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Generally, two similar matrices have same characteristic polynomial. Infact the converse need not be true as two different Jordan forms can have same characteristic polynomial.

Is there any restriction can make the one side implication in to an an equivalence. I feel that 'Matrices over $\Bbb Z_p$, where $p$ is a prime, admits the equivalence' though I couldn't justify the claim.

Messi Lio
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    Definitely not true for matrices in characteristic $p$, for the very same reason (they have Jordan forms too, over an algebraic closure). – Marc van Leeuwen Sep 07 '22 at 10:02
  • Thank you, what about $2\times 2$ matrices over $\Bbb Z_3$ and $\Bbb Z_7$? – Messi Lio Sep 07 '22 at 10:05
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    Nilpotent matrices have the same characteristic polynomial as the zero matrix, but unless they are the zero matrix, they are not equivalent. You can find nonzero nilpotents over any ring as soon as as the size is at least$~2$. – Marc van Leeuwen Sep 07 '22 at 10:09

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Of course there are restrictions that will allow for equivalence; for instance restriction to diagonal matrices does the job. Somewhat less uselessly, it is true for $n\times n$ matrices with $n$ distinct eigenvalues, or more generally for matrices of that size whose minimal polynomial has degree$~n$ (and therefore coincides with the characteristic polynomial). A different generalisation is that it is valid, is for diagonalisable matrices. The common theme here is that given the characteristic polynomial, one arranges to rule out all Jordan types but one,

Marc van Leeuwen
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  • Sir, I have seen in an answer, two matrices in $M_2(\Bbb Z_7)$ are similar if and only if they have same characteristic polynomial. I was checking the reasoning and their generalizations. – Messi Lio Sep 07 '22 at 10:16