Matrix exponential of a 2x2 matrix composed of a antihermitian matrix and a symmetric matrix As per the title, I'd like to calculate the exponential of a matrix which has an antihermitian component and a symmetric component (although this fact may not be useful). More specifically
$$\mathbf{M}=\begin{pmatrix}ia&b+ic\\b+ic&id\end{pmatrix}$$
I have in front of me the solutions for the case that $b = 0$ (antihermitian) and the case for $a,c,d = 0$ (symmetric), but not for $\mathbf M$. Does anyone know of a convenient form for calculating this?
Perhaps I should elaborate; computing by various methods is not a problem, but I'm wondering if there is a convenient set of formulae for this as there are for, say, the real case. To further elaborate, I've used a formula for convenience in the case of $\exp (-i\mathbf{H}t)$, where $\mathbf{H}t$ is Hermitian many times to solve the Schrödinger equation. This is really just to expedite calculations, since it's a waste of my time to do it the long way. I'm now working with dissipative systems a lot, and one way to handle them is with what is referred to as a non-hermitian Hamiltonian. That's where this question comes in.
 A: You can always perform a Schur decomposition on $\mathbf M$, $\mathbf M=\mathbf Q\mathbf T\mathbf Q^\ast$, where $\mathbf T$ is upper triangular, $\mathbf Q$ is unitary, and ${}^\ast$ is the conjugate transpose. Then you can compute the exponential of a $2\times 2$ triangular matrix easily:
$$\exp\begin{pmatrix}a&b\\0&c\end{pmatrix}=\begin{pmatrix}\exp\,a&\frac{b}{a-c}(\exp\,a-\exp\,c)\\0&\exp\,c\end{pmatrix}$$
and then multiply the unitary factors to obtain the exponential of your original matrix.
A: You can also bite the bullet and use the good ol' eigenvalue decomposition $M = VJV^{-1}$.
$$ e^M = Ve^JV^{-1}$$
Since your matrix $2\times 2$, this should not cause too much of a trouble.
Moreover, every nonhermitian(nonsymmetric) matrix can be decomposed into a Hermitian and anti-Hermitian (symmetric and anti-symmetric) part via
$$
S = \frac{M+M^*}{2}, A = \frac{M-M^*}{2}
$$
and $A+S = M$. Hence, there is not much of specialty. Your feeling about that fact is indeed correct. If this is not enough, you can look for the alternative ways for which there are at least 18 more. :) Well, let say at least 17 more since J.M. also gave another method.
