Let $\tau$ be a linear map on a finite dimensional complex vector space. Clearly, if $\lambda$ is an eigenvalue of $\tau$ then $\lambda^n$ is an eigenvalue of $\tau^n$, for any natural (integer, on condition $\tau$ is invertible) number $n$. It easily follows from Jordan theorem, that *every* eigenvalue of $\tau^n$ has to be of the form $\lambda^n$.


I have to convince students who have only basic knowledge about linear algebra that the above statement is true.


Is there any elementary explanation of this fact *without* using Jordan theorem?



Eigenvalues of powers of linear mappings
Perhaps you can use:
$$\det(\lambda - \tau^n) = \det((-1)^n\prod\_{\omega\_i:\text{ nth roots of } \lambda} \omega\_i - \tau)= (-1)^n\prod\_{\omega\_i:\text{ nth roots of } \lambda} \det(\omega\_i - \tau) $$
by the multiplicativity of the determinant. This righthand side is only zero if for an $i$
$$\det(\omega\_i - \tau)=0$$


