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Let $A=\begin{bmatrix}A_{11} & A_{12}\\0 & A_{22}\end{bmatrix}$ be a square matrix. Prove that $$e^{tA}=\begin{bmatrix}e^{tA_{11}} & F(t)\\0 & e^{tA_{22}}\end{bmatrix}$$ where $$F(t)=\int_0^te^{(t-s)A_{11}}\cdot A_{12}\cdot e^{sA_{22}}ds.$$

I do not know where to start. Can you give me any hint?

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If you differentiate $e^{tA}$ w.r.t. $t$ in the two obvious ways and equate the results you get that $$ \pmatrix{A_{11}e^{tA_{11}}&F'(t)\cr0&A_{22}e^{tA_{22}}\cr}=\frac d{dt}e^{tA}=Ae^{tA}=\pmatrix{A_{11}&A_{12}\cr0&A_{22}\cr}\pmatrix{e^{tA_{11}}&F(t)\cr0&e^{tA_{22}}\cr}. $$ Therefore $F(t)$ is the solution of the initial value problem: $F(0)=0$, and $$ F'(t)=A_{11}F(t)+A_{12}e^{tA_{22}}. $$ It sure looks like the given formula satisfies that matrix differential equation.

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HINT: $e^{tA} = I + tA + \frac{t^2A^2}{2} + \frac{t^3A^3}{3!} + \dots$

Note that $A^k = \begin{pmatrix}A_{11}^k & F\\0 & A_{22}^k\end{pmatrix}$, where $F=\sum_{i=1}^k A_{11}^iCA_{22}^{k-i}.$

Hope you can go forward.

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  • $\begingroup$ I tried but got stuck. I do not know how to get the integral from $\displaystyle\sum_{k=0}^{\infty}\dfrac{t^k}{k!}\sum_{i=1}^kA_{11}^{i-1}A_{12}A_{22}^{k-i}$ $\endgroup$ – Cachorro Sep 20 '17 at 13:38

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