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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?

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The map $\tau^n-\lambda$ can be written as a composite of maps $\tau-\mu$, where $\mu$ runs over the $n$'th roots of $\lambda$. If $\tau^n-\lambda$ is not invertible, then one of the factors must fail to be invertible. – Neil Strickland Jun 20 '13 at 11:10
@Neil, thanks for this comment. Now I feel stupid :-) I thought that showing the polynomial $\tau^n - \lambda$ decomposes on $\tau - \mu$ is not so trivial. – Michal R. Przybylek Jun 20 '13 at 12:03
BTW, I would accept your comment if I could (it seems that Thibaut's answer reassembles what you have written in a slightly different language, so I'll accept it). – Michal R. Przybylek Jun 20 '13 at 12:11
@Michal, there is indeed some subtelty involved in the decomposing of $\tau^n - \lambda$, e.g. the decomposition of $A^n - B^n$ is not straightforward if $A$ and $B$ are noncommuting matrices. But luckily: $\tau$ and $\lambda \mathbf{1}$ do commute. – Thibaut Demaerel Jun 20 '13 at 12:22
Side note: this is of course a special case of the spectral mapping theorem for unital associative complex algebras, whose proof is essentially the argument in Neil Strickland's answer (replace the power function by any polynomial). – Yemon Choi Jun 20 '13 at 22:09
up vote 3 down vote accepted

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 $i$ $$\det(\omega_i - \tau)=0$$

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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.

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@Neil, that's another nice approach --- thank you. – Michal R. Przybylek Jun 21 '13 at 21:14

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.

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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.

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But why should an eigenvector of $A^n$ be an eigenvector of $A$? – Thibaut Demaerel Jun 20 '13 at 12:56
@Thibaut $A$ is an operator, not a vector. $v$ is a vector. – Piotr Migdal Jun 20 '13 at 15:25
@Piotr, you have shown one (the trivial) direction. What about the other? (it is in Mark's answer) – Michal R. Przybylek Jun 20 '13 at 21:28
@Michal Edited. – Piotr Migdal Jun 21 '13 at 13:41
@Piotr, could you write down the whole proof? – Michal R. Przybylek Jun 21 '13 at 14:51

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