# Power of a matrix in REAL jordan form

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 $$(J^\mathfrak{Re})^k = ?$$

• What do you mean by "the power matrix (of the real form)"? – Servaes Mar 5 at 20:53
• post edited, power matrix example added – incud Mar 5 at 21:03

$$J^{\Re} = \sigma I + \omega K$$ 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 $$(\sigma + j \omega)^k$$.