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I'm working on a problem involving applying FLT to matrices, so any information about how to do this or prove this is true would be helpful. I've been doing some research and experimenting a little, but right now I'm trying to do a little proof. My specific question is this:

Suppose you have a matrix $A$ and a prime number $p$. If $A^p=A$ mod $p$, then is $A$ diagonalizable? I've already shown that the reverse direction is true.

Any help would be appreciated!

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The matrix $A=\begin{pmatrix} 2 & 2 \cr 0 & 2 \end{pmatrix}$ satisfies $A^p\equiv A$ mod $p$ for $p=2$, but is not diagonalizable. There are generalisations of Fermat's little theorem, but they involve the trace of matrices, see http://www.math.binghamton.edu/mazur/papers/pub5.pdf.

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  • $\begingroup$ right. didn't think of that. $\endgroup$ – Will Jagy Apr 19 '14 at 20:02
  • $\begingroup$ @Dietrich Burde: If $A=\begin{pmatrix} 2 & 2 \cr 0 & 2 \end{pmatrix}$, then $A^2=\begin{pmatrix} 4 & 8 \cr 0 & 4 \end{pmatrix} = \begin{pmatrix} 0 & 0 \cr 0 & 0 \end{pmatrix}$ mod $2$. Unless I'm missing something, I don't believe they are equivalent... $\endgroup$ – Peter Apr 19 '14 at 20:12
  • $\begingroup$ Maybe I'm not understanding when is the correct time/placement to modulo? $\endgroup$ – Peter Apr 19 '14 at 20:15
  • $\begingroup$ We have $A^2\equiv 0 \equiv A$ mod $2$. Modulo $2$ both $A$ and $A^2$ are zero. $\endgroup$ – Dietrich Burde Apr 19 '14 at 20:29
  • $\begingroup$ I think I understand. Thanks! $\endgroup$ – Peter Apr 19 '14 at 20:47

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