A problem on matrix exponential

Question

I know that $$e^{M}=\sum_{k=0}^{+\infty} \frac{M^k}{k!}$$ But from this I stuck

Answer 1

Using the power series definition is probably the neatest approach for this problem in any case. Verify (by induction, perhaps) that for $k \geq 1$, $$ \begin{bmatrix} A & B\\0 & 0 \end{bmatrix}^k = \begin{bmatrix} A^k & A^{k-1}B\\0 & 0 \end{bmatrix}. $$ Thus, conclude that $$ \exp\begin{bmatrix} A & B\\0 & 0 \end{bmatrix} = \begin{bmatrix} \sum_{k=0}^\infty A^k/k! & \left(\sum_{k=1}^\infty A^{k-1}/k! \right)B\\0 & I \end{bmatrix}. $$ With that, it suffices to verify that $\sum_{k=1}^\infty x^{k-1}/k!$ is the power series corresponding to the function $f(x) = \int_0^1 \exp(tx)\,dt$.

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Below is an incomplete attempt at an alternative approach

As an alternative, we could use the IVP definition of the matrix exponential, which is to say that the function $\Phi:\Bbb R \to \Bbb R^{(n+m) \times (n+m)}$ given by $\Phi(x) = \exp(Mx)$ satisfies $\frac{d\Phi}{dx} = M\Phi$ and $\Phi(0) = I$.

With that in mind, define $$ \Phi(x) = \begin{bmatrix} \exp(Ax) & x \int_0^1 \exp(txA)\,dt \cdot B\\ 0 & I \end{bmatrix}. $$ Note that $\frac d{dx}\exp(xA) = A\exp(xA)$, and using integration by parts compute $$ \begin{align*} \frac d{dx}\int_0^1 \exp(txA)\,dt &= \int_0^1 \frac{\partial}{\partial x} \exp(txA)\,dt \\ & = \int_0^1 tA \exp(txA)\,dt \\ & = \left[(t/x)\exp(txA) \right]_{t = 0}^1 - \int_0^1 \frac 1x \exp(txA)\,dt \\ & = \frac 1x\left[\exp(xA) - \int_0^1\exp(txA)\,dt\right] \end{align*} $$ Thus, with the product rule, we have $$ \frac d{dx}x\int_0^1 \exp(txA)\,dt = 1\cdot \int_0^1 \exp(txA)\,dt + x \cdot \frac 1x \left[\exp(xA) - \int_0^1\exp(txA)\,dt\right] \\ = \exp(xA). $$ With that, we find $$ \frac d{dx}\Phi(x) = \begin{bmatrix} A \exp(xA) & \exp(xA) B\\ 0 & 0 \end{bmatrix} $$