Given a $2\times 2$ matrix in Jordan canonical form, whose eigenvalues are a couple of complex conjugate values
$$ J = \left[ \begin{array}{cc} \sigma+j\omega & 0 \\ 0 & \sigma – j\omega \end{array} \right], $$
Its exponential matrix should be
$$ e^{Jt} = \left[ \begin{array}{cc} e^{\sigma t} e^{j\omega t} & 0 \\ 0 & e^{\sigma t} e^{j\omega t} \end{array} \right], $$
and its power matrix should be
$$ J^k = \left[ \begin{array}{cc} \binom{k}{1} (\sigma+j\omega)^k & 0 \\ 0 & \binom{k}{1} (\sigma-j\omega)^k \end{array} \right]. $$
However, we can write the original matrix in Jordan real form
$$ J^{\mathfrak{Re}} = \left[ \begin{array}{cc} \sigma & \omega \\ -\omega & \sigma \end{array} \right]. $$
The exponential matrix of the real form is
$$ e^{J^{\mathfrak{Re}} t } = (e^{\sigma t}) \left[ \begin{array}{cc} \cos(\omega t) & \sin(\omega t) \\ -\sin(\omega t) & \cos(\omega t) \end{array} \right], $$
but what is its power matrix
$$ (J^\mathfrak{Re})^k = ? $$
Best Answer
$$J^{\Re} = \sigma I + \omega K$$ where $$I = \pmatrix{1 & 0\cr 0 & 1\cr},\ K = \pmatrix{0 & 1\cr -1 & 0\cr}$$ Note that $K^2 = -I$. The map $\sigma + j \omega \to \sigma I + \omega K$ is a ring homomorphism from $\mathbb C$ to the $2 \times 2$ real matrices. Thus $$ (J^\Re)^k = \sigma_k I + \omega_k K$$ where $\sigma_k$ and $\omega_k$ are the real and imaginary parts of $(\sigma + j \omega)^k$.