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Let $E$ be a $n\times n$ matrix over field $F$. And assume its minimal polynomial (assume it is irreducible) is with degree $m$, then how to show $m|n$?

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  • $\begingroup$ Worth recording that the minimal polynomial is not necessarily irreducible; see the deleted answer. $\endgroup$ – Will Jagy Dec 3 '17 at 0:56
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Hint: Let $M$ be the matrix and $P$ be the minimal polynomial, $F[X]/(P)$ is a field $L$ and $F^n$ is an $L$-vector space where the action is induced by the action of $F[X]$ on $F^n$ defined by $X.v=M(v)$, let $p$ be the dimension of the $L$-vector space $F^n$, $n=dp$ where $d$ is $[L:F]$. In fact if $(e_1,..,e_p)$ is a basis of the $L$-vector space $F^n$, $(e_i,X.e_i=(..,M(e_i),...,X^{p-1}.e_i,...)$ is a basis of the $F$-vector space $F^n$.

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The minimal polynomial is the largest invariant factor, and characteristic polynomial is the product of all invariant factors. So if the largest is irreducible, then the other invariant factors are the same, so the characteristic polynomial is a power of the minimal polynomial, which implies $m|n$.

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