Give an example of a linear map $T: \Bbb R^2 → \Bbb R^2$, such that $T^{12}$ has real eigenvalues, while $T^n$, with $n= 1,2,...,11,$ do not. I am doing my revision for exam, but I stuck in a past paper question.
Give an example of a linear transformation $T: R^2 → R^2$, such that $T^{12}$ has real eigenvalues, while $T^n$, with $n= 1,2,\dots,11$ do not. Write the matrix which represents $T$.
(Edited)
Please teach me how to tackle with that question.
 A: Hint: a rotation that is neither $180^\circ$ nor $0^\circ$ has no real eigenvalues.
A: By diagonalization, you can find that any diagonal matrix $A$ can be represented as
$$A = P D P^{-1},$$
where $D$ is a diagonal matrix in which each element is an eigenvalue, and then $P$ is a nonsingular matrix (i.e. its columns are linearly independent).
Now, the power of a diagonalization is that you can easily figure out the power of a matrix (pun somewhat intended):
$$A^n = P D^n P^{-1}$$
Hence, if you have $\lambda_i$ is the $i$th eigenvalue of $A$, then $\lambda_i^n$ is the $i$th eigenvalue of $A^n$.
So if you want $A^{12}$ to have real eignvalues, but $A^1, A^2, ... A^{11}$ to have only complex eignenvalues, all you need is to find any nonsingular $2\times 2$ matrix $P$ and two values $\lambda_i$ so that $\lambda_i^n$ is real if $n = 12$ and complex if $n < 12$.
To find those $\lambda_i$ values, you can note that the nth roots of unity are complex, usually. If you take any two complex twelfth-roots of unity $e^{2\pi i {k \over 12}}$ as your two eigenvalues, then $A$ will have only complex eigenvalues, while $A^{12}$ will have only real eigenvalues.
While the other answer does provide one way of looking at the problem (rotations), here I showed how to think of the problem in general.
Hope this helps!
