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!


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.

  • $\begingroup$ 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. $\endgroup$ Apr 19 '14 at 20:29
  • $\begingroup$ I think I understand. Thanks! $\endgroup$
    – Peter
    Apr 19 '14 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.