Compute ${e^{At}}$ for the matrix $A=\left[ {\begin{array}{*{20}{c}} B&I\\ 0&B \end{array}} \right]$ What is the easiest approach for computing ${e^{At}}$ for the following block matrix?
$$A = \left[ {\begin{array}{*{20}{c}}
B&I\\
0&B
\end{array}} \right]$$
my attempt: I was trying to write the matrix in diagonal form, and maybe from there computations are easy.
 A: Use the formula $e^{At} = \sum_{k=0}^\infty {(At)^n \over n!}$.
Prove the following formula for $A^n$ by recurrence.
$$
A^n = \pmatrix{B^n & nB^{n-1}\cr 0 & B^n \cr}
$$
A: Here is a simple way leading to the final result:
$$\exp(A)=\begin{pmatrix}\exp(B)&\exp(B)\\0&\exp(B)\end{pmatrix}$$
Why that ? Because one can decompose $A$ into the sum
$$A=\underbrace{\begin{pmatrix}B&0\\0&B\end{pmatrix}}_C+\underbrace{\begin{pmatrix}0&I_n\\0&0\end{pmatrix}}_D$$
of 2 commuting matrices ($CD=DC$) and it is known that in such a case:
$$\exp(A)=\exp(C) \exp(D)$$
$$\exp(A)=\begin{pmatrix}\exp(B)&0\\0&\exp(B)\end{pmatrix}\begin{pmatrix}I_n&I_n\\0&I_n\end{pmatrix}\tag{2}$$
giving (1).
Why can we say that the second matrix in the RHS of (2) is the good one ? Because $D$ is nilpotent with degree 2, meaning that $D^2=D^3=...=D^n=...=0$ giving:
$$\exp(D)=I_{2n}+D+\underbrace{(D^2/2!+...)}_0 = \begin{pmatrix}I_n&0\\0&I_n\end{pmatrix}+\begin{pmatrix}0&I_n\\0&0\end{pmatrix}.$$
Remark: decompositions involving nilpotent matrices are common and useful in many parts of linear algebra.
