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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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    In general there is no relationship between this degree and the rank (consider diagonal matrices). The degree of the minimal polynomial tells you something about how many eigenvalues $A$ can have, and it also tells you something about how large the corresponding Jordan blocks can be; see http://en.wikipedia.org/wiki/Jordan_normal_form .2012-08-17
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    @QiaochuYuan Thanks. Could you please convert it to an answer. And also, (in the answer) are there certain known cases where this degree is equal to the rank?2012-08-17
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    As long as you don't have an eigenvalue that belongs to two different Jordan blocks...2012-08-17
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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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