Matrix exponential of a particular block matrix Can anyone compute the following matrix exponential
$$\Phi = \exp\begin{pmatrix} -\alpha I_3 & \alpha A \\ I_3 & O_3 \\
\end{pmatrix}$$
where $A$ is an arbitrary $3 \times 3$ matrix, $I_3$ is the $3 \times 3$ identity and $O_3$ is the $3 \times 3$ zero matrix?
Thanks in advance!

Edit: This comes about as the solution to a linear ODE. I tried evaluating it in MAPLE for the 2D case and it gives a result, but MAPLE can only do calculations component-wise so it's difficult to extract a concise expression in terms of the matrix A. For the 2D case, the answer seems to be in the form
$$\Phi=M_1e^{\omega_1}+M_2e^{-\omega_1}+M_3e^{\omega_2}+M_4e^{-\omega_2}$$
where $$\omega_{1}=\alpha/2+1/2\,\sqrt {\alpha\, \left( \alpha+2\,\sqrt { \left( {\it 
Tr} \left( A \right)  \right) ^{2}-4\,{\it \det} \left( A
 \right) }+2\,{\it Tr} \left( A \right)  \right) }
$$
$$\omega_{2}=-\alpha/2+1/2\,\sqrt {\alpha\, \left( \alpha-2\,\sqrt { \left( {\it 
Tr} \left( A \right)  \right) ^{2}-4\,{\it \det} \left( A
 \right) }+2\,{\it Tr} \left( A \right)  \right) }
$$
and all the $M_i$ are 4 by 4 matrices. It is not clear to me if $\Phi$ will have this structure in 3 dimensions, however.
 A: Write out the eigenvector equation
$$
Mx = \lambda x \implies \pmatrix{-\alpha I & \alpha A\\ I & 0} \pmatrix{x_1\\x_2} = \lambda \pmatrix{x_1\\x_2}. 
$$
That is, we have the system
$$
\begin{cases}
- \alpha x_1 + \alpha Ax_2 = \lambda x_1\\
x_1 = \lambda x_2.
\end{cases}
$$
Substituting the second equation into the first yields
$$
- \alpha \lambda x_2 + \alpha Ax_2 = \lambda^2 x_2 \implies\\
Ax_2 = \frac{\lambda(\lambda + \alpha)}{\alpha}x_2.
$$
So: if $\alpha$ is non-zero, then for every eigenvector $v$ of $A$ associated with $\mu$ and the solutions to $\lambda(\lambda+ \alpha)/\alpha = \mu$, the vector $x = (x_1,x_2)$ with $x_1 = \lambda v$ and $x_2 = v$ is an eigenvector of $M$.
If $\alpha \neq 0$ and $A$ is diagonalizable, this allows us to compute the exponential of $M$ directly.

Yet another approach: suppose that $J = S^{-1}AS$ is the Jordan form of $A$. It follows that
$$
M_2 = \pmatrix{S & 0\\0 & S}
\pmatrix{-\alpha I & \alpha A\\ I & 0}
\pmatrix{S & 0\\0 & S}^{-1} = 
\pmatrix{-\alpha I  & \alpha J \\ I & 0}.
$$
If $J = \operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$, then we can see that $M_2$ is permutation similar to the block-diagonal matrix
$$
\operatorname{diag}\left[\pmatrix{-\alpha  & \alpha \lambda_1\\ 1 & 0},
\pmatrix{-\alpha  & \alpha \lambda_2\\ 1 & 0},
\pmatrix{-\alpha & \alpha \lambda_3\\ 1 & 0}\right].
$$
The blocks can be exponentiated separately.

(My previous answer below)
One approach is to simply use eigenvalues to calculate the Jordan form and proceed in the standard fashion. The key here, however, is that it is easy to find the eigenvalues of $M$ using only the eigenvalues of $A$. Denote
$$
M = \pmatrix{-\alpha I & \alpha A\\ I & 0}.
$$
Because the blocks of $M$ commute, we find that
$$
\det(M - \lambda I) = \det((-\alpha - \lambda I)(-\lambda I) - (\alpha A) (I))
\\
= \det(\lambda(\alpha + \lambda)I - \alpha A).
$$
So, if $p$ is the characteristic polynomial of $A$ (i.e. $p(\lambda) = \det(A - \lambda I)$), then the characteristic polynomial of $M$ is $p(\frac{\lambda(a + \lambda)}{\alpha})$. So, for each eigenvalue $\lambda$ of $A$, both $\lambda$ and $\lambda - \alpha$ are eigenvalues of $M$.

In fact, we could say a bit more. Suppose that $\lambda$ is an eigenvalue of $A$. We find that
$$
\operatorname{rank}(\lambda I - M) = 
\operatorname{rank} \pmatrix{-\alpha I - \lambda I & \alpha A\\ I & -\lambda I}.
$$
If it holds that $\alpha$ is not an eigenvalue of $A$, then the nullity of $M - \lambda I$ is equal to the nullity of the Schur complement
$$
(M - \lambda I)/(-\alpha I - \lambda) = -\lambda I + \frac{\alpha}{\lambda + \alpha}A \\
\to \frac{\lambda(\lambda + \alpha)}{\alpha} I - A
$$
A: That follows is valid for every dimension $n$.
We consider $U=\begin{pmatrix}-aI_n&au\\I_n&0_n\end{pmatrix}$ where $u\in M_n(\mathbb{C}),a\in\mathbb{C}^*$; we use the notation $a+u$ for $aI_n+u$.
If we put $w=\sqrt{a(aI_n+4u)}$, and if $-a/4\notin spectrum(A)$ then $e^U=$

It remains to correctly choose the square root $w$; there are several square roots and, moreover, we must choose one which commutes with $u$.
We assume that, for every $\lambda\in spectrum(u), a(a+4\lambda)\notin (-\infty,0]$. Then $a(a+4u)$ admits a unique square root whose spectrum lies in the open right half plane; moreover, this square root is a polynomial in $u$ (cf. Functions of matrices, Higham). Note that, in these conditions, to divide by $w$ has a (obvious) mathematical meaning.
