Exercise 5.A.7

Proof. Taking $\lambda (x,y)=(-3y,x)$ gives $\mathit{\lambda x}=-3y$ and $\mathit{\lambda y}=x$. It follows that $y\ne 0$, because otherwise $x$ would be $0$ and we’re not looking for the $0$ vector. Substituting $x$ in first equation, we get ${\lambda }^{2}y=-3y$. Diving by $y$ on both sides we have that $\lambda =i\sqrt{3}$ or $\lambda =-i\sqrt{3}$. Since we’re working on ${ℝ}^2$, it follows that there are no eigenvalues. Consider ${ℂ}^2$ instead. To prove $i\sqrt{3}$ is indeed an eigenvalue, choose any vector of the form $(i\sqrt{3}x,x)$. Then

For $-i\sqrt{3}$, consider the vectors of the form $(x,-i\frac{\sqrt{3}}{3}x)$. Then

Since $\dim <!--\; FUNCTION\; APPLICATION\; -->{ℂ}^2=2$, there are only two eigenvalues. □