Determine the Matrix Exponential \color{red} e^{\color{blue}A} =
\begin{pmatrix} E_{11} & E_{12}\\
E_{21} & E_{22} \end{pmatrix}
of the matrix \color{blue}
\begin{pmatrix} A & B\\
0 & D \end{pmatrix}
.
\color{red} E_{11}
=
e^{A}
\color{red} E_{12}
=
Be^{A}
\color{red} E_{21}
=
0
\color{red} E_{22}
=
e^{D}
Write
{\color{blue}
\begin{pmatrix} A & B\\
0 & D \end{pmatrix}}
= \underbrace{\begin{pmatrix} A & 0 \\
0 & D \end{pmatrix}}_{=B} +
\underbrace{\begin{pmatrix} 0 & B \\ 0 & 0 \end{pmatrix}}_{= C}
.
It holds that BC = CB.
Due to commutativity, it follows e^A = e^{B+C} = e^B e^C.
For the diagonal matrix B is
e^B = \begin{pmatrix} e^{A} & 0 \\ 0 & e^{D} \end{pmatrix}
.
For
e^C we consider the series:
e^C = E_2 + C+ \frac 12 C^2 + \ldots =
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
+ \begin{pmatrix} 0 & B \\ 0 & 0 \end{pmatrix}+ \frac 12 \begin{pmatrix} 0 & B \\ 0 & 0 \end{pmatrix}^2
+ \ldots \\
= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
+ \begin{pmatrix} 0 & B \\ 0 & 0 \end{pmatrix}+ \frac 12 \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
+ (\text{only zero matrices remain}) = \begin{pmatrix} 1 & B \\ 0 & 1 \end{pmatrix}.
Hence
e^A = e^{B+C} = e^B e^C
= \begin{pmatrix} e^{A} & 0 \\ 0 & e^{D} \end{pmatrix} \begin{pmatrix} 1 & B \\ 0 & 1
\end{pmatrix}
= \begin{pmatrix} e^{A} & B e^{A} \\ 0 & e^{D}\end{pmatrix}.