }UV{\mathrm{\parallel }}_F\le \mathrm{\parallel }U{\mathrm{\parallel }}_F\mathrm{\parallel }V{\mathrm{\parallel }}_F$. Use this submultiplicative Frobenius norm below.

The cycle trace is compact and disjoint from $\sigma (A)$, so $R(z)=(zI-A{)}^{-1}$ is continuous and bounded on it. By [F4], each eigenvalue of $A$ has an open neighborhood contained in $\mathrm{\Omega }$ and disjoint from the trace on which the cycle index remains one (intersect the index neighborhood with $\mathrm{\Omega }$, which is open and contains that eigenvalue); the finite union of these neighborhoods is an open index-one region $U_1$. Every sufficiently small $H$ has $\sigma (A+H)\subset U_1$: otherwise choose $H_j\to 0$ and ${\lambda }_j\in \sigma (A+H_j)\setminus U_1$. Normalize corresponding eigenvectors to Euclidean norm one; Cauchy–Schwarz gives $\mathrm{\mid }{\lambda }_j\mathrm{\mid }\le \mathrm{\parallel }A+H_j{\mathrm{\parallel }}_F$, so a subsequence converges to some $\lambda$. Continuity of the determinant, a polynomial in the entries, gives $\det (\lambda I-A)=0$, so $\lambda \in \sigma (A)\subset U_1$. Since $U_1$ is open, ${\lambda }_j$ is eventually in $U_1$, a contradiction.

Put $M={\sup }_{z\in \mathrm{\Gamma }}\mathrm{\parallel }R(z){\mathrm{\parallel }}_F$. If $\mathrm{\parallel }H{\mathrm{\parallel }}_F<1\mathrm{/}(2M)$, then $\mathrm{\parallel }R(z)H{\mathrm{\parallel }}_F<1\mathrm{/}2$ uniformly. The Neumann expansion gives $(zI-A-H{)}^{-1}=R(z)+R(z)HR(z)+{\sum }_{k\ge 2}(R(z)H{)}^{k}R(z),$ and its final sum is bounded uniformly by ${\sum }_{k\ge 2}{M}^{k+1}\mathrm{\parallel }H{\mathrm{\parallel }}_F^k\le 2M^3\mathrm{\parallel }H{\mathrm{\parallel }}_F^2$.

Apply [F2] to $A+H$ and $A$, whose spectra lie in $U_1$ by step 1.2. Subtract the contour formulas and use the uniform expansion in step 2.1. The integral of its remainder is $O(\mathrm{\parallel }H{\mathrm{\parallel }}_F^2)$ because $\mathrm{\Gamma }$ has finite length. Therefore $f(A+H)-f(A)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}f(z)R(z)HR(z)dz+O(\mathrm{\parallel }H{\mathrm{\parallel }}_F^2),$ which is the real Fréchet derivative in [F1] and gives the stated derivative formula.

For $\tilde{A}=\left(\begin{array}{cc}A& E\\ 0& A\end{array}\right)$, direct block multiplication gives $(zI-\tilde{A}{)}^{-1}=\left(\begin{array}{cc}R(z)& R(z)ER(z)\\ 0& R(z)\end{array}\right)\mathrm{.}$ The determinant of a block upper-triangular matrix is the product of the determinants of its diagonal blocks, so $\sigma (\tilde{A})=\sigma (A)$ and the cycle has index one on this spectrum. Applying [F2] to $\tilde{A}$ and reading its four blocks therefore gives the asserted block-function identity, with the upper-right block equal to the derivative integral from step 3.1.

The terms linear in $E$ in the absolutely convergent series for $\exp (A+E)$ are ${\sum }_{p,q\ge 0}{A}^{p}E{A}^q\mathrm{/}(p+q+1)!$; terms containing at least two factors $E$ are $O(\mathrm{\parallel }E{\mathrm{\parallel }}_F^2)$ by the binomial bound, submultiplicativity from step 1.1, and absolute convergence [F3]. Indeed their norm sum is at most $e^{\mathrm{\parallel }A{\mathrm{\parallel }}_F}(e^{\mathrm{\parallel }E{\mathrm{\parallel }}_F}-1-\mathrm{\parallel }E{\mathrm{\parallel }}_F)=O(\mathrm{\parallel }E{\mathrm{\parallel }}_F^2)$. Thus this is also $D\exp (A)[E]$. Expanding $e^{(1-t)A}E{e}^{tA}$ into its two uniformly absolutely convergent series and integrating termwise, the coefficient of $A^{p}E{A}^q$ is $\frac{1}{p!q!}{\int }_0^1(1-t{)}^p{t}^{q}dt=\frac{1}{(p+q+1)!},$ where the integral identity follows by repeated integration by parts. Hence the time integral equals the same derivative series.