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Given two distinct elements $A,B\in\text{GL}(n,q)$. Regarding $A,B$ as matrices, and assume they both have the same irreducible characteristic polynomial over $\text{GF}(q)$. Then must $A,B$ be conjugate in $\text{GL}(n,q)$?

If the higher dimension is intricate, how about the lower dimension, say $n=2$?

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This is linear algebra rather than group theory. If they have the same irreducible characteristic polynomial, then that must also be their minimal polynomial, and they are both conjugate to its companion matrix. –  Derek Holt Feb 3 '13 at 14:01
    
Thank you, Derek. I think this is more straightforward! –  Easy Feb 5 '13 at 2:59
    
So it will be even true when the field is infinite, isn't it? –  Easy Feb 5 '13 at 3:04

1 Answer 1

Yes, they are conjugate. One way to prove it is as follows:

  • Since the char. poly. is irred., its roots are distinct, and so each of $A$ and $B$ have $n$ distinct eigenvalues over the algebraic closure $k$ of $GF(q)$. General theory then shows that they can both be diagonalized in $GL_n(k)$, and since their eigenvalues are the same, they are conjugate in $GL_n(k)$.

  • It is a general fact for any field that if two matrices are conjugate in $GL_n$ of the algebraic closure, they are already conjugate in $GL_n$ of the field itself.

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I am convinced with your first point, but the second point sounds mysterious to me however believable. Where can I find the reference? –  Easy Feb 3 '13 at 13:20
    
+1 Matt E. Good to see you around. I still remember you very thoughtful and warm answer to one of my earliest posts on math/meta math.se! I was away for almost a year, but have returned! –  amWhy Feb 3 '13 at 13:30
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@user60079 When $k$ is infinite, there is a slick proof. Let $A$ and $B$ be our matrices in $GL_n(k)$. Let $V$ be the set of $n \times n$ matrices $X$ such that $AX=XB$. So $V$ is a linear subspace of $k^{n^2}$. We want to show that there is an invertible element in $V$ if and only if there is one in $V \otimes \bar{k}$. But $\det$ is a polynomial, so if $k$ is infinite and $\det$ is $0$ everywhere on $V$, then it is still zero on $V \otimes \bar{k}$. –  David Speyer Feb 5 '13 at 20:22
    
@DavidSpeyer, sorry, I didn't notice your comment. I don't quite get your last sentence. Can you explain it a bit more? Thanks –  Easy Feb 25 '13 at 2:04

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