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Given the matrix $A∈M_2(\mathbb{Z_7})$

$$ \begin{bmatrix} 1 & 2 \\ 2 & 5 \\ \end{bmatrix} $$

Is it diagonalizable? I think it is because I calculated the eigenvalues which are $4$ and $2$. Since the eigenvalues are two and the order of the matrix is two then the matrix is diagonalizable, right?

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    $\begingroup$ If $4$ and $2$ are the eigenvalues then it is diagonalisable. If you also have the eigenvectors, then you can diagonalise it explicitly. $\endgroup$ Feb 9 '18 at 21:21
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    $\begingroup$ In general, if $A\in M_n(k)$ and you have $n$ distinct eigenvalues, then $A$ is diagonalisable. However, here $k=\mathbb{Z}_7$ is not algebraically closed, so a priori it could be that you don't have enough eigenvalues which is not the case here. $\endgroup$ Feb 9 '18 at 21:23
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Yes, your argument is correct. Perhaps that this will sound pedantic, but you could also have added that, in $\mathbb{Z}_7$, $4\neq2$. That is, you found two distinct eigenvalues,

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    $\begingroup$ Not pedantic at all... $\endgroup$ Feb 9 '18 at 21:49

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