I'm taking an introductory differential equations course, and we learned the matrix method for systems. We're currently covering complex eigenvalues, and I was asked to prove the following problem. However, I have no idea of how to start, and I was hoping you could provide hints as to what the solution is.
What I know: If $\mathbf{x_1}+i\mathbf{x_2}$ is a solution for the system $\mathbf{\dot x}=A\mathbf{x}$, then its real and imaginary parts $\mathbf{x_{1}},\mathbf{x_{2}}$ are solutions to the system.
Suppose the matrix $\mathbf A$ with real entries has eigenvalues $\lambda=\alpha+i\beta$ and $\bar\lambda=\alpha-i\beta$. Suppose also that $\mathbf Y_{0}=\binom{x_{1}+iy_{1}}{x_{2}+iy_{2}}$ is an eigenvector for the eigenvalue $\lambda$. Show that $\mathbf{\bar Y_{0}}=\binom{x_{1}-iy_{1}}{x_{2}-iy_{2}}$ is an eigenvector for the eigenvalue $\bar\lambda$. In other words the complex conjugate of an eigenvector for $\lambda$ is an eigenvector for $\bar\lambda$.
Best Answer
As $Y_0$ is an eigenvector you have $AY_O = \lambda Y_0$. Now take the conjugates of both sides, which for matrix means conjugating all entries you have:
$$\overline{AY_0} = \overline{\lambda Y_0} \iff \bar{A }\bar{Y_0} = \bar{\lambda} \bar{Y_0} \iff A\overline{Y_0} = \bar{\lambda}\bar{Y_0}$$
Therefore $\bar{\lambda}$ is an eigenvalue and $\overline{Y_0}$ is an eigenvector corresponding to it.