Eigenvalues of powers of linear mappings 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?
 A: 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 $i$
$$\det(\omega_i - \tau)=0$$
A: While I agree with the above comments and answers, here is one more approach that may be a little more low-tech.  There is an invertible matrix $A$ such that, $\tau = AUA^{-1}$ where $U$ is upper triangular and $A$ is a product of determinant 1 elementary matrices. The eigenvalues for $U$ are the entries along the diagonal. Also, $\tau^n = AU^n A^{-1}$ and $U^n$ remains upper triangular and it's eigenvalues are in 1-1 correspondence with the nth powers of the eigenvalues for $U$.  
The last time I taught linear algebra, we dealt with the matrix $A$ when showing that if $0$ is an eigenvalue for $\tau$, $det(\tau)=0$, so there's a chance it would be a little familiar to your students.
A: The only gap in Piotr Migdal's answer is the fact that for every eigenvalue $\lambda$ of $A^n$, there is a $\lambda$-eigenvector of $A^n$ which is an eigenvector of $A$. Here is a proof. Let $\lambda$ be an eigenvalue of $A^n$. Let $U$ be the set of all $\lambda$-eigenvectors of $A^n$ (and $0$). It is a subspace. If $A^nv=\lambda v$, then $A^n Av=AA^n v=\lambda Av$, so $AU\subseteq U$. Then $A$ has an eigenvector $u$ in $U$. The end of the proof is as in Migdal's answer. Let $\lambda$ be an eigenvalue of $A^n$. Then (by the above) there exists a non-zero vector $v$ such that $A^nv=\lambda v$ and $Av=\mu v$ for some $\mu$. Then $A^n v=\mu^nv=\lambda v$, so $\lambda=\mu^n$ - every eigenvalue of $A^n$ is the $n$th power of an eigenvalue of $A$.   
 Comment  This proof does use the Fundamental Theorem of algebra: we need to know that every subspace invariant under $A$ contains an eigenvector of $A$. For non-algebraically closed fields, the result is not true of course. 
A: For $v$ being an eigenvector $A v = \lambda v$ we get
$$A^n v = A^{n-1} \lambda v = A^{n-2} \lambda^2 v  = \ldots = \lambda^n v.$$
Any it holds for any linear operator, not only for your map.
For the converse, you can take a n-th root of an operator using analytic functions, and it maps eigenvalues to their roots.
