2
$\begingroup$

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?

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ 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. $\endgroup$
    – Neil Strickland
    Commented Jun 20, 2013 at 11:10
  • $\begingroup$ @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. $\endgroup$
    – Michal R. Przybylek
    Commented Jun 20, 2013 at 12:03
  • $\begingroup$ 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). $\endgroup$
    – Michal R. Przybylek
    Commented Jun 20, 2013 at 12:11
  • $\begingroup$ @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. $\endgroup$
    – 5th decile
    Commented Jun 20, 2013 at 12:22
  • $\begingroup$ 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). $\endgroup$
    – Yemon Choi
    Commented Jun 20, 2013 at 22:09

4 Answers 4

Reset to default
3
$\begingroup$

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

Improve this answer
$\endgroup$
3
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ @Neil, that's another nice approach --- thank you. $\endgroup$
    – Michal R. Przybylek
    Commented Jun 21, 2013 at 21:14
3
$\begingroup$

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.

Improve this answer
$\endgroup$
0
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ But why should an eigenvector of $A^n$ be an eigenvector of $A$? $\endgroup$
    – 5th decile
    Commented Jun 20, 2013 at 12:56
  • $\begingroup$ @Thibaut $A$ is an operator, not a vector. $v$ is a vector. $\endgroup$
    – Piotr Migdal
    Commented Jun 20, 2013 at 15:25
  • $\begingroup$ @Piotr, you have shown one (the trivial) direction. What about the other? (it is in Mark's answer) $\endgroup$
    – Michal R. Przybylek
    Commented Jun 20, 2013 at 21:28
  • $\begingroup$ @Michal Edited. $\endgroup$
    – Piotr Migdal
    Commented Jun 21, 2013 at 13:41
  • $\begingroup$ @Piotr, could you write down the whole proof? $\endgroup$
    – Michal R. Przybylek
    Commented Jun 21, 2013 at 14:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .