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Let $\mathsf{F}$ be any field. Let $A$ be an $n \times n$ matrix over $\mathsf{F},$ whose rank is $r \le n.$ Let $\mu \in \mathsf{F}[x]$ be the minimal polynomial of $A.$

What does $\deg(\mu)$ tell about $A$? Is it related to the rank of $A$?


Edit: As noted by Qiaochu, $\deg(\mu)$ is not equal to rank. Are there known cases (or conditions) where $\deg(\mu)$ is actually equal to the rank of $A$?

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    Also $deg(\mu) \leq r+1$ : [Link](http://math.stackexchange.com/questions/670939/a-is-mn%c3%97nc-with-rank-r-and-mt-is-the-minimal-polynomial-of-a-prove-deg-mt)2014-08-12

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If $A$ has distinct $n$ non-zero eigen values in an extension field of $F$, $\mu$ is the characteristic polynomial of $A$. Hence deg $\mu = n =$ rank $A$.