AMM problem 12451

This is AMM problem 12451 that I solved with my friends Satvik Saha and Sohom Gupta.

Problem Statement -
Let $A$ and $B$ be complex $n\times n$ and $n\times m$ matrices respectively. Let $0_{m,n}$ denote the $m\times n$ zero matrix and let $I_m$ denote the $m\times m$ identity matrix. Prove 

$$\exp \left[\begin{array}{cc}A& B\\ 0_{m,n}& 0_{m,m}\end{array}\right]=\left[\begin{array}{cc}\exp (A)& ({\int }_0^1\exp (tA)dt)\cdot B\\ 0_{m,n}& I_m\end{array}\right]$$

where $\exp$ be the matrix exponential function.

Solution -
Setting 

$$P=\left[\begin{array}{cc}A& B\\ 0_{m,n}& 0_{m,m}\end{array}\right]$$

we observe that the (unique) solution of the IVP ${\mathbf{z}}^{\prime }=P\mathbf{z}$ with $\mathbf{z}(0)={\mathbf{z}}_0$ is precisely $\mathbf{z}(s)=\exp (sP){\mathbf{z}}_0$ for all ${\mathbf{z}}_0\in {\mathbb{C}}^{m+n}$.

Thus, it suffices to check that for all ${\mathbf{x}}_0\in {\mathbb{C}}^n,{\mathbf{y}}_0\in {\mathbb{C}}^m$, the curves

$$\begin{array}{rl}\mathbf{x}(s)& =\exp (sA)\cdot {\mathbf{x}}_0+({\int }_0^1\exp (stA)dt)\cdot sB\cdot {\mathbf{y}}_0,\\ \mathbf{y}(s)& ={\mathbf{y}}_0\end{array}$$

solve the system of differential equations

$$\begin{array}{rl}{\mathbf{x}}^{\prime }& =A\mathbf{x}+B\mathbf{y},\\ {\mathbf{y}}^{\prime }& =\mathbf{0}\end{array}$$

and then put $s=1$.

This is easily verified; with $\mathbf{x}(s)$ as above, we have

$$\begin{array}{rl}{\mathbf{x}}^{\prime }(s)& =A\exp (sA){\mathbf{x}}_0+({\int }_0^1(I_n+stA)\exp (stA)dt)B{\mathbf{y}}_0\\ & =A\exp (sA){\mathbf{x}}_0+\exp (sA)B{\mathbf{y}}_0\end{array}$$

via integration by parts, and

$$\begin{array}{rl}A\mathbf{x}(s)& =A\exp (sA){\mathbf{x}}_0+({\int }_0^{1}sA\exp (stA)dt)B{\mathbf{y}}_0\\ & =A\exp (sA){\mathbf{x}}_0+(\exp (sA)-I_n)B{\mathbf{y}}_0\\ & ={\mathbf{x}}^{\prime }(s)-B\mathbf{y}(s)\end{array}$$

hence completing the proof. $◻$