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I am trying to solve this exercise here:

Let \phi denote the Frobenius map x |-> x^p on the finite field F_{p^n}. Determine the Jordan canonical form (over a field containing all the eigenvalues) for \phi considered as an F_p-linear transformation of the n-dimensional F_p-vector space F_{p^n}.

So, this is how I start:

Suppose that F_{p^n}=F_p(a1,a2,a3, ..., an) (those n elements will be powers of one element, but it doesn't matter). Now, since the Frobenius map is an isomorphism of F_{p^n} to itself, then \phi permutes a1, a2, ..., an.

Since a1, a2, ..., a3 form a basis of the n-dimensional F_p-vector space F_{p^n}, then the matrix of \phi in respect with that basis will be just a permutation matrix.

So the problem becomes equivalent with: "find the jordan canonical form of a permutation matrix".

Am I doing some obvious mistake here? Would the latter be something straightforward? I admit I can't see it...

Any help would be greatly appreciated.

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# Jordan Can. Form of Frobenius map

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