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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!

Peter
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  • oh, Little theorem. Not sure, but be aware of this, which can be done in any dimension: http://math.stackexchange.com/questions/153041/matrices-with-a3b3-c3/153118#153118 – Will Jagy Apr 19 '14 at 19:58
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    But also $A \equiv 0 \pmod 2,$ as all four entries are even. – Will Jagy Apr 19 '14 at 20:11

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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.

Dietrich Burde
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  • right. didn't think of that. – Will Jagy Apr 19 '14 at 20:02
  • @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... – Peter Apr 19 '14 at 20:12
  • Maybe I'm not understanding when is the correct time/placement to modulo? – Peter Apr 19 '14 at 20:15
  • We have $A^2\equiv 0 \equiv A$ mod $2$. Modulo $2$ both $A$ and $A^2$ are zero. – Dietrich Burde Apr 19 '14 at 20:29
  • I think I understand. Thanks! – Peter Apr 19 '14 at 20:47
  • Not to beat a dead horse, but can you think of other matrices that serve as a counterexample for this? So far I have $\begin{pmatrix} p & p \ 0 & p \end{pmatrix}$ and $\begin{pmatrix} p & 0 \ p & p \end{pmatrix}$ in mod $p$. – Peter Apr 20 '14 at 20:46
  • @DietrichBurde but the matrix $0$ is diagonal (and hence diagonalizable), isn't it? – Ben Grossmann Apr 28 '14 at 13:48
  • @Omnomnomnom No, the matrix $A$ is not diagonalizable, because it is not zero. You mean the matrix modulo $2$, which is something different. – Dietrich Burde Apr 28 '14 at 14:35
  • @DietrichBurde When we perform operations on the matrix $A$ modulo $2$, do we not simply use the value of $A$ modulo $2$? Why would we make a point of stating $A \equiv B$, but $A \neq B$? – Ben Grossmann Apr 28 '14 at 15:08
  • Again - the matrix $A\in M_2(\mathbb{C})$ is not diagonalizable and is not zero. It just satisfies $A^p\equiv A\equiv 0$ mod $p$ for $p=2$. – Dietrich Burde Apr 28 '14 at 15:36