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 

$$\exp\begin{bmatrix}A & B \\ 0_{m, n} & 0_{m, m}\end{bmatrix} = \begin{bmatrix}\exp(A) & \left(\int_0^1 \exp(tA) \:dt\right)\cdot B \\0_{m, n} & I_{m}\end{bmatrix}$$

where $\exp$ be the matrix exponential function.

Solution -
Setting 

$$P = \begin{bmatrix} A & B \\ 0_{m, n} & 0_{m, m}\end{bmatrix}$$

we observe that the (unique) solution of the IVP $\mathbf{z}' = P\mathbf{z}$ 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{align*} \mathbf{x}(s) &= \exp(sA)\cdot\mathbf{x}_0 + \left(\int_0^1 \exp(stA) \:dt\right)\cdot sB\cdot \mathbf{y}_0, \\ \mathbf{y}(s) &= \mathbf{y}_0 \end{align*}

solve the system of differential equations

\begin{align*}\mathbf{x}' &= A\mathbf{x} + B\mathbf{y}, \\ \mathbf{y}' &= \mathbf{0}\end{align*}

and then put $s = 1$.

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

\begin{align*}\mathbf{x}'(s) &= A\exp(sA)\, \mathbf{x}_0 + \left(\int_0^1 (I_n + stA)\,\exp(stA) \:dt\right)B\mathbf{y}_0 \\ &= A\exp(sA)\, \mathbf{x}_0 + \exp(sA)\,B\mathbf{y}_0\end{align*}

via integration by parts, and

\begin{align*}A\mathbf{x}(s) &= A\exp(sA)\, \mathbf{x}_0 +  \left(\int_0^1 sA\,\exp(stA) \:dt\right)B\mathbf{y}_0 \\&= A\exp(sA)\, \mathbf{x}_0 + (\exp(sA) - I_n)\, B\mathbf{y}_0 \\&= \mathbf{x}'(s) - B\mathbf{y}(s)\end{align*}

hence completing the proof. $\square$