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Can you find a minimal polynomial of $A^n$ if you know the minimal polynomial of $A$?

  • I'm talking about minimal polynomials of matrices.
  • I'm asking in the general sort of way, I know that in some cases you can use algebraic tricks
    or some other cleverness.
  • $n$ is just some integer, it's not connected to the matrix in any way.
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What is $n$? Is it the size of the matrix, or just some integer? – Olivier Bégassat Nov 11 '12 at 23:05
If $f(x)$ is the minimal polynomial of $A$, then the subring of the matrix ring generated by $A$ is isomorphic to $K[\alpha]/f(\alpha)$. Your question is then the same as asking for the minimal polynomial of $\alpha^n$ over $K$. In particular the answer should depend only on $n$ and on the minimal polynomial of $A$. On the other hand, we should be able use Jordan canonical form to reduce to the case where the minimal polynomial has the form $(x-\lambda)^m$. – Jeff Tolliver Nov 12 '12 at 0:01
There are plenty of question related to this on the site. Do a search first. – Bombyx mori Nov 12 '12 at 0:07
@OlivierBégassat $n$ is just some integer, I edited my question to make it clear. – ante.ceperic Nov 12 '12 at 7:16
up vote 2 down vote accepted

This can be answered in general by examining the Jordan structure of $A$. Recall that the eigenvalues of $A^n$ are precisely $\lambda^n$ for each eigenvalue $\lambda$ of $A$. Multiplicity is preserved.

If the minimal polynomial contains a factor of the form $x^k$ then this is transformed to $x^{\left\lceil\frac{k}{n}\right\rceil}$.

Factors of the form $(x-\lambda)^k$ are preserved as $(x-\lambda^n)^k$. If it happens that we have $\lambda_1^n = \lambda_2^n$ for distinct eigenvalues $\lambda_1$ and $\lambda_2$, then we take on the factor with the largest exponent.

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