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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

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  • $\begingroup$ $\pm i \sqrt{3}$. $\endgroup$
    – copper.hat
    Oct 27 '16 at 2:57
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Your calculation is correct. There aren't any real eigenvalues, but there are the complex eigenvalues $\lambda=\pm\sqrt{3}i$.

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  • $\begingroup$ But we are in $\mathbb{R^2}$. So the aren't . unless we are in the complex $\endgroup$
    – Mahidevran
    Oct 27 '16 at 2:59
  • $\begingroup$ The eigenvalues can be complex even though the transformation is in $T(\mathbb{R}^2)$. $\endgroup$
    – pi66
    Oct 27 '16 at 3:00
  • $\begingroup$ So we get that $\lambda = \sqrt 3i$ and $\lambda = -\sqrt 3i$? are the only eigenvalues? $\endgroup$
    – Mahidevran
    Oct 27 '16 at 3:02
  • $\begingroup$ Yes, that's right. $\endgroup$
    – pi66
    Oct 27 '16 at 3:03

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