# Find eigenvalues of $T(x,y)=(-3y,x)$

Suppose $T \in L(\mathbb{R^2})$ is defined as $T(x,y) = (-3y,x)$. Find the eigenvalues of $T$.

attempt: Let $\lambda$ be an eigenvalues of $T$ such that $T(x,y) = \lambda (x,y)$. Then we have $T(x,y) = \lambda (x,y)$ so $(-3y,x) = (\lambda x, \lambda y)$ thus $\lambda x = -3y$ and $x = \lambda y$. So we have $\lambda (\lambda y) = -3y$ implies we have $\lambda^2 = -3$. which has no solution. So there arent any eigenvalues.

Can someone please verify this? Any feedback would help. Thank you

• $\pm i \sqrt{3}$. Oct 27 '16 at 2:57

Your calculation is correct. There aren't any real eigenvalues, but there are the complex eigenvalues $\lambda=\pm\sqrt{3}i$.
• But we are in $\mathbb{R^2}$. So the aren't . unless we are in the complex Oct 27 '16 at 2:59
• The eigenvalues can be complex even though the transformation is in $T(\mathbb{R}^2)$.
• So we get that $\lambda = \sqrt 3i$ and $\lambda = -\sqrt 3i$? are the only eigenvalues? Oct 27 '16 at 3:02