Showing the complex conjugate of an eigenvector to a given eigenvalue is an eigenvector for the conjugate eigenvalue.
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 $\bar\lambda$.
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2 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$Define $$x=\begin{bmatrix}x_1\\ x_2\end{bmatrix}\\y=\begin{bmatrix}y_1\\ y_2\end{bmatrix}$$
Then, you already know that, $$A(x+iy) = \lambda(x+iy)=(\alpha + i\beta)(x+iy)=\alpha x + i\alpha y + i\beta x - \beta y$$ Splitting this into real and imaginary parts (and using the fact that $A$ is a real matrix), we have
$$Ax + i(Ay) = (\alpha x - \beta y) + i(\alpha y + \beta x)$$
from which you can conclude
- $Ax = \alpha x - \beta y$
- $Ay = \beta x + \alpha y$
On the other hand, you have
$$A(x-iy) = Ax - i(Ay)$$ and
$$\bar\lambda (x-iy) = (\alpha - i\beta)(x-iy) = (\alpha x - \beta y) + i(-\beta x - \alpha y)$$
Now, it should be just a matter of putting everything together.
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