Eigenvalues of a linear map over a finite field

[snipez90]

Homework Statement

Let F be a finite field of characteristic p. As such, it is a finite
dimensional vector space over Z_p.
(a) Prove that the Frobenius morphism T : F -> F, T(a) = a^p is a
linear map over Z_p.
(b) Prove that the geometric multiplicity of 1 as an eigenvalue of T
is 1.
(c) Let F have dimension 2 over Z_7. Prove that 2 is not an eigenvalue
of T.

Homework Equations

Fermat's little theorem (lagrange's theorem applied to multiplicative group)

The Attempt at a Solution

I got a) and b), which are essentially straightforward applications of Fermat's little theorem. For c), I'm trying to show that T has 2 eigenvalues (neither of which is 2 of course) since it's a well-known theorem that a linear map cannot have more than dim(F) eigenvalues. Again 1 is an eigenvalue by Fermat's theorem as before, but I can't find another eigenvalue. Is there a better approach? Thanks in advance.

 

[Dick]

What's T^7(a)?

 

[snipez90]

Hmm, I think I made this way harder than it should be. We already have a^7 = a. If a^7 = 2a, we would have a = 0, which we could have forbidden since 0 is never considered an eigenvector.

But supposing a =/= 0 is an eigenvector with eigenvalue x, calculating T^7(a) gives a(x^7 - 1) = 0 which implies x^7 - 1 = 0, and x cannot be 2. Is this what you were getting at Dick? Thanks.

 

[Dick]

Hmm, I think I made this way harder than it should be. We already have a^7 = a. If a^7 = 2a, we would have a = 0, which we could have forbidden since 0 is never considered an eigenvector.

But supposing a =/= 0 is an eigenvector with eigenvalue x, calculating T^7(a) gives a(x^7 - 1) = 0 which implies x^7 - 1 = 0, and x cannot be 2. Is this what you were getting at Dick? Thanks.

Almost, but not quite. a^7 is not necessarily equal to a. And T^7(a) isn't a^7. It's a^49. Since T(a)=a^7. Can you fix that up? You need to use that F has dimension 2. Can you use that to show a^49=a?

 

[snipez90]

Bah, I wondered why the hypothesis that F has dim 2 over Z_7 seemed confusing to me. Of course an n-dimensional vector space over Z_p has p^n elements. Here V has 49 elements. V\{0} is multiplicative, so by Lagrange, a^48 = 1, whence a^49 = a, and the original proof carries out correctly when modified. Thanks.

 

[Dick]

Bah, I wondered why the hypothesis that F has dim 2 over Z_7 seemed confusing to me. Of course an n-dimensional vector space over Z_p has p^n elements. Here V has 49 elements. V\{0} is multiplicative, so by Lagrange, a^48 = 1, whence a^49 = a, and the original proof carries out correctly when modified. Thanks.

Bingo.

 

[snipez90]

Okay my friend pointed out that part b) might not be as simple as I thought it to be. I assumed that F had dimension 1 over Z_p in part b), which implies a^p = a for any a in V = F/Z_p which means that the eigenspace has the same dimension as V.

But if F is n-dimensional, then F has p^n elements. Then you deduce a^(p^n) = a, but how do you go from that to a^p = a?

 

[Dick]

Okay my friend pointed out that part b) might not be as simple as I thought it to be. I assumed that F had dimension 1 over Z_p in part b), which implies a^p = a for any a in V = F/Z_p which means that the eigenspace has the same dimension as V.

But if F is n-dimensional, then F has p^n elements. Then you deduce a^(p^n) = a, but how do you go from that to a^p = a?

You don't. You know p roots of a^p-a=0. So it factors completely. How many roots can it have?

 

[spamiam]

You were very much right in using Fermat's little theorem. By using it, you should be able to prove that

$$\mathbb{F}_p \subseteq \text{ker}(T-I)$$

and using Dick's suggestion, you can then show equality.